Bessel Equation

Bessel equation

Jan Derezi´nski

Department of Mathematical Methods in Physics Faculty of Physics University of Warsaw Pasteura 5, 02-093 Warszawa, Poland June 8, 2020

1 Introduction

Consider the Laplace operator on R2

2 2 ∆2 = ∂x + ∂y and the Helmholtz −∆2F (x, y) = EF (x, y). (1.1) Introduce the polar coordinates

x = r cos φ, y = r sin φ, p y r = x2 + y2, φ = arctan . x The 2-dimensional Laplacian in polar coordinates is 1 1 −∆ = −∂2 − ∂ − ∂2. (1.2) 2 r r r r2 φ We make an ansatz F (x, y) = v(r)u(φ). The equation (1.1) becomes 1 1 − ∂2 − ∂ − ∂2 v(r)u(φ) = Ev(r)u(φ). (1.3) r r r r2 φ We divide both sides by v(r)u(φ) and multiply by r2. We put functions depend- ing on r on one side and on φ on the other side. We obtain

r2∂2 + r∂ + Ev(r) ∂2u(φ) r r = − φ = C, (1.4) v(r) u(φ)

1 where C does not depend on r or φ. We obtain an equation for u:

2 ∂φu(φ) = Cu(φ), (1.5) solved by linear combinations of eimφ and e−imφ with C = m2. Condition u(φ) = u(φ + 2π) implies that m ∈ Z. Thus we obtain 2 2 2 2 r ∂r + r∂r − m + Er v(r) = 0. (1.6) √ Now if E < 0, then we change the variables z = −Er obtaining the modiﬁed Bessel equation

2 2 2 2 z ∂z + z∂z − m − z v = 0. (1.7) √ Now if E > 0, then we change the variables z = Er obtaining the (standard) Bessel equation

2 2 2 2 z ∂z + z∂z − m + z v = 0. (1.8)

Certain distinguished solutions of (1.7) are denoted Im, Km, and of (1.8) are ± denoted Jm and Hm. We call them jointly the Bessel family. They are probably the best known and the most widely used special functions in mathematics and its applications. The parameter m in the above analysis was an integer. However, if we consider the Laplacian on the surface of the cone described by 0 ≤ φ < α, then mα the condition u(φ) = u(φ + α) leads to 2π ∈ Z. Hence non-integer values of the parameter m are also relevant in applications. We will see that the Helmholtz equation in any dimension d

−∆dF = EF (1.9) leads in spherical coordinates to the Bessel equation. The d-dimensional Lapla- cian in spherical coordinates is given by

2 d − 1 1 −∆ = −∂ − ∂ − ∆ d−1 , d r r r r2 S where r is the radial coordinate and ∆Sd−1 is the Laplace-Beltrami operator on d−1 the sphere S . Eigenvalues of −∆Sd−1 for d = 2, 3,... are l(l + d − 2), l = 0, 1, 2,... (1.10) where l corresponds to the order of spherical harmonics. Setting F = v(r)u(Ω), where Ω are the angular coordinates, we obtain the radial part of the Helmholtz equation

2 2 2 r ∂r + (d − 1)r∂r − l(l + d − 2) + Er v(r). (1.11) By the same scaling argument as above, we can reduce ourselves to the case E = ±1. We will see that (1.11) is equivalent to (1.7) or (1.8).

2 The operator in (1.7)/(1.8) can be transformed as

d d 1− 2 2 2 2 2 −1+ 2 r r ∂r + r∂r ∓ r − m r (1.12) d d = r2∂2 + (d − 1)r∂ ∓ r2 − m − + 1 m + − 1 . (1.13) r r 2 2 Setting d m := l + − 1, (1.14) 2 we can rewrite (1.13) as

2 2 2 r ∂r + (d − 1)r∂r ∓ r − l(l + d − 2), (1.15) which is the radial part of the Helmholtz equation in dimension d for spherical harmonics of order l, see (1.11). (1.15) is sometimes called the d-dimensional Bessel equation. Note that in even dimensions the parameter m takes integer values and in odd dimensions it takes half-integer values, see (1.14). Note special cases of (1.13):

−m 2 2 2 2 m r r ∂r + r∂r ∓ r − m r 2 2 2 = r ∂r + (1 + 2m)r∂r ∓ r , (1.16)

1 1 2 2 2 2 2 − 2 r r ∂r + r∂r ∓ r − m r 1 = r2 ∂2 ∓ 1 + 1/4 − m2 . r r2

Here are some other operators related to the Bessel equation: Set r := tδ, 1 1−δ so that ∂r = δ t ∂t. Then

1 1 2δ 2 2 2 2 − 2δ r r ∂r + r∂r ∓ r − m r (1.17) 1 1 2 −2 2 2 −2 2δ 2 − 2 = t δ t ∂t + δ t∂t ∓ t − m t (1.18) 1 1 = δ−2t2 ∂2 ∓ (δtδ−1)2 + − m2δ2 . (1.19) t 4 t2 t If we set r = e , so that r∂r = ∂t, then

2 2 2 2 2 2t 2 r ∂r + r∂r ∓ r − m = ∂t ∓ e − m . (1.20)

2 Modiﬁed Bessel equation 2.1 Integral representations The modiﬁed Bessel equation is given by the operator

2 2 2 2 Im(z, ∂z) := z ∂z + z∂z − z − m .

3 γ Theorem 2.1 Bessel–Schl¨aﬂitype representations Let ]0, 1[3 τ 7→ t(τ) be a contour such that z z t(1) (t − t−1) + m exp (t + t−1) t−m = 0, (2.21) 2 2 t(0)

Then Z z exp (t + t−1) t−m−1dt (2.22) γ 2 is a solution of the modiﬁed Bessel equation

Proof. We diﬀerentiate the integral with respect to the parameter z: Z 2 2 2 2 z −1 −m−1 (z ∂z + z∂z − z − m ) exp (t + t ) t dt γ 2 Z z 2 2 z z = t + t−1 + t + t−1 − z2 − m2 exp (t + t−1) t−m−1dt γ 2 2 2 Z z 2 2 z z = t − t−1 + t + t−1 − m2 exp (t + t−1) t−m−1dt γ 2 2 2 Z z −1 z −1 −m = ∂t (t − t ) + m exp (t + t ) t dt γ 2 2 z z t(1) = (t − t−1) + m exp (t + t−1) t−m = 0. 2 2 t(0) 2

γ Theorem 2.2 Poisson type representations. Let ]0, 1[3 τ 7→ t(τ) be a contour such that t(1) 2 m+ 1 zt (1 − t ) 2 e = 0. t(0) Then Z m 2 m− 1 zt z (1 − t ) 2 e dt γ is a solution of the modiﬁed Bessel equation.

Proof. We use (1.16).

Z 1 2 2 m− 2 zt z∂z + (1 + 2m)∂z − z (1 − t ) e dt γ Z 2 m− 1 2 zt = (1 − t ) 2 (zt + (1 + 2m)t − z)e dt γ Z 2 m+ 1 zt = − ∂t(1 − t ) 2 e dt = 0. γ

4 2.2 Modified Bessel function The modified Bessel equation has a regular-singular point at 0 with the indicial equation λ(λ − 1) + λ − m2 = 0. Its indices at 0 are equal to ±m. Therefore, we should look for a solution of the modified Bessel equation in the form ∞ 2 2 2 2 X m+n 0 = z ∂z + z∂z − z − m cnz (2.23) n=0 ∞ ∞ X 2 2 n X n+2 = cn (m + n) (m + n − 1) + m + n − m z − cnz . (2.24) n=0 n=0 This leads to the recurrence relation

cn(2m + n)n = cn−2. (2.25)

The initial condition c−1 = 0 together with (2.25) implies that cn = 0 for n odd. For even subscripts, we can rewrite (2.25) in the form

c2n(2m + 2n)2n = c2(n−1). (2.26)

With c0 = 1, this is solved by 1 c = . 2n 22n(m + 1) ··· (m + n)n!

1 Multiplying this with Γ(m+1) , we obtain 1 c = . 2n 22nΓ(m + n + 1)n! The resulting function is called the modiﬁed Bessel function:

∞ z 2n+m X I (z) = 2 . m n!Γ(m + n + 1) n=0 It is a solution of the modified Bessel equation with the parameter ±m. Note 1 that Γ(m+1) 6= 0 for m 6= −1, −2,... For m 6= −1, −2,... the function Im is the unique solution of the modified Bessel equation satisfying z m 1 I (z) ∼ , z ∼ 0, m 2 Γ(m + 1) which can be treated as a definition of the modified Bessel function. (By f(z) ∼ f(z) g(z), z ∼ 0, we understand that g(z) is analytic around zero and at zero equals 1).

5 If m 6∈ Z, then I−m(z) and Im(z) are linearly independent and span the space of solutions of the modiﬁed Bessel equation. We have

±iπ ±iπm Im(e z) = e Im(z), (2.27)

Im(z) = Im(z). (2.28)

In particular, Im(x) is real for x > 0, m ∈ R.

2.3 Integral representations of modified Bessel function Theorem 2.3 (Bessel-Schläfli-type representations.) Let Rez > 0. Then Z 1 z −1 −m−1 Im(z) = exp (t + t ) t dt (2.29) 2πi ]−∞,0+,−∞[ 2 1 z m Z z2 = exp s + s−m−1ds. (2.30) 2πi 2 ]−∞,0+,−∞[ 4s We also have Z 1 z −1 −m−1 I−m(z) = exp (t + t ) t dt. (2.31) 2πi [(0−0)+] 2 Here, the contour starts at 0 from the negative side on the lower sheet, encircling 0 in the positive direction and ends at 0 from the negative side on the upper sheet. One of concrete realizations of (2.29) is the Schläflirepresentation: Z π Z ∞ 1 z cos φ 1 −z cosh β−mβ Im(z) = e cos(mφ)dφ − sin(mπ) e dβ. (2.32) 2π −π π 0 Proof. To see this note that by Thm 2.1, the RHS of (2.29) is a solution of the modified Bessel equation. Besides,

1 z m Z z2 exp s + s−m−1ds 2πi 2 ]−∞,0+,−∞[ 4s is holomorphic around zero. By the Hankel identity 1 1 Z = exp (s) s−m−1ds. Γ(m + 1) 2πi ]−∞,0+,−∞[

z m 1 Therefore, the RHS of (2.29) behaves at zero as ∼ 2 Γ(m+1) . Therefore, it coincides with Im(z), at least for z 6∈ {..., −2, −1}. By continuity, it is Im also for z ∈ {..., −2, −1}. To see (2.31), we make a substitution t = s−1 in (2.29) noting that dt = −s−2ds, and then we change the orientation of the contour. To see (2.32), we take a contour consisting of three pieces: −e−β : β ∈ ] − ∞, 0], eiφ : φ ∈ [−π, π], −e−β : β ∈ [0, −∞[. 2

6 Theorem 2.4 (Poisson-type representations.) We have the Poisson representation

m Z 1 1 z 2 m− 1 zt 1 Im(z) = √ (1 − t ) 2 e dt, m > − . (2.33) 1 2 −1 2 πΓ m + 2

The following representation is due to Hankel:

1 m Z Γ( 2 − m)z m− 1 m− 1 zt I (z) = √ (t − 1) 2 (t + 1) 2 e dt. (2.34) m 2πi π 2 [1,−1−,1+]

Proof. By Thm 2.2, the RHS of (2.33) satisﬁes the modiﬁed Bessel equation. Then we use the fact that √ Z 1 1 2 m− 1 Γ(m + 2 ) π (1 − t ) 2 dt = , (2.35) −1 Γ(m + 1)

z m 1 to see that the RHS of (2.33) behaves at zero as ∼ 2 Γ(m+1) . Similarly, to show (2.34) we use √ 1 Z 1 1 π m− 2 m− 2 (t − 1) (t + 1) dt = 1 . (2.36) 2πi [1,−1−,1+] Γ( 2 − m)Γ(m + 1) 2

2.4 Modiﬁed Bessel function for integral parameters

For m ∈ Z the Bessel-type integrals (2.29), (2.30) and (2.32) simplify:

Theorem 2.5 For m ∈ Z we have

Im(z) = I−m(z).

Z 1 z −1 dt Im(z) = exp (t + t ) m+1 (2.37) 2πi [0+] 2 t 1 z m Z z2 ds = exp s + m+1 (2.38) 2πi 2 [0+] 4s s 1 Z π = ez cos φ cos(mφ)dφ. (2.39) 2π −π

Proof. We will give two proofs.

7 The ﬁrst is based on the power series. It is enough to assume that m = 0, 1,... .

∞ z 2n+m X I (z) = 2 (2.40) m n!(n + m)! n=0

∞ z 2(n+m)−m X = 2 (2.41) (n + m)!(n + m − m)! n=0

∞ z 2n−m X = 2 (2.42) n!(n − m)! n=m

∞ z 2n−m X = 2 = I (z). (2.43) n!Γ(n − m + 1) −m n=0 The second uses the contour integrals (2.29) and (2.31): we note that if m ∈ Z, one can be deformed into the other. 2

Theorem 2.6 (Generating function.) ∞ z X exp (t + t−1) = tmI (z). (2.44) 2 m m=−∞ Proof. Again, we will give two proofs. The ﬁrst uses power series: ∞ X X tm(z/2)m+2n tmI (z) = m n!(n + m)! m=−∞ n≥0, n+m≥0 X X (z/2t)n(tz/2)m+n = n!(n + m)! n≥0 n+m≥0 = ez/2tetz/2. The second notes that (2.44) is the Laurent series in t of a function holomorphic in C\{0}, and (2.37) is just the formula for the coeﬃcient in the Laurent series. 2

2.5 MacDonald function We deﬁne the Macdonald function: Z ∞ 1 z −1 ±m−1 Km(z) := exp − (s + s ) s ds. (2.45) 2 0 2 Other names: the Basset function or the modiﬁed Bessel function of the second kind. The integral (2.45) is absolutely convergent. Substitution s = t−1 shows that m can be replaced by −m (and thus Km = K−m).

8 Theorem 2.7 Km solves the modiﬁed Bessel equation. We have

Km(z) = Km(z).

Km(x) is real for x > 0, and m ∈ R or m ∈ iR. Proof. We can write −iπm Z e z −1 −m−1 Km(z) = exp (t + t ) t dt (2.46) 2 ]−∞−i0,0] 2 eiπm Z z = exp (t + t−1) t−m−1dt, (2.47) 2 ]−∞+i0,0] 2 which are integrals satisfying Theorem 2.1. (Note that the contours lie on the boundary of the Riemann surface of the principal branch of t−m−1, each projecting onto ] − ∞, 0], the ﬁrst is on the lower sheet and the second on the upper sheet). 2

Theorem 2.8 π K (z) = K (z) = (I (z) − I (z)). (2.48) −m m 2 sin πm −m m Proof. We add the appropriate multiples of (2.46) and (2.47):

iπm −iπm −4i sin(πm)Km(z) = −2e Km(z) + 2e Km(z) Z z = exp (t + t−1) t−m−1dt ]−∞−i0,0] 2 Z z + exp (t + t−1) t−m−1dt ]0,−∞−i0] 2 Z z = exp (t + t−1) t−m−1dt ]−∞,0+,−∞[ 2 Z z − exp (t + t−1) t−m−1dt (0−0)+ 2

= 2πi(Im(z) − I−m(z)). where, as we recall, (0 − 0)+ is the contour starting at 0 from the negative side on the lower sheet, encircling 0 in the positive direction and ending at 0 from the negative side on the upper sheet. This proves (2.48). 2 It is useful to note that for positive z the contour in (2.45) can be turned by π 2 , losing however its absolute convergence at 0 or ∞. We thus obtain

±i π m Z ∞ e 2 zi −1 ±m−1 Km(z) := exp − (s − s ) s ds. (2.49) 2 0 2

9 Theorem 2.9 Setting s = eθ in (2.45) and t = eφ in 2.49) we obtain Z ∞ 1 −z cosh θ −mθ Km(z) = e e dθ, (2.50) 2 −∞ ∓i π m ∞ e 2 Z = e∓iz sinh φe−mφdφ. (2.51) 2 −∞

2.6 Analytic continuation of the MacDonald function Theorem 2.10 1 I (z) = ∓iK (e∓iπz) ± ieiπmK (z). (2.52) m π m m Proof. Write π K(e−iπz) = eiπmI (z) − e−iπmI (z) 2 sin πm −m m We subtract from this eiπm times (2.48) obtaining π K(e−iπz) − eiπmK(z) = −e−iπmI (z) + eiπmI (z) = iπI (z). 2 sin πm m m m This proves (2.52). 2 ∓iπ The function Km(z) is exponentially decaying whereas ∓iKm(e z) are exponentially exploding. It will be useful to have such “exploding partners” ∓iπ for Km(z). Unfortunately, the natural domain for ∓iKm(e z) is C\[0, ∞[, which is inconvenient if we are primarily interested in [0, ∞[. It seems useful to introduce the following function (which has no name and apparently does not appear in the literature):

i X (z) := K (eiπz) − K (e−iπz). (2.53) m 2 m m Theorem 2.11 We have the identities 1 I (z) = X (z) − sin(πm)K (z), (2.54) m π m m π X (z) = I (z) + I (z). (2.55) m 2 −m −m 2.7 Asymptotics of the MacDonald function Theorem 2.12 For | arg z| < π − ,

Km(z) √ lim −z = 1. |z|→∞ e √ π 2z

10 1 −1 Proof. We use the steepest descent method. Set φ(t) := − 2 (t + t ). We compute 1 φ0(t) = − (1 − t−2), φ00(t) = −t−3. 2 00 Hence φ has a critical point at t0 = 1 with φ(t0) = −1 and φ (t0) = −1. Thus Z ∞ 1 −m−1 Km(z) = t exp(zφ(t))dt 2 0 1 Z ∞ φ00(t ) ' exp zφ(t ) + z 0 (t − t )2 dt 2 0 2 0 −∞ √ 1 Z ∞ z 1 2π = e−z exp (t − 1)2 dt = e−z √ . 2 −∞ 2 2 z 2

Corollary 2.13 As x → ∞ we have

1 x Im(x) ∼ √ e . (2.56) 2πx Next we derive the precise asymptotics of the MacDonald function Theorem 2.14 ∞ 1 1 √ −z X n ( 2 − m)n( 2 + m)n Km(z) ∼ πe (−1) 1 . (2.57) n+ 2 n=0 n!(2z) Proof. To derive (2.57), at least formally, we ﬁrst we transform the equation:

1 1 z 2 2 2 2 2 − 2 −z e z z ∂z + z∂z − z − m z e (2.58) 1 = z2∂2 − 2z2∂ − m2 + . (2.59) z z 4 Acting with (2.59) on ∞ X −n cnz (2.60) n=0 we obtain ∞ X 1 n(n + 1)c z−n + 2nc z−n+1 − m2 − c z−n . (2.61) n n 4 n n=0 This yields the recurrence relation 1 2nc = − (n − 1)n − m2 + c (2.62) n 4 n−1 1 1 = − n − − m n − + m c . (2.63) 2 2 n−1 Therefore, ( 1 − m) ( 1 + m) c = (−1)n 2 n 2 n . (2.64) n n!2n 2

11 2.8 More integral representations Theorem 2.15 (Poisson-type representations.)

Km(z) √ ±m 1 Z z πΓ(∓m + 2 ) zt 2 ±m− 1 = e (1 − t ) 2 dt. (2.65) 2 2πi ]−∞,−1+,−∞[ √ ∞ z m π Z 1 1 −sz 2 m− 2 = 1 e (s − 1) ds, m > − ; (2.66) 2 Γ(m + 2 ) 1 2 −m 1 Z ∞ z Γ(m + 2 ) −isz 2 −m− 1 = √ e (s + 1) 2 ds, m > 0. (2.67) 2 2 π −∞ Proof. The RHS of (2.65) solves the modiﬁed Bessel equation by Thm 2.2. Let s us check its behavior for Rez → ∞. We set t = z − 1: Z 1 zt ±m− 1 ±m− 1 e (t + 1) 2 (−t + 1) 2 dt (2.68) 2πi ]−∞,−1+,−∞[ m− 1 −z Z 2 2 e 1 −s ±m− 1 ∼ e s 2 ds (2.69) z z 2πi ]−∞,−1+,−∞[ 2±m e−z = √ . (2.70) z 1 2zΓ( 2 ∓ m)

−z √ Therefore, the RHS of (2.65) behaves as e √ π . Therefore, it coincides with 2z Km(z). (2.67) follows from (2.65) by setting t = is. 2

2.9 MacDonald function for integer parameters Pn 1 Theorem 2.16 Set Hn := k=1 k . Then for m = 0, 1, 2,... z K (z) = (−1)m+1 log + γ I (z) m 2 m

m−1 2k−m m ∞ 2k+m 1 X z (m − k − 1)! (−1) X Hk + Hm+k z + (−1)k + . 2 2 k! 2 k!(m + k)! 2 k=0 k=0 Proof. Set d 1 1 φ(z) := = − ∂ log Γ(z). dz Γ(z) Γ(z) z Then φ(−n) = (−1)nn!, n = 0, 1, 2,...,

γ − H φ(n + 1) = n , n = 0, 1, 2,... n!

12 Besides,

∞ z X φ(m + k + 1)z m+2k ∂ I (z) = log I (z) + . m m 2 m k! 2 k=0 Hence for m = 0, 1, 2,...

∞ m+2k z X Hm+k z ∂nIn(z) = log + γ Im(z) − , (2.71) n=m 2 (m + k)!k! 2 k=0

m−1 2k−m z X m−k−1 (m − k − 1)!z ∂nIn(z) = log + γ I−m(z) + (−1) (2.72) n=−m 2 k! 2 k=0

∞ −m+2k X H−m+k z − . (2.73) (−m + k)!k! 2 k=m The last sum can be written as

∞ m+2k X Hk z − . k!(k + m)! 2 k=m We use the De L’Hopital rule:

π d (I − I (z)) K (z) = dm −m m m 2 d dm sin πm (−1)m d d = − In(z) − In(z) 2 dn n=−m dn n=m 2

Corollary 2.17 As x → 0, we have 1 xm I (x) ∼ , m 6= −1, −2,... ; (2.74) m Γ(m + 1) 2  − ln x − γ if m = 0,  2  Γ(m) 2 m Km(x) ∼ 2 x if Rem ≥ 0, m 6= 0; (2.75)   Γ(−m) x m 2 2 if Rem ≤ 0, m 6= 0. Clearly, for integer m 1 I (z) = Z (z). (2.76) m π m

13 2.10 Relationship to hypergeometric type functions

The modiﬁed Bessel function and the hypergeometric functions 0F1 and 1F1 are closely related:

1 z m z2 I (z) = F 1 + m; m Γ(m + 1) 2 0 1 4 1 z m 1 = e−z F m + ; 2m + 1; 2z . Γ(m + 1) 2 1 1 2

The MacDonald function is closely related to the hypergeometric function 2F0: √ 2π 1 1 1 K (z) = √ e−z F + m, − m; −; − . m z 2 0 2 2 2z

2.11 Recurrence relations Theorem 2.18

2∂zIm(z) = Im−1(z) + Im+1(z), (2.77)

2mIm(z) = zIm−1(z) − zIm+1(z). (2.78)

2∂zKm(z) = −Km−1(z) − Km+1(z), (2.79)

2mKm(z) = −zKm−1(z) + zKm+1(z). (2.80)

Proof. Recall that Z z −1 −m −m−2 2∂zIm(z) = exp (t + t ) (t + t )dt. γ 2

Z z −1 −m 0 = 2 ∂t exp (t + t ) t dt γ 2 Z z = −2m exp (t + t−1) t−m−1dt γ 2 Z z Z z +z exp (t + t−1) t−mdt − z exp (t + t−1) t−m−2dt. γ 2 γ 2

iπm By (2.46), e Km satisfy the same recurrence relations. 2

Corollary 2.19 m ∂ (zmI (z)) = zmI (z), or ∂ + I (z) = I (z), z m m−1 z z m m−1 m ∂ z−mI (z) = z−mI (z), or ∂ − I (z) = I (z). z m m+1 z z m m+1

14 Hence 1 n ∂ zmI (z) = zm−nI (z), (2.81) z z m m−n 1 n ∂ z−mI (z) = z−m−nI (z). (2.82) z z m m+n

Analoguous identities hold for Km(z).

2.12 Half-integral parameters 1 For m = 2 we have 1 − 1 2 2 z 2 I 1 z 2 = z ∂ − 1 . 2 z

− 1 ±z Hence, z 2 e is annihilated by I 1 . 2

1 1 2 z 2 1 sinh z 2 I 1 (z) = = sinh z (2.83) 2 1 2 Γ(1 + 2 ) z πz 1 1 2 z − 2 1 2 I 1 (z) = cosh z = cosh z, (2.84) − 2 1 2 Γ(1 − 2 ) πz 1 π 2 −z K± 1 (z) = e , (2.85) 2 2z 1 π 2 z X± 1 (z) = e . (2.86) 2 2z One way to derive (2.83) and (2.84) is to use r r 2n+ 1 π n n+1 π 2 2 n!Γ(1/2 + n + 1) = 2 n!2 (1/2) = (2n + 1)!, 2 n+1 2 r r 2n− 1 π n n π 2 2 n!Γ(−1/2 + n + 1) = 2 n!2 (1/2) = (2n)!. 2 n 2

15 For k = 0, 1, 2,... we have

1 2 2 1 1 k sinh z 2 +k I 1 +k(z) = z ∂z (2.87) 2 π z z 1 ∞ 2 2 k X 1 +2j−k (−j)k = (−2) z 2 π (2j + 1)! j=0

1 k = Z 1 +k(z) − (−1) K 1 +k(z) , π 2 2 1 2 2 1 1 k cosh z 2 +k I− 1 −k(z) = z ∂z (2.88) 2 π z z 1 ∞ 1 2 2 k X − 1 +2j−k ( 2 − j)k = (−2) z 2 π (2j)! j=0

1 k = Z 1 +k(z) + (−1) K 1 +k(z) , π 2 2 1 −z π 2 1 1 k e 2 +k K±( 1 +k)(z) = z − ∂z (2.89) 2 2 z z 1 ∞ 1 n π 2 k X − 1 −k+n n ( 2 − 2 )k = 2 z 2 (−1) , 2 n! n=0 k π = (−1) I− 1 −k(z) − I 1 +k(z) , 2 2 2 1 z π 2 1 1 k e 2 +k X±( 1 +k)(z) = z ∂z (2.90) 2 2 z z 1 ∞ 1 n π 2 k X − 1 −k+n ( 2 − 2 )k = (−2) z 2 2 n! n=0 π = I− 1 −k(z) + I 1 +k(z) . 2 2 2 Note that in the sum for (2.87) the terms with j = 0, . . . , k − 1 vanish.

2.13 Wronskians Recall that the Wronskian is defined as W (f, g) := fg0 − f 0g. The Wronskian of two solutions of the modified Bessel equation satisfies 1 ∂ + W (z) = 0. z z

1 Hence W (z) is proportional to z . Using 1 z ±m 1 z ±m−1 I (z) ∼ ,I0 (z) ∼ , ±m Γ(±m + 1) 2 ±m 2Γ(±m) 2

16 we can compute the Wronskian of Im and I−m, and then other Wronskians: 2 sin πm W (I ,I ) = − , (2.91) m −m πz 1 W (K ,I ) = , (2.92) m m z 2 W (K ,X ) = . (2.93) m m z 3 Standard Bessel equation 3.1 Bessel equation Replacing z with ±iz in the modiﬁed Bessel equation leads to the standard Bessel equation, given by the operator

2 2 2 2 Jm(z, ∂z) := z ∂z + z∂z + z − m .

3.2 Integral representations Theorem 3.1 Bessel–Schl¨aﬂirepresentations Let γ be a contour satisfying γ(1) z −1 z −1 1 (t + t ) + m exp (t − t ) m = 0, (3.94) 2 2 t γ(0) Then Z z −1 dt C exp (t − t ) m+1 (3.95) γ 2 t is a solution of the Bessel equation. Theorem 3.2 Poisson type representations Let γ be a contour satisfying

γ(1) 2 m+ 1 izt (1 − t ) 2 e = 0. γ(0) Then Z m 2 m− 1 izt z (1 − t ) 2 e dt γ is a solution of the Bessel equation.

3.3 Bessel function The Bessel function is deﬁned as

±iπ m Jm(z) = e 2 Im(∓iz) ∞ n z 2n+m X (−1) = 2 n!Γ(m + n + 1) n=0 1 −iπ m iπ m = e 2 K(−iz) − e 2 K(iz) iπ

17 ±iπ ±imπ Jm(e z) = e Jm(z).

Theorem 3.3 Let Rez > 0. (Bessel-Schl¨aﬂi-type representations) Z 1 z −1 dt Jm(z) = exp (t − t ) m+1 2πi ]−∞,0+,−∞[ 2 t 1 z m Z z2 ds = exp s − m+1 . 2πi 2 ]−∞,0+,−∞[ 4s s

(Poisson-type representations)

m Z 1 1 z 2 m− 1 izt 1 Jm(z) = √ (1 − t ) 2 e dt, m > − , 1 2 −1 2 πΓ m + 2 m Z 1 1 z m− 1 m− 1 Jm(z) = √ Γ − m (t − 1) 2 (t + 1) 2 . 2πi π 2 2 [1,−1−,1+]

3.4 Relationship to hypergeometric type functions

1 z m z2 J (z) = F 1 + m; − m Γ(m + 1) 2 0 1 4 1 z m 1 = e−iz F m + ; 2m + 1; 2iz . Γ(m + 1) 2 1 1 2

3.5 Bessel function for integer parameters

Let m ∈ Z. Theorem 3.4

∞ n z 2n+m X (−1) J (z) = (−1)mJ (z) = 2 . m −m n!(n + m)! n=0

Theorem 3.5 Z 1 z −1 dt Jm(z) = exp (t − t ) m+1 2πi [0+] 2 t 1 z m Z z2 ds = exp s − m+1 2πi 2 [0+] 4s s 1 Z π = cos(z sin φ − mφ)dφ. π 0

18 3.6 Hankel functions There are two Hankel functions. Both are analytic continuations of the Mac- Donald function – one to the lower and the other to the upper part of the complex plane:

± 2 ∓i π (m+1) H (z) = e 2 K (∓iz). m π m In the literature the usual (and less natural) notation for Hankel functions is

(1) + (2) − Hm (z) = Hm(z),Hm (z) = Hm(z). (3.96) Note the identities

π ±i π (m+1) ± K (z) = e 2 H (±iz), (3.97) m 2 ± ±mπi ± H−m(z) = e Hm(z), (3.98) 1 J (z) = H+(z) + H−(z) , (3.99) m 2 m m 1 J (z) = emπiH+(z) + e−mπiH−(z) , (3.100) −m 2 m m ie∓mπiJ (z) − iJ (z) H±(z) = ± m −m . (3.101) m sin mπ .

π Theorem 3.6 The following asymptotic formulas are true for −π ± 2 + δ < π arg z < π ± 2 − δ, δ > 0: ± Hm(z) lim 1 = 1, imπ iπ z→∞ 2 2 ±iz ∓ 2 ∓ 4 πz e e Here is a more precise asymptotics:

√ ∞ 1 1 ± 2 ±iz∓i πm ∓i π X ( 2 − m)n( 2 + m)n H (z) ∼ √ e 2 4 . (3.102) m πz n!(±2iz)n n=0

3.7 Integral representations of Hankel functions Let us first consider Bessel-Schläfli-type representations. For Rez > 0, Z + 1 z −1 dt Hm(z) = − exp (t − t ) m+1 , πi ]−∞,(0+1·0)−[ 2 t Z − 1 z −1 dt Hm(z) = exp (t − t ) m+1 . πi ]−∞,(0+1·0)+[ 2 t By ] − ∞, (0 + 1 · 0)−[ we understand the contour starting at −∞, encircling 0 clockwise and reaching zero from the positive direction. The contour is located on the upper halfplane.

19 Similarly, by ] − ∞, (0 + 1 · 0)+[ we understand the contour starting at −∞, encircling 0 counterclockwise and reaching zero from the positive direction. The contour is located on the lower halfplane. Note that z z 1 lim (t + t−1) + m exp (t − t−1) = 0, t→0+1·0 2 2 tm where by t → 0+1·0 we denote the convergence to zero through positive values of t (sometimes denoted by t → 0+). Hence the contours ]−∞, (0+1·0)+[ and ] − ∞, (0 + 1 · 0)−[ satisfy (3.94). + If 0 < arg z < π, then a good contour in the representation of Hm is [i∞, 0]. − If −π < arg < 0, then for Hm one can use [−i∞, 0]. This leads to the representations valid for ±Imz ≥ 0: ∞ π π 1 Z z ds ±i 2 m ± ∓i 2 m ± −1 e Hm(z) = e H−m(z) = ± exp ±i (s + s ) m+1 (3.103) πi 0 2 s 1 Z ∞ = ± exp (±iz cosh(t) − mt) dt. (3.104) πi −∞ Poisson type integral representations valid for Imz ≥ 0 and all m:

1 Γ( − m)z m Z 1 1 ± 2 izt m− 2 m− 2 Hm(z) = √ e (t − 1) (t + 1) dt. πi π 2 ]i∞,∓1∓,i∞[

1 Poisson type representations valid for m ≥ − 2 . ∞ π z m 2 Z 1 ±i 2 m ± ±isz 2 m− 2 e Hm(z) = ± √ 1 e (s − 1) ds. 2 i πΓ( 2 + m) 1 3.8 Hankel functions for integer parameters

i2 z H±(z) = J (z) ± log + γ J (z) (3.105) m m π 2 m

m−1 2k−m ∞ k 2k+m i X z (m − k − 1)! i X (−1) (Hk + Hm+k)z ∓ ∓ . (3.106) π 2 k! π k!(m + k)! 2 k=0 k=0

3.9 Neumann function Neumann function is deﬁned as 1 + − Ym(z) = 2i Hm(z) − Hm(z)

cos πmJm(z)−J−m(z) = sin πm . We then have

+ − Hm(z) = Jm(z) + iYm(z),Hm(z) = Jm(z) − iYm(z).

20 Theorem 3.7 For m ∈ Z we have

2 z Ym(z) = π log( 2 ) + γ Jm(z)

m−1 ∞ k 1 P (m−k−1)! z 2k−m 1 P (−1) z m+2k − π k! ( 2 ) − π k!(m+k)! ( 2 ) Hk + Hm+k . k=0 k=0

3.10 Recurrence relations Theorem 3.8 We have the identities

2∂zJm(z) = Jm−1(z) − Jm+1(z),

2mJm(z) = zJm−1(z) + zJm+1(z).

Sometimes, more convenient are the following forms of the recurrence relations: Corollary 3.9 m ∂ (zmJ (z)) = zmJ (z), or ∂ + J (z) = J (z), z m m−1 z z m m−1 m −∂ z−mJ (z) = z−mJ (z), or −∂ + J (z) = J (z). z m m+1 z z m m+1 Besides, 1 n ∂ zmJ (z) = zm−nJ (z), z z m m−n 1 n − ∂ z−mJ (z) = z−m−nJ (z). z z m m+n

± Analogous identities hold for Hm(z), and Ym(z).

3.11 Half-integer parameters

1 2 2 J 1 (z) = = sin z, 2 πz 1 2 2 J− 1 (z) = cos z, 2 πz 1 2 ± 2 ±iz H 1 (z) = ∓i e , 2 πz

1 2 ± 2 ±iz H− 1 (z) = e 2 πz

21 For k = 0, 1, 2,... we have

1 2 2 1 1 k sin z 2 +k J 1 +k(z) = z − ∂z (3.107) 2 π z z 1 ∞ 2 2 k X j 1 +2j−k (−j)k = 2 (−1) z 2 , π (2j + 1)! j=k 1 2 2 1 1 k cos z 2 +k J− 1 −k(z) = z ∂z (3.108) 2 π z z 1 ∞ 1 2 2 k X j − 1 +2j−k ( 2 − j)k = (−2) (−1) z 2 , π (2j)! j=0 1 ±iz 2 2 1 1 k e ± 2 +k H 1 +k(z) = ∓i z − ∂z (3.109) 2 π z z 1 ∞ 1 n 2 2 k X n − 1 −k+n ( 2 − 2 )k = ∓i 2 (±i) z 2 , π n! n=0 1 ±iz 2 2 1 1 k e ± 2 +k H− 1 −k(z) = z ∂z (3.110) 2 π z z 1 ∞ 1 n 2 2 k X n − 1 −k+n ( 2 − 2 )k = (−2) (±i) z 2 . π n! n=0

3.12 Wronskians of solutions of the Bessel equation The Wronskian of two solutions of the Bessel equation satisﬁes 1 ∂ + W (z) = 0. z z

1 Hence W (z) is proportional to z . Using

1 z ±m 0 1 z ±m−1 J±m(z) ∼ Γ(±m+1) ( 2 ) ,J±m(z) ∼ Γ(±m) ( 2 ) , we can compute the Wronskian of Jm(z) and J−m(z): 2 sin πm 4i 2 W (J ,J ) = − ,W (H−,H+) = − ,W (J ,Y ) = . m −m πz m m πz m m πz 3.13 Putting together the Hankel and Macdonald function ± Km and Hm are naturally analytic functions on the Riemann’s surface of the logarithm.√ However,√ it is sometimes convenient to treat them together, as functions of w, resp. −w, where w ∈ R. It is important to indicate precisely how the analytic continuation of the square root is performed—whether we bypass the branch point at zero from above or from below. This is encoded by adding

22 ±i0 to the variable. More precisely, we will use the following notation: ( √ √ Km w , w > 0, Km w ∓ i0 := √ π ±iπm ±√ (3.111a) Km(∓i −w = ±i 2 e Hm −w , w < 0; √ √ √ ( ± 2 ∓iπm ± Hm ± i w = ∓i π e Km w , w > 0, Hm −w ± i0 := ±√ (3.111b) Hm −w , w < 0. Note the functions (3.111a) and (3.111b) should be understood as distributions, possibly with a singularity at 0.

4 Helmholtz equation in 2 dimensions 4.1 Polar coordinates

Introduce in R2 polar coordinates x = r cos φ, y = r sin φ, p y r = x2 + y2, φ = arctan . x Remark 4.1 Change from Cartesian coordinates to polar cordinates can be interpreted as a unitary transformation U : L2(R2) → L2([0, ∞[×[0, 2π], rdrdφ) deﬁned as (Uf)(r, φ) := f(r cos φ, r sin φ). We have

2 −1 −2 2 ∆ = ∂r + r ∂r + r ∂φ. It is also convenient to introduce another variety of polar coordinates, setting w = eiφ:

r −1 x = 2 (w + w ), r −1 y = 2i (w − w ).

We have ∂φ = iw∂w, and so

2 −1 −2 2 ∆ = ∂r + r ∂r − r (w∂w) . (4.112)

4.2 Plane waves We look for solutions of the Helmholtz equation (∆ + 1)F = 0, where F is a distribution in S0(R2). We can assume that 1 ZZ F (x, y) = Fˆ(ξ, η)ei(xξ+yη)dξdη. 2π

23 We obtain

(−ξ2 − η2 + 1)Fˆ(ξ, η) = 0, so that suppFˆ ⊂ {(x, y) | x2 + y2 = 1}. Thus the Helmholtz equation is solved by 1 Z Ψ (x, y) = ei(x cos ψ+y sin ψ)g(ψ)dψ. (4.113) g 2π g is an arbitrary distribution on the circle S1. In particular, if we set g = 2πδψ, then the corresponding solutions are the plane waves i(x cos ψ+y sin ψ) fψ(x, y) := e , or in polar coordinates ir cos(φ−ψ) fψ(r, φ) = e .

4.3 Circular waves

We will denote by H the Hilbert space of functions on R2 of the form (4.113) with g ∈ L2(S1). We will treat them as “nice” solutions of the Helmholtz equation. We have the scalar product

Z 2π

(Ψg1 |Ψg2 ) := g1(ψ)g2(ψ)dψ. 0

Note that if we set g (ψ) = i−meimψ, then √1 g is an orthonormal basis in m 2π m L2(S1). We deﬁne the mth circular wave:

Z 2π 1 −m i(x cos ψ+y sin ψ) imψ fm := Ψgm = i e e dψ. (4.114) 2π 0 √ 2πfm form an orthonormal basis of H.

Proposition 4.2 In polar coordinates we have

imφ fm(r, φ) = Jm(r)e . (4.115)

Proof. Recall that Z 1 r −1 −m−1 Jm(r) = exp (t − t ) t dt (4.116) 2πi [0+] 2 1 Z 2π = eir sin φe−imφdφ. (4.117) 2π 0

24 Hence Z 2π 1 ir cos(φ−ψ) imψ −im π fm(r, φ) = e e e 2 dψ (4.118) 2π 0 1 Z 2π = eir sin ξe−imξeimφdξ (4.119) 2π 0 imφ = Jm(r)e , (4.120)

π where we set ξ = φ − ψ − 2 . 2 Clearly, by the inverse Fourier transformation, a plane wave can be decom- posed into circular waves:

∞ X −imψ m fψ = fme i . (4.121) m=−∞

Note also that (4.121) corresponds to the formula for the generating function.

∞ r (t−t−1) X m e 2 = t Jm(r). (4.122) m=−∞

i(φ−ψ+ π ) Indeed, when we insert t = e 2 in (4.122), we obtain (4.121).

4.4 Helmholtz equation on a disk Consider the disc K = {x2 + y2 < 1} and the Helmholtz equation on K with the Dirichlet 2 ∆DF + k F = 0,F (x, y) = 0, (4.123) and Neumann boundary conditions

2 ∆NF + k F = 0, (x∂x + y∂y)F (x, y) = 0. (4.124)

We are looking for F ∈ L2(K) and k ≥ 0 that solve these eigenvalue problems Recall that 1 ∂2 ∆ = ∂2 + ∂ + φ . (4.125) r r r r2 ± Solutions of the radial equation are spanned by Jm(kr) and Hm(kr), m = 0, 1, 2,... . We check which solutions are square integrable. We have Jm(kr) ∼ rm, m = 0, 1,... , therefore they are all square integrable. On the othere hand, ± −m Hm(kr) ∼ r , m = 1, 2,... , therefore they are not square integrable. ± The function H0 (kr) ∼ log(r), is square integrable. We however discard it as well—it corresponds to a “Dirac delta charge” at the origin. In fact, if we apply the Gauss law and compute the ﬂux through the circle around 0 we obtain Z 2π 1 rdφ∂r ln(r) = 2πr = 2π. 0 r

25 In ther words ∆ ln(r) = 2πδ0. Hence this solution corresponds to a “charge” equal to 2π at zero. Hence acceptable solutions are given by

imφ F (r, φ) = e Jm(kr), m ∈ Z. (4.126) The boundary conditions give

Jm(k) = 0, (4.127) 0 Jm(k) = 0. (4.128)

Thus the frequencies are the zeros of the Bessel function Jm, respectively of the 0 derivative of the Bessel function Jm.

4.5 Translation operator

For a function f on R and t ∈ R we deﬁne

(Utf)(x) := f(x + t).

Ut can be understood as a family of operators acting on various spaces of functions on R. It satisﬁes UtUs = Ut+s,Ut = 1. For instance, Ut can be interpreted as operator on L2(R), in which case they are unitary. We also have the operator ∂x. We have d U f = ∂ U f. dt t x t Therefore, we write t∂x Ut = e and call ∂x the generator of translations. In quantum mechanics customarily 1 instead of ∂x one uses the momentum operator p = i ∂x, which is Hermitian in the sense of L2(R).

4.6 Rotation operator on L2(R2) For a function f on R2 and for θ ∈ R we deﬁne

(Rθf)(x, y) := f(cos θx − sin θy, sin θx + cos θy).

We obtain a family of operators satisfying Rθ1 Rθ2 = Rθ1+θ2 , R0 = 1. In particular, understood as operators on L2(R2), they are unitary. Deﬁne L = x∂y − y∂x. We will show that d R f = LR f. (4.129) dθ θ θ

26 Introduce notation

x˜ := x cos θ − y sin θ, y˜ := x sin θ + y cos θ.

d R f(x, y) = (x sin θ + y cos θ)∂ f(˜x, y˜) dθ θ x˜ +(−x cos θ + y sin θ)∂y˜f(˜x, y˜),

LRθf(x, y) = x(sin θ∂x˜ + cos θ∂y˜)f(˜x, y˜)

−y(cos θ∂x˜ − sin θ∂y˜)f(˜x, y˜), which shows (4.129). Therefore, we write

θL Rθ = e .

In quantum mechanics customarily one uses the angular momentum operator 1 2 2 i L, which is Hermitian in the sense of L (R ). In polar coordinates the operator Rθ acts as

iθ (Rθf)(r, φ) = f(r, φ + θ),Rθf(r, w) = f(r, we ).

Here is L in the polar coordinates:

L = ∂φ = iw∂w.

2 4.7 The group R o SO(2) Consider translations and rotations on R2:

u(x1,y1)(x, y) = (x − x1, y − y1), (4.130)

rθ(x, y) = (x cos θ + y sin θ, −x sin θ + y cos θ). (4.131)

Transformations of the form (x1, y1, θ) := u(x1,y1)rθ form a group:

(x2, y2, θ2)(x1, y1, θ1)

=(x2 + cos θ2x1 + sin θ2y1, y2 − sin θ2x1 + cos θ2y1, θ2 + θ1) (4.132)

We have an obvious complex form of this group, where we write zi = xi + iyi, −iθi wi = e . Then (4.132) corresponds to

(z2, w2)(z1, w1) = (z2 + w2z1, w2w1). (4.133)

For a function f on R2 set U f(x) = f u−1 (x, y) , (4.134) (x1,y1) (x1,y1) −1 Rθf(x, y) = f rθ (x, y) (4.135)

27 We have

(U(x1,y1)f)(x) := f(x + x1, y + y1), (4.136)

(Rθf)(x, y) := f(cos θx − sin θy, sin θx + cos θy). (4.137)

These operators can be understood on various spaces of functions on R2, e.g. on L2(R2) We have a representation 2 2 ) R o SO(2) 3 (x, y, θ) 7→ U(x, y, θ) := U(x, y)R(θ) ∈ U(L (R ). The Laplacian ∆ is an invariant operator: U(x, y, θ)∆U(x, y, θ)−1 = ∆. (4.138)

2 4.8 Lie algebra R o so(2)

The operators ∂x, ∂y and L form a Lie algebra commuting with ∆:

[∂x,L] = ∂y, (4.139)

[∂y,L] = −∂x. (4.140) We have

−1 ∂x = cos φ∂r − r sin φ∂φ, −1 ∂y = sin φ∂r + r cos φ∂φ. (4.141) Therefore,

+ iφ −1 −1 A := ∂x + i∂y = e (∂r + ir ∂φ) = w(∂r − r w∂w), − −iφ −1 −1 −1 A := ∂x − i∂y = e (∂r − ir ∂φ) = w (∂r + r w∂w),

Note the relations [A+,A−] = 0, (4.142) [A±,L] = ±A±, (4.143) ∆ = A+A− = A−A+, (4.144) [∆,L] = [∆,A±] = 0. (4.145)

4.9 Action of translations and rotations on solutions of the Helmholtz equation

The operators U(x,y) preserve H and are unitary:

U(x0,y0)Ψg = Ψ ˆ , (4.146) U(x0,y0)g ˆ i cos ψx0+i sin ψy0 U(x0,y0)g(ψ) = e g(ψ), (4.147)

∂xΨg = Ψi cos ψg, (4.148)

∂yΨg = Ψi sin ψg (4.149)

28 In other words, plane waves diagonalize translations:

i(x0 cos ψ+y0 sin ψ) Ux0,y0 fψ = e fψ.

∂xfψ = i cos ψfψ, ∂yfψ = i sin ψfψ. (4.150)

The operators Rθ preserve the space H and are unitary:

R Ψ = Ψ , (4.151) θ g Uˆθ g

Rˆθg(ψ) = g(ψ + θ), (4.152)

LΨg = Ψ∂ψ g. (4.153)

Thus circular waves diagonalize rotations:

imθ Rθfm = e fm.

Lfm = imfm. By (4.143) A± raises/lowers L by 1. This is expressed in the following proposition:

Proposition 4.3

+ A fm = fm+1, (4.154) − A fm = fm−1. (4.155)

Proof.

± ±1 −1 m A fm = w (∂r ∓ r w∂w)Jm(r)w ±1 −1 m = w (∂rJm(r) ∓ r mJm(r))w m±1 = Jm±1(r)w = fm±1.

We used m ∂ + J (z) = J (z), (4.156) z z m m−1 m −∂ + J (z) = J (z). (4.157) z z m m+1 2

4.10 Graf addition formula Theorem 4.4 Assume that R, r, ρ and Φ, φ, ψ are related as s q reiφ + ρeiψ R = (reiφ + ρeiψ)(re−iφ + ρe−iψ), eiΦ = . re−iφ + ρe−iψ

29 Then ∞ imΦ X inψ i(m−n)φ Jm(R)e = Jn(ρ)e Jm−n(r)e . n=−∞ If m ∈ Z, then there are no restrictions on the parameters in the formula. If m is nonintegral and all variables real, then one has to assume that ρ < r (or, π equivalently, |Φ − φ| < 2 ). We can then replace the Bessel function in Jm(R) (i) and Jm−n(r) with Hm or Ym

Proof. We put ψ˜ = ψ − φ, Φ˜ = Φ − φ. Then the problem is reduced to the case φ = 0.

∞ X inψ Jn(ρ)e Jm−n(r) n=−∞ ∞ 1 X Z r = exp (t − t−1)t−m−1J (ρ)(teiψ)n 2πi 2 n n=−∞ γ 1 Z r ρ = exp (t − t−1) + teiψ − (teiψ)−1 t−m−1dt 2πi γ 2 2 Z 1 R −1 −m−1 imΦ imΦ = exp (s − s ) s dse = e Jm(R). 2πi γ 2

In the ﬁrst step we used the integral representation of Jm−n(r), in the second step we used the generating function to sum up Jn(ρ), ﬁnally, in the last step we used r + ρeiψ = ReiΦ and turned the contour. 2 Substituting x1 = r cos φ, y1 = r sin φ,

x2 = ρ cos ψ, y2 = ρ sin ψ, x = R cos Φ, y = R sin Φ, we obtain (x1, y1) + (x2, y2) = (x + y) and the addition formula can be rewritten as m p 2 2 x + iy Jm( x + y ) px2 + y2 q n q m−n X 2 2 x2 + iy2 2 2 x1 + iy1 = Jn( x + y ) Jm−n( x + y ) . 2 2 px2 + y2 1 1 px2 + y2 n∈Z 2 2 1 1

Writing U(x, y) for U(x,y), we can rewrite the Graf addition formula as m−n X p 2 2 x + iy U(x1, y1)fm (x2, y2) = fn(x2, y2)Jm−n( x + y ) . px2 + y2 n∈Z (4.158)

30 Let Unm(x, y) denote the matrix of U(x, y) in the basis of circular waves. By (4.158), m−n p 2 2 x + iy Unm(x, y) = Jm−n( x + y ) . (4.159) px2 + y2 The matrix of the operator U(x, y, θ) in the basis of the circular waves is m−n p 2 2 x + iy imθ Unm(x, y, θ) := Jm−n( x + y ) e . (4.160) px2 + y2 We have the representation property:

∞ X Uk,m (x2, y2, θ2)(x1, y1, θ1) = Uk,n(x2, y2, θ2)Un,m(x1, y1, θ1). m=−∞

5 Distributions in d = 1 5.1 Homogeneous distributions of order −1 and 0 1 1 The function x is not in Lloc, therefore it does not deﬁne a regular distribution. However, it can be naturally interpreted as a distribution as follows Z 1 Z a Z ∞ 1 Z a 1 P φ(x)dx := + φ(x)dx + φ(x) − φ(0)dx. x −∞ a x −a x (We are writing P to indicate that it is not the usual integral). Equivalently, 1 1 1 1 := + . x 2 (x + i0) (x − i0) The Sochocki formula is relationship between three kinds of order −1 distributions: 1 1 = ∓ iπδ(x). x ± i0 x

Z e−ixkdx = 2πδ(k), (5.161) Z ∓i θ(±x)e−ixkdx = , (5.162) k ∓ i0 Z 1 sgn(x)e−ixkdx = −2i , (5.163) k Z δ(x)e−ixkdx = 1, (5.164)

Z e−ikx dx = ∓2πiθ(±k), (5.165) x ± i0 Z e−ikx P dx = −πisgn(k), (5.166) x

31 ( Z −2πiθ(ξ)e−iλξ Imλ < 0, e−iξs(s − λ)−1d s = (5.167) 2πiθ(−ξ)e−iλξ Imλ > 0; Z P e−iξs(s − λ)−1d s = −πisgn(ξ)e−iλξ, Imλ = 0. (5.168)

5.2 Homogeneous distributions of integral order Deﬁne for n = 0, 1, 2,... 1 1 1 1 := + . xn+1 2 (x + i0)n+1 (x − i0)n+1

Clearly 1 1 ∂n = (−1)nn! . x x xn+1

Z xne−ixkdx = 2πinδ(n)(k), (5.169)

Z (−i)n+1n! xnθ(±x)e−ixkdx = ± , (5.170) (k ∓ i0)n+1 Z 1 xnsgn(x)e−ixkdx = 2(−i)n+1n! , (5.171) kn+1 Z δ(n)(x)e−ixkdx = inkn, (5.172)

Z e−ikx 2π(−i)n+1 dx = ± knθ(±k), (5.173) (x ± i0)n+1 n! Z e−ikx π(−i)n+1 P dx = knsgn(k). (5.174) xn+1 n!

5.3 Homogeneous distributions of arbitrary order I

For any λ ∈ C (±ix + 0)λ := lim(±ix + )λ. →0 is a tempered distribution. If Reλ > −1, then it is simply the distribution given by the locally integrable function

±isgn(x) π λ λ e 2 |x| . (5.175)

The functions λ λ x± := (±x) θ(±x) (5.176)

32 deﬁne distributions only for Reλ > −1. We can extend them to λ ∈ C except for λ = −1, −2,... by putting

λ 1 −i π λ λ i π λ λ x := − e 2 (∓ix + 0) + e 2 (±ix + 0) . (5.177) ± 2i sin πλ λ Instead of x± it is often more convenient to consider xλ ρλ (x) := ± (5.178) ± Γ(λ + 1)

Γ(−λ) −i π λ λ i π λ λ = e 2 (∓ix + 0) − e 2 (±ix + 0) . (5.179) 2πi λ Note that using (5.178) and (5.179) we have deﬁned ρ± for all λ ∈ C. λ Theorem 5.1 The distributions ρ± satisfy the recurrence relations

λ λ−1 ∂xρ±(x) = ±ρ± (x). At integers we have xn ρn = ± , n = 0, 1,... ; (5.180) ± n! −n−1 n n ρ± = (±1) δ (x), n = 0, 1,.... (5.181) Their Fourier transforms are below: Z −iξx λ −λ−1 e ρ±(x)dx = (±iξ + 0) , Z −iξx λ −λ−1 e (∓iξ + 0) dξ = 2πρ± (x).

Proof. (5.181) follows from (∓1)n+1n! ρ−n−1(x) = (x ± i0)−n−1 − (x ∓ i0)−n−1 (5.182) ± 2πi (±1)n = − ∂n (x ± i0)−1 − (x ∓ i0)−1 . (5.183) 2πi x 2

Theorem 5.2 Let −n − 1 < Reλ, λ 6∈ {..., −2, −1}. Then for any a > 0, Z Z ∞ λ λ P x+φ(x)dx = x φ(x)dx a n−1 Z a X xj + xλ φ(x) − φ(j)(0) dx j! 0 j=0 n−1 j X X (−1)l + aλ+j+1φ(j)(0) . (5.184) (j − l)!(λ + 1) ··· (λ + 1 + l) j=0 l=0

33 If −n − 1 < Reλ < −n, we can even go with a to inﬁnity

n−1 Z Z ∞ X xj P xλ φ(x)dx = xλ φ(x) − φ(j)(0) dx. (5.185) + j! 0 j=0

Proof. We use induction. Suppose that the formula is true for λ Z Z λ−1 λ −λP x+ φ(x)dx = P x+∂xφ(x)dx (5.186) Z ∞ λ = x ∂xφ(x)dx (5.187) a n Z a X xj + xλ∂ φ(x) − φ(j)(0) dx x j! 0 j=0 n−1 j X X (−1)l + aλ+j+1φ(j+1)(0) . (j − l)!(λ + 1) ··· (λ + 1 + l) j=0 l=0

Then we integrate by parts, obtaining the identity for λ − 1. 2

5.4 Homogeneous distributions of arbitrary order II We also can deﬁne

λ 1 λ λ |x| = π (−ix + 0) + (ix + 0) , (5.188) 2 cos( 2 λ) λ 1 λ λ |x| sgn(x) = π − (−ix + 0) + (ix + 0) . (5.189) 2i sin( 2 λ) The Fourier transforms:

Z λ+1 −λ−1 λ −ixk 1 Γ( 2 ) x |k| e dk = π 2 , (5.190) λ 2 Γ(− 2 ) Z λ+2 −λ−1 λ −ixk 1 Γ( 2 ) x |k| sgn(k)e dk = −iπ 2 sgn(x), (5.191) 1−λ 2 Γ( 2 ) Especially symmetric expressions for Fourier transforms are obtained if we

34 introduce 2 λ λ 1−1x 2 ηλ (x) := Γ + (5.192) ev 2 2 2 −1 λ 1 − λ λ λ = (2π) Γ − + 2 2 (ix + 0) + (−ix + 0) (5.193) 2 2 λ 2 2 λ = √ Γ 1 + ρλ (x) + ρλ (x) , (5.194) π 2 + −

2 λ+1 λ −1x 2 1 ηλ (x) := Γ + 1 (5.195) odd 2 2 x −1 λ − λ − 1 λ λ = i(2π) Γ − 2 2 2 (ix + 0) − (−ix + 0) (5.196) 2 λ − 1 2 2 2 1 λ = √ Γ + ρλ (x) − ρλ (x) . (5.197) π 2 2 + − We then have the following relations:

λ λ−1 λ λ−1 ∂xηev = ληodd , ∂xηodd = ηev , (5.198) λ λ+1 λ λ+1 xηev(x) = (λ + 1)ηodd (x), xηodd(x) = ηev (x); (5.199) λ −λ−1 λ −λ−1 Fηev = ηev , Fηodd = −iηodd ; (5.200) √ (−1)m 2 η−1−2m(x) = δ(2m)(x), m = 0, 1,... ; (5.201) ev 2m 1 2 √m (−1)m 2 η−2m(x) = δ(2m−1)(x), m = 1, 2,.... (5.202) odd m 1 2 2 m 5.5 Anomalous distributions of degree −1

θ(±k) −1 We introduce the distributions k = ±k± : Z Z ∞ −1 (1) k+ φ(k)dk := − log(k)φ (k)dk (5.203) 0 Z a φ(k) − φ(0) Z ∞ φ(k) = dk + log(a)φ(0) + dk (5.204) 0 k a k Z ∞ φ(k) = lim + φ(0) ln() , (5.205) &0 k Z Z 0 −1 (1) k− φ(k)dk := − log(−k)φ dk (5.206) −∞ Z 0 φ(k) − φ(0) Z −a φ(k) = − dk + log(a)φ(0) − dk (5.207) −a k −∞ k Z − φ(k) = lim − + φ(0) ln() . (5.208) &0 −∞ k

35 We have 1 = −k−1 + k−1, k − + We also deﬁne 1 = k−1 + k−1. |k| − +

1 Proposition 5.3 Here are the Fourier transform of various forms of |k| and the logarthm: Z −1 −ixk k± e dk = − log(±ix + 0) − γ (5.209) iπ = − log |x| ∓ sgn(x) − γ, (5.210) 2 Z 1 P e−ixkdk = −2 log |x| − 2γ, (5.211) |k| Z 1 log |x|e−ixkdx = −πP − 2πγδ(k), (5.212) |k| Z θ(∓k) log(±ix + 0)e−ixkdx = −2π − 2πγδ(k), (5.213) |k| Z −ixk −1 log(x ∓ i0)e dx = −2πk∓ + (−2πγ ∓ iπ)δ(k), (5.214) Z −ixk −iλk −1 log(x − λ)e dx = e − 2πk∓ + (−2πγ ∓ iπ)δ(k) , ±Imλ > 0. (5.215)

Proof. We start from one of the formulas for the Euler constant. We change the variable from k to yk, with y > 0: Z ∞ −γ = e−k log(k)dk 0 Z ∞ = e−yk log(ky)d(ky) 0 Z ∞ Z ∞ = y log(y) e−kydk + y e−ky log(k)dk 0 0 Z 1 Z ∞ 1 −ky −ky = y log(y) − ∂k(e − 1) log(k)dk − ∂ke log(k)dk y 0 1 Z 1 e−ky − 1 Z ∞ e−ky = log(y) + dk + dk. 0 k 1 k The rhs is analytic in y on the right halfplane. It is constant on the positive halﬂine. So it is constant on the whole halfplane. Therefore, we can replace y with ix. This proves (5.209), which implies (5.210) and (5.211).

36 By inverting the Fourier transform we obtain (5.212) and (5.214). We can also get (5.214) from (5.212): Z Z log(x ∓ i0)e−ixkdx = log |x| ∓ iπθ(−x)e−ixkdx (5.216) 1 1 = −πP − 2πγδ(k) ± π (5.217) |k| (k + i0) 1 = −2πP + (−2πγ ∓ iπ)δ(k). (5.218) k∓ 2 Here is an alternative approach: 1 1 1 = lim − δ(k) , (5.219) ν&0 1+ν k± k± ν 1 1 2 = lim − δ(k) (5.220) |k| ν&0 |k|1+ν ν

Here is the computation of the Fourier transform by this method:

Z e−ikx Z e−ikx 1 P dk ≈ 1+ν dk − (5.221) k± k± ν 1 = Γ(ν)(±ik + 0)−ν − , (5.222) ν 1 1 ≈ − γ 1 − ν log(±ik + 0) − , (5.223) ν ν ≈ − log(±ik + 0) − γ. (5.224)

5.6 Anomalous distributions of integral degree Deﬁne (∓1)n k−n−1 := ∂nk−1, (5.225) ± n! k ± sgn(k) := k−n−1 + (−1)nk−n−1. (5.226) kn+1 + −

37 Theorem 5.4 Using Hn deﬁned in (.449), we have

(n) 1 n δ (x) 1+n + (∓1) Hn (5.227) x± n! 1 (∓1)n δ(n)(x) = lim − , (5.228) ν→0 1+n−ν x± ν n! 1 δ(n)(x) + (−1)n + 1H (5.229) |x|1+n n n! 1 (−1)n + 1 δ(n)(x) = lim − , (5.230) ν→0 |x|1+n−ν ν n! sgn(x) δ(n)(x) + (−1)n − 1H (5.231) |x|1+n n n! sgn(x) (−1)n − 1 δ(n)(x) = lim − . (5.232) ν→0 |x|1+n−ν ν n!

−n−1 Proof. It is enough to consider only x+ .

Z Z ∞ n+1 ν −n−1+ν ∂x x φ(x) P x+ φ(x)dx = P dx (5.233) 0 ν(ν − 1) ··· (ν − n) Z ∞ xν φ(n+1)(x) = + dx (5.234) 0 (−ν)(1 − ν) ··· (n − ν) Z ∞ (xν − 1)φ(n+1)(x) = + dx (5.235) 0 (−ν)(1 − ν) ··· (n − ν) Z ∞ φ(n+1)(x) + dx (5.236) 0 (−ν)(1 − ν) ··· (n − ν) Z ∞ log(x)φ(n+1)(x) = − dx (5.237) 0 n! 1 φ(n)(0) + (5.238) ν (1 − ν) ··· (n − ν) Z −n−1 = P x+ φ(x)dx (5.239) 1 φ(n)(0) φ(n)(0) + + H + O(ν). (5.240) ν n! n n! 2 The above proof is taken from Hörmander, sect. 3.2. Note that Hörmander −n−1 treats (5.228) as the standard regularization of x± . We prefer the definition (5.225).

38 Theorem 5.5

n−1 Z Z ∞ 1 X xj xn P x−n−1φ(x)dx = φ(x) − φ(j)(0) − φ(n)(x) dx + xn+1 j! n! 0 j=0 Z 1 1 φ(n)(x) φ(n)(0) Z ∞ 1 φ(n)(x) + − dx + dx 0 x n! n! 1 x n! φ(n)(0) − H . (5.241) n! n Proof. Let a > 0. If we assume that Reν > −1, then we can use (5.184) with n replaced with n + 1: Z 1 P n+1−ν φ(x)dx (5.242) x+ Z ∞ 1 = n+1−ν φ(x)dx a x n Z a 1 X xj + φ(x) − φ(j)(0) dx xn+1−ν j! 0 j=0 n j X X 1 − a−n+j+ν φ(j)(0) . (5.243) (j − l)!(n − l − ν) ··· (n − ν) j=0 l=0

The last term of (5.243) is

n−1 X 1 − aν φ(n)(0) (5.244) (n − l)!(n − l − ν) ··· (n − ν) l=0 1 − aν φ(n)(0) (5.245) (−ν) ··· (n − ν) H = − φ(n)(0) n (5.246) n! 1 1 1 + φ(n)(0) + log(a)φ(n)(0) (5.247) ν n! n! H + φ(n)(0) n + O(ν) (5.248) n! 1 1 1 = φ(n)(0) + log(a)φ(n)(0) + O(ν). (5.249) ν n! n!

39 Thus we have proven that Z Z ∞ −n−1 1 P x+ φ(x)dx = n+1 φ(x)dx a x n Z a 1 X xj + φ(x) − φ(j)(0) dx xn+1 j! 0 j=0 n−1 j X X 1 − a−n+jφ(j)(0) (5.250) (j − l)!(n − l) ··· n j=0 l=0 φ(n)(0) φ(n)(0) − H + log(a) (5.251) n! n n! Then we take a → ∞, noting that Z a x−1dx = log(a). 1 2

Proposition 5.6 The Fourier transform: Z (∓ix)n P k−n−1e−ixkdk = − log(±ix + 0) − γ (5.252) ± n! (∓ix)n iπ = − log |x| ∓ sgn(x) − γ (5.253) n! 2 Proof. We use (5.228): Z ∞ e−ixk (∓ix)n Z ∞ e−ixk (∓ik)n n+1 dk + Hn = lim 1+n−ν dk − (5.254) 0 k n! ν&0 0 k νn! (∓ik)n = lim Γ(−n + ν)(±ix + 0)n−ν − (5.255) ν&0 νn! n n ! (−1) 1 n (∓ik) = lim − γ + Hn (±ix) 1 − ν log(±ix + 0) − ν&0 n! ν νn! (∓ix)n = − γ + H − log(±ix + 0). (5.256) n! n

Note that the terms containg Hn cancel. 2

5.7 Infrared regularized distributions Theorem 5.7 Let n + 1 > 2α > n. Then θ(k) θ(k) = lim (5.257) k2α m→0 (k2 + m2)α n−1 j j X Γ(α − − 1 )Γ( + 1 ) − 2 2 2 2 (−1)jδ(j)(k) . (5.258) 2m2α−j−1Γ(α)j! j=0

40 Proof. Clearly,

n−1 Z θ(k)φ(k) Z ∞ 1 X kj P dk = φ(k) − φ(j)(0) dk (5.259) k2α k2α j! 0 j=0 is the limit as m → 0 of

n−1 Z ∞ 1 X kj φ(k) − φ(j)(0) dk. (5.260) (k2 + m2)α j! 0 j=0

Now

Z ∞ j j 1 j 1 k Γ(α − 2 − 2 )Γ( 2 + 2 ) 2 2 α dk = 2α−j−1 (5.261) 0 (k + m ) 2m Γ(α) 2

Theorem 5.8 Let 2p + 1 > 2α > 2p − 1, p = 1, 2,... . Then 1 1 = lim (5.262) |k|2α m→0 (k2 + m2)α

p−1 3 −2α+2l+1 l X π 2 m (−1) − δ(2l)(k) . (5.263) Γ(α) sin π(α − 1 )22ll!Γ( 3 − α + l) l=0 2 2

Proof. Clearly,

p−1 Z φ(k) Z 1 X k2l P dk = φ(k) − φ(2l)(0) dk (5.264) k2α k2α (2l)! l=0 is the limit as m → 0 of

p−1 Z 1 X k2l φ(k) − φ(2l)(0) dk. (5.265) (k2 + m2)α (2l)! l=0 Now

1 Z k2l m−2α+2l+1Γ(α − l − 1 )Γ(l + 1 ) dk = 2 2 (5.266) (2l)! (k2 + m2)α Γ(α)(2l)! 3 −2α+2l+1 l π 2 m (−1) = . (5.267) 1 2l 3 Γ(α) sin π(α − 2 ) 2 l!Γ( 2 − α + l) 2

41 Theorem 5.9 Let n = 0, 1,... . Then

θ(k) θ(k) = lim (5.268) n+1 n + 1 k m→0 (k2 + m2) 2 2 n−1 j j X m−n+jΓ( n − )Γ( + 1 ) − 2 2 2 2 (−1)jδ(j)(k) (5.269) 2Γ(α)j! j=0

 1 1 1 (2p+1)  Hp+1( ) + log(m) (−1)δ (k), n = 2p + 1; − 2 2 (2p+1)! (5.270) 1 1 (2p)  2 Hp − log(2) + log(m) (2p)! δ (k), n = 2p.

Proof. Clearly,

Z θ(k)φ(k) P dk (5.271) k2α is the limit as m → 0 of

n−1 Z ∞ 1 X kj φ(k) − φ(j)(0) dk (5.272) (k2 + m2)n+1 j! 0 j=0 Z 1 kn φ(n)(0) φ(n)(0) − − Hn . (5.273) 2 2 n + 1 0 (k + m ) 2 2 n! n!

Now for n = 2p + 1 we have

Z 1 k2p+1 2 2 p+1 dk (5.274) 0 (k + m ) p X k2j 1 Z 1 k = − + dk (5.275) 2j(k2 + m2)j 0 k2 + m2 j=1 0 p X 1 1 t = − + arctan (5.276) 2j(1 + m2)j 2 m j=1 1 = − H − log(m) + o(m0). (5.277) 2 p Then we use 1 1 1 H − H = H . (5.278) 2p+1 2 p 2 p+1 2

42 For n = 2p we compute Z 1 k2p dk (5.279) 2 2 p+ 1 0 (k + m ) 2 p−1 k2j+1 1 Z 1 k X = − 1 + 1 dk (5.280) 2 2 j+ 2 0 2 2 2 j=0 (2j + 1)(k + m ) 0 (k + m ) p X 1 p 2 = − 1 + log 1 + 1 + m − log(m) (5.281) 2 j+ 2 j=1 (2j + 1)(1 + m ) 1 1 = − H + log(2) − log(m) + o(m0). (5.282) 2 p 2 Then we use 1 1 1 H − H = H . (5.283) 2p 2 p 2 2 p 2 Using (2.67) we derive

−ikx 1 −2α+1 1 Z e 2π 2 m m|x|α− 2 dk = Kα− 1 (m|x|). (5.284) (k2 + m2)α Γ(α) 2 2 1 Note that (5.284) is bounded if α > 2 , has a logarithmic singularity at zero 1 2α−1 1 if α = 2 , and has a singularity |x| if α < 2 . Therefore, it is no longer a 2 2 regular distribution if α < 0. However, by applying (−∂x+m ) suﬃciently many times to (5.284) we can interpret it as a distribution for all α. (For α = −n, 2 2 n n = 0, −1, −2,... we simply obtain (−∂x + m ) ). Suppose now that the assumptions of Thm 5.8 are satisﬁed. Let us compute the Fourier transform of the linear combination of the deltas:

p−1 3 −2α+2l+1 l Z X π 2 m (−1) δ(2l)(k)e−ikxdk (5.285) Γ(α) sin π(α − 1 )22ll!Γ( 3 − α + l) l=0 2 2 p−1 3 −2α+2l+1 X π 2 m = x2l. (5.286) Γ(α) sin π(α − 1 )22ll!Γ( 3 − α + l) l=0 2 2 Now the rhs of (5.284), using the identity (2.48), can be written as

3 −2α+1 1 π 2 m m|x|α− 2 I 1 (m|x|) (5.287) 1 −α+ 2 Γ(α) sin π(α − 2 ) 2 3 −2α+1 1 π 2 m m|x|α− 2 − I 1 (m|x|) (5.288) 1 α− 2 Γ(α) sin π(α − 2 ) 2 Now (5.287) is equal to (5.286) modulo O(m−2α+2p+1). (5.287) is equal to

1 1 π 2 Γ( − α)|x|2α−1 2 (5.289) Γ(α) 2 modulo O(m2). This is a conﬁrmation of the correctness of Thm 5.8.

43 5.8 Distributions on halﬂine We will denote by C∞[0, ∞[ smooth function having all right-sided derivatives at 0. We set

∞ ∞ (2m+1) CN [0, ∞[ := {φ ∈ C [0, ∞[: φ (0) = 0, m = 0, 1,... }, (5.290) ∞ ∞ (2m) CD [0, ∞[ := {φ ∈ C [0, ∞[: φ (0) = 0, m = 0, 1,... }. (5.291)

0 0 SN[0, ∞[, SD[0, ∞[ have obvious deﬁnitions. We set SN[0, ∞[, SD[0, ∞[ to be their duals. Note that ∂x and the multiplication by x map SN[0, ∞[ into SD[0, ∞[ and 0 0 vice versa, as well as SN[0, ∞[ into SD[0, ∞[ and vice vera. The cosine transformation with the kernel r 2 F (x, k) := cos(xk) N π

0 maps SN[0, ∞[ into into itself. We have Likewise, the sine transformation with the kernel

r 2 F (x, k) := sin(xk) N π

0 maps SD[0, ∞[ into into itself. Let Iφ(x) := φ(−x). I maps S(R), as well as extends to a map of S0(R) into itself. We will write

Sev(R) := {φ ∈ S(R): Iφ = φ}, (5.292) 0 0 Sev(R) := {λ ∈ S (R): Iλ = λ}, (5.293) Sodd(R) := {φ ∈ S(R): Iφ = −φ}, (5.294) 0 0 Sodd(R) := {λ ∈ S (R): Iλ = −λ}. (5.295)

If φ ∈ SN[0, ∞[, we set ( φ(x) x ≥ 0; φev(x) := φ(−x) x ≤ 0.

ev Note that φ ∈ Sev(R). 0 If λev is an even distribution in S (R), then we can associate with it a 0 distribution in SN[0, ∞[ by Z ∞ Z 1 ev λN(x)φ(x)dx := λev(x)φ (x)dx. 0 2

Similarly, if φ ∈ SD[0, ∞[, we set ( φ(x) x ≥ 0; φodd(x) := −φ(−x) x ≤ 0.

44 odd Note that φ ∈ Sodd(R). 0 If λodd is an odd distribution in S (R), then we can associate with it a 0 distribution in SD[0, ∞[ by We set Z ∞ Z 1 odd λD(x)φ(x)dx := λodd(x)φ (x)dx. 0 2

The usual Fourier transform F preserves Sev(R) and Sodd(R). The Fourier transform on even distributions is closely related to the cosine transform and on odd distributions to the sine transform:

0 FNλN = (Fλ)N, λ ∈ Sev(R), (5.296) 0 FDλD = i(Fλ)D, λ ∈ Sodd(R). (5.297)

An example of an even distribution is ηev. Let ηN denote the corresponding 0 distributtion in SN[0, ∞[. Likewise, an example of an odd distribution is ηodd. Let ηD denote the 0 corresponding distributtion in SD[0, ∞[. We have

λ −λ−1 λ −λ−1 FNηN = ηN , FDηD = ηD ; (5.298)

6 Distributions in arbitrary dimension 6.1 Sphere Sd−1

Consider the Euclidean space Rd. Introduce two varieties of spherical coordinates on a d − 1-dimensional sphere

(θd−2, . . . , θ1, φ) ∈ [0, π] × · · · × [0, π] × [0, 2π[,

(wd−2, . . . , w1, φ) ∈ [0, π] × · · · × [0, π] × [0, 2π[, d−1 with wj = cos θj, The spherical measure on S is d−2 sin θd−2dθd−2 ··· sin θ1dθ1dφ 2 (d−3)/2 = (1 − wd−2) dwd−2 ··· dw1dφ. Theorem 6.1 The area of the d − 1-dimensional sphere is

d 2π 2 Sd−1 = d , Γ( 2 ) or, in a more elementary form, 2πm+1 S = , m = 0, 1,... ; (6.299) 2m+1 m! 2πm S2m = 1 , m = 0, 1,.... (6.300) ( 2 )m

45 Proof. Method I. We compute in two ways the Gaussian integral: in the Cartesian coordinates Z −x2−···−x2 d e 1 d dx1 ··· dxd = π 2 , and in spherical coordinates: Z ∞ −r2 d−1 1 d Sd−1 e r dr = Γ . (6.301) 0 2 2 Method II. We compute the area of the sphere in the spherical coordinates: Z π Z π Z 2π d−2 Sd−1 = sin φd−1dφd−1 ··· sin φ2dφ2 dφ1 0 0 0 Then we use √ Z π k−1 Z 2π k−1 πΓ( 2 ) sin φkdφk = k , k = 2, . . . , d − 1; dφ1 = 2π. 0 Γ( 2 ) 0 2

6.2 Homogeneous functions in arbitrary dimension

Theorem 6.2 Let −d < λ < 0. Then on Rd Z λ+d −λ−d λ −ixξ d Γ( 2 ) ξ |x| e dx = π 2 . (6.302) λ 2 Γ(− 2 ) Proof. We use the spherical coordinates: Z |x|λe−ixξdx (6.303) Z ∞ Z π λ+d−1 −ir|ξ| cos φd−1 λ+d−1 d−2 = dr dφd−1r e r sin φd−1Sd−2 (6.304) 0 0 π Z 2 −λ−d −λ−d d−2 = Γ(λ + d) (i|ξ| cos φd−1 + 0) + (−i|ξ| cos φd−1 + 0) sin φd−1dφd−1Sd−2 0 π Z 2 λ + d −λ−d −λ−d d−2 = Γ(λ + d)2 cos π |ξ| cos φd−1 sin φd−1dφd−1Sd−2. 2 0 Then we apply

d−1 2π 2 Sd−2 = d−1 , Γ( 2 ) π −λ−d+1 d−1 Z 2 1 Γ( )Γ( ) cos−λ−d φ sind−2 φ dφ = 2 2 , d−1 d−1 d−1 2 λ 0 Γ(− 2 ) − 1 λ+d−1 λ + d λ + d + 1 Γ(λ + d) = π 2 2 Γ Γ 2 2 λ + d π cos π = , 2 λ+d+1 −λ−d+1 Γ( 2 )Γ( 2 )

46 and we obtain (6.302) 2 In order to express (6.302) in a more symmetric way, deﬁne

2 λ 1 x 2 ηλ(x) := , λ > −d. λ+d 2 Γ( 2 ) We extend it to λ ≤ −d by setting

m λ−2m (−2) m λ η (x) := λ ∆ η (x). (6.305) − 2 m Then

Fηλ = η−λ−d, (6.306) x2ηλ = (λ + d)ηλ+2, (6.307) ∆ηλ = ληλ−2. (6.308)

6.3 Renormalizing the |k|−d function

Deﬁne the distribution |k|−d on Rd: Z Z Z P |k|−dφ(k)dk := |k|−dφ(k) − φ(0)dk + |k|−dφ(k)dk.. |k|<1 |k|>1

Theorem 6.3 We have an alternative deﬁnition of |k|−d:

d 2π 2 |k|−d = lim |k|−d+ν − δ(k) . (6.309) ν&0 d νΓ( 2 ) Here is its Fourier transform:

d Z Γ( 2 ) −d −ikx r 1 d 1 d P |k| e dk = − log + ψ − γ 2π 2 2 2 2 2 = − log r − γ, d = 1; r = − log − γ, d = 2; 2 1 1 = − log r − γ + H , d = 2m + 1; 2 m 2 r 1 = − log − γ + H , d = 2(m + 1). 2 2 m Proof. Z Z d −d+ν −1+ν 2π 2 |k| dk = |k| d|k|Sd−1 = d . (6.310) |k|<1 |k|<1 νΓ( 2 ) This proves (6.309).

47 Z |k|−d+ν e−ikxdk (6.311)

d ν r −ν π 2 Γ( ) = 2 (6.312) 2 d ν Γ( 2 − 2 ) d r π 2 ν d 2 ≈ 1 − ν log 1 + ψ − γ 2 d 2 2 ν Γ( 2 ) d 2π 2 1 r 1 d 1 ≈ − log + ψ − γ . (6.313) d ν 2 2 2 2 Γ( 2 ) 2

7 Bessel potentials

In this section we analyze some distinguished solutions of

(1 + 2)αD(x) = δ(x), (7.314) where 2 is the Laplacian on the pseudo-Euclidean space Rq,d−q (q minuses and d − q pluses). Formally, D is given by the inverse Fourier transform

Z eipx dp D(x) = , (7.315) (1 + p2)α (2π)d where p2 is the square of p wrt the scalar product of Rq,d−q. However usually we need to regularize the integrand of (7.315) Note that if D solves (7.314), and m > 0, then

(m2 + 2)αmd−2αD(mx) = δ(x). (7.316)

The following identities will be useful: Z ∞ 1 1 −sA α−1 α = e s ds, (7.317) A Γ(α) 0 ∓i π α Z ∞ 1 e 2 ±itA α−1 α = e t dt. (7.318) (A ± i0) Γ(α) 0 We will also need the Fourier transform of the Gaussian function on the Eu- clidean space Rd, and of the Fresnel function on the pseudo-Euclidean space Rq,d−q (with q minuses):

d Z 2 2 − sp ipx 2π 2 − x dpe 2 e = e 2s , (7.319) s d Z 2 2 ±i tp ipx q2π 2 ±i π d ∓i x dpe 2 e = (∓i) e 4 e 2t . (7.320) t

48 7.1 Euclidean and anti-Euclidean signature

Consider the Euclidean space Rd, with |x| denoting the Euclidean norm. Theorem 7.1 ipx d Z e dp 2 |x|α− 2 2 α d = d Kα− d (|x|), (7.321) (p + 1) (2π) Γ(α)(4π) 2 2 2 ipx d−1 d Z e dp π(∓i) |x|α− 2 = H± (|x|). (7.322) 2 α d d α− d (−p + 1 ± i0) (2π) Γ(α)(4π) 2 2 2 Proof. By (7.317) and then by (2.45), Z eipxdp (1 + p2)α α Z ∞ Z |x| 1 α−1 −(1+p2) |x|s ipx = ds dps e 2 e 2 Γ(α) 0 α− d d Z ∞ |x| 2 π 2 α− d −1 −(s+ 1 ) |x| = dss 2 e s 2 2 Γ(α) 0 d d 2π 2 |x|α− 2 = Kα− d (|x|). Γ(α) 2 2 By (7.318), and then by (3.103), Z eipxdp (1 − p2 ± i0)α α ∓i π α Z ∞ Z |x| e 2 ±it(1−p2) |x| α−1 ipx = dt dpe 2 t e 2 Γ(α) 0 (α− d ) ∓i π (α+ d ) d Z ∞ |x| 2 e 2 2 π 2 ±i |x| (t+ 1 ) α− d −1 = dte 2 t t 2 2 Γ(α) 0 d−1 d d π(∓i) π 2 |x|α− 2 = H± (|x|). α− d Γ(α) 2 2

7.2 General signature

Consider now a pseudo-Euclidean space of general signature Rq,d−q. Theorem 7.2 Z eipx dp (7.323) (1 + p2 ± i0)α (2π)d √ q 2 d α− 2 2(∓i) x ∓ i0 p 2 = d Kα− d x ∓ i0 (7.324) Γ(α)(4π) 2 2m 2 √ q 2 α− d πi(∓i) −x ± i0 2 p = ± H± −x2 ± i0. (7.325) d α− d Γ(α)(4π) 2 2 2

49 Remark 7.3 In (7.324) and (7.325) we use the notation explained in (3.111a) and (3.111b). Note that (7.324) works best for x2 > 0, because then we can ignore ∓i0. Likewise, (7.325) is best suited for x2 < 0, because then we can ignore ±i0.

Proof of Thm 7.2. using (7.318) and (2.49) we obtain

Z eipxdp (1 + p2 ± i0)α ∓i π α ∞ e 2 Z Z i 2 ± 2 t(1+p ) α−1 ipx = α dt dpe t e 2 Γ(α) 0 π d d q ∓i (α− ) Z ∞ 2 (∓i) e 2 2 π 2 ± i (t− x ) α− d −1 = dte 2 t t 2 . (7.326) α− d 2 2 Γ(α) 0 √ For x2 ≥ 0, we change the variable t = ±is x2, so that (7.326) becomes √ d d √ 2 (α− ) q Z ∞ 2 x 2 (∓i) π 2 − x (s+ 1 ) α− d −1 dse 2 s s 2 2 Γ(α) √ 0 q d 2 d 2 α− 2 √ 2(∓i) π x 2 = Kα− d x , Γ(α) 2 2 √ For x2 ≤ 0, we change the variable t = s −x2. By (7.318) and (3.103) we transform (7.326) into

√ d π d d √ 2 (α− ) q ∓i (α− ) Z ∞ 2 −x 2 (∓i) e 2 2 π 2 ±i −x (s+ 1 ) α− d −1 dse 2 s s 2 2 Γ(α) √ 0 q d 2 d 2 α− 2 2(∓i) π −x p 2 = ∓ i Kα− d ∓ i −x Γ(α) 2 2 √ q d 2 α− d πi(∓i) π 2 −x 2 p = ± H± −x2. α− d Γ(α) 2 2 √ √ √ Then we notice that x2 and ∓i −x2 can be joined in x2 ∓ i0. 2

7.3 The Minkowski signature

R1,d−1 is called the Minkowski space. (We use the signature “mostly pluses”). This case is especially important and rich. Let us state the special case of Thm 7.2 for the Minkowski signature as a separate theorem. We also introduce special notation.

50 Theorem 7.4 Z eipx dp DF/F(x) := (7.327) α (1 + p2 ± i0)α (2π)d √ 2 d α− 2 2i x ∓ i0 p 2 = ∓ d Kα− d x ∓ i0 (7.328) Γ(α)(4π) 2 2 2 √ 2 α− d π −x ± i0 2 p = H± −x2 ± i0. (7.329) d α− d Γ(α)(4π) 2 2 2

Suppose that x0 denotes the ﬁrst coordinate of R1,d−1, which we assume to be timelike (having a negative coeﬃcient in the scalar product). The remaining, spacelike coordinates will be denoted ~x. Then we set

∨ 1,d−1 2 0 J := {x ∈ R : x ≤ 0, x ≥ 0}, ∧ 1,d−1 2 0 J := {x ∈ R : x ≤ 0, x ≤ 0}. Theorem 7.5 Z eipx dp D∨/∧(x) := (7.330) α (1 + p2 ∓ i0sgnp0)α (2π)d √ 2 α− d π −x + i0 2 p =θ(±x0) H+ −x2 + i0 d α− d Γ(α)(4π) 2 2m 2 √ 2 α− d ! −x − i0 2 p + H− −x2 − i0 (7.331) α− d 2m 2 is a distribution whose support is contained in J ∨/∧. Inside J ∨/∧ we have the identity √ 2 d α− 2 ∨/∧ 2π −x p 2 Dα (x) = d Jα− d −x . (7.332) Γ(α)(4π) 2 2 2

We also have F F ∨ ∧ Dα(x) + Dα(x) = Dα (x) + Dα (x). (7.333)

Proof. Let us now prove that the support of (7.330) with the minus signed is contained in J ∨. By the Lorentz invariance it suﬃces to prove that it is zero for x0 < 0. We write

0 0 Z eipxdp Z e−ip x +i~p~xdp0d~p = α (p2 + 1 − i0sgnp0)α ~p2 + 1 − (p0 + i0)2

Next we continuously deform the contour of integration, replacing p0 + i0 by p0 + iR, where R ∈ [0, ∞[. We do not cross any singularities of the integrand 0 0 and note that e−ix (p +iR) goes to zero (remember that x0 < 0).

51 Next we note that Z 1 1 + eipxdp (7.334) (1 + p2 + i0)α (1 + p2 − i0)α √ d 2 α− d ππ 2 −x + i0 2 p = H+ −x2 + i0 α− d Γ(α) 2 2 √ 2 α− d ! −x − i0 2 p + H− −x2 − i0 . (7.335) α− d 2 2

Taking into account the support properties, we obtain (7.331). Finally, using (3.99) we obtain (7.332). 2 We also introduce special notation for some solutions of

(1 + 2)αD(x) = 0. (7.336)

PJ ∨ ∧ Dα (x) := Dα (x) − Dα (x), (7.337) (+)/(−) F ∨/∧ Dα (x) := −iDα(x) + iDα (x) (7.338) F ∧/∨ = iDα(x) − iDα (x). (7.339)

Here are expicit formulas for these solutions: √ 2 α− d π −x + i0 2 p DPJ(x) =sgn(x0) H+ −x2 + i0 α d α− d Γ(α)(4π) 2 2m 2 √ 2 α− d ! −x − i0 2 p + H− −x2 − i0 (7.340) α− d 2m 2

p 2 0 d α− 2 (±) 2 x ± isgnx 0 p 2 0 Dα = d Kα− d x ± isgnx 0 . (7.341) Γ(α)(4π) 2 2 2

Note the identities

PJ (+) (−) Dα (x) = −iDα (x) + iDα (x), (7.342) F F (+) (−) Dα(x) − Dα(x) = iDα (x) + iDα (x). (7.343)

PJ ∨ ∧ ∨ ∧ The support of Dα is contained in J ∪ J . In the interior of J ∪ J we have the identity √ 2 d α− 2 PJ 0 2π −x p 2 Dα (x) =sgn(x ) d Jα− d −x . (7.344) Γ(α)(4π) 2 2 2

52 7.4 Fourier transforms and Bessel type functions

We will have two generic notations for elements of Rd: for k ∈ Rd, p := |k|, and for x ∈ Rd, r := |x|.

d Theorem 7.6 Let Reα > 2 . Then Z d 2 2 −α d d−2α Γ(α − 2 ) (k + m ) dk = π 2 m , (7.345) Γ(α) d Z d 2 mr α− 2 −ikx 2 2 −α 2 d−2α e (k + m ) dk = π m Kα− d (mr). (7.346) Γ(α) 2 2

Proof. (7.345) is

Z ∞ d d 2 2 −α d−1 −1 d−2α Γ( 2 )Γ(α − 2 ) Sd−1 (p + m ) p dp = Sd−12 m . 0 Γ(α) To prove (7.346) we use

1 Z ∞ A−α = e−tAtα−1dt. Γ(α) 0 We obtain Z (1 + k2)−αe−ikxdk

α Z ∞ Z r 1 α−1 −(1+k2) rt −ikx = dt dkt e 2 e 2 Γ(α) 0 α− d d Z ∞ r 2 π 2 α− d −1 −(t+t−1) r = dtt 2 e 2 . 2 Γ(α) 0 2 Note that the rhs of (7.346) is locally integrable for α > 0. Therefore, (7.346) is true also for Reα > 0, if the Fourier transform is interpreted appropriately. We can actually extend (7.346) to all α ∈ C, in the sense of distributions. d In the range 0 < Reα < 2 both (7.346) and (7.347) are regular distributions Using the asymptotics of the MacDonald function we easily see that as m & 0, the rhs of (7.346) converges to

d 2α−d d Γ( 2 − α) r π 2 . (7.347) Γ(α) 2

In the distributional sense this convergence is true for all α ∈ C except for d 1 1 α = 2 + n, n = 0, 1,... , because then k2α = kd+2n , which is the anomalous case and an additional renormalization is needed.

53 7.5 Averages of plane waves on sphere

Consider the Euclidean space Rd. Let us take the average of plane waves over the unit sphere Sd−1. Let dΩ denote the standard measure on the sphere.

Z e−ikxdΩ(k) (7.348) Z π −ipr cos θ d−2 = e sin θdθSd−1 (7.349) −π Z 1 −iprw 2 d−3 = e (1 − w ) 2 dwSd−1 (7.350) −1 d − d −1 = (2π) 2 J d (pr)(pr) 2 . (7.351) 2 −1 The Fourier transform of a radial function is radial and we have the identity Z Z −ikx d − d +1 d−1 f(|k|)e dk = (2π) 2 f(p)J d (rp)(rp) 2 p dp. 2 −1

Here are the low dimensional cases: Z f(|k|)e−ikxdk Z ∞ = 2 f(p) cos(pr)dp, d = 1; 0 Z ∞ = 2π f(p)pJ0(pr)dp, d = 2; 0 Z ∞ sin(pr) = 4π f(p)p2 dp, d = 3. 0 pr In particular, in dimension 1 we obtain

Z ∞ 1/2 α− 1 2 −α 2π r 2 2 (1 + p ) cos(pr)dp = K 1 (r). (7.352) α− 2 0 Γ(α) 2

1 Setting m = α − 2 , we obtain the Poisson representation (2.65). In dimension d = 2 we obtain

Z ∞ α−1 2 −α 2π r 2π (1 + p ) J0(pr)pdp = Kα−1(r). 0 Γ(α) 2 In dimension d = 3 we obtain

Z ∞ 3 α− 3 2 −α sin(pr) 2π 2 r 2 4π (1 + p ) pdp = K 3 (r), α− 2 0 pr Γ(α) 2 which could be also deduced from (7.352) by diﬀerentiating wrt r.

54 7.6 General signature

Suppose that the scalar product on Rd has a signature with q minuses. Theorem 7.7 Z e−ikx(m2 + k2 ± i0)−αdk (7.353)

 √ d d q α− 2 √ d−2α 2(∓i) m x2 2 2 π 2 m K d m x x ≥ 0;  Γ(α) 2 α− 2 = d (7.354) q √ α− √ d πi(∓i) m −x2 2 ±  2 d−2α 2 2 ±π m Γ(α) 2 H d m −x x ≤ 0. α− 2 Proof. We use ∓i π α ∞ e 2 Z (A ± i0)−α = e±itAtα−1dt. (7.355) Γ(α) 0 (7.353) for x2 ≥ 0 is

√ π √ 2 α ∓i α Z ∞ Z 2 x e 2 ±it(1+k2) x α−1 −ikx dt dke 2 t e (7.356) 2 Γ(α) √ 0 d π d d √ 2 (α− ) q ∓i (α− ) Z ∞ 2 x 2 (∓i) e 2 2 π 2 ±i x (t− 1 ) α− d −1 = dte 2 t t 2 (7.357) 2 Γ(α) √ 0 d d √ 2 (α− ) q Z ∞ 2 x 2 (∓i) π 2 − x (s+ 1 ) α− d −1 = dte 2 s s 2 , (7.358) 2 Γ(α) 0 which is the ﬁrst case of (7.354). (7.353) for x2 ≤ 0 is

√ π √ 2 α ∓i α Z ∞ Z 2 −x e 2 ±it(1+k2) −x α−1 −ikx dt dke 2 t e (7.359) 2 Γ(α) 0 √ d π d d √ 2 (α− ) q ∓i (α− ) Z ∞ 2 −x 2 (∓i) e 2 2 π 2 ±i −x (t+ 1 ) α− d −1 = dte 2 t t 2 , (7.360) 2 Γ(α) 0 which is the second case of (7.354). 2

7.7 Averages of plane waves on hyperboloid

Consider the Minkowski space R1,d−1. Let us take the average of plane waves d−1 over the unit future hyperboloid H+ . Let dΩ denote the standard measure on d−1 H+ . Let x be a future oriented vector.

55 Z e−ikxdΩ(k) (7.361) Z π −ipr cosh θ d−2 = e sinh θdθSd−1 (7.362) −π Z 1 −iprw 2 d−3 = e (w − 1) 2 dwSd−1 (7.363) −1 1 d d −iπ(m+ 2 ) 2 − − 2 −1 = e (2π) H d (pr)(pr) . (7.364) 2 −1 8 Integrals of Bessel functions 8.1 Scalar products Theorem 8.1 We have the following indeﬁnite integrals:

Z ∞ y xKm(ax)Km(bx)dx = 2 2 aKm−1(ay)Km(by) − bKm(ay)Km−1(by) , y a − b Re(a + b) > 0, Z ∞ 2 2 2 y 2 my y 2 xKm(ax) dx = Km(ay) + Km(ay)Km−1(ay) − Km−1(ay) , y 2 a 2 Rea > 0.

Proof. Let us prove the ﬁrst identity. Using Km = K−m, we write y aKm−1(ay)Km(by) − bKm(ay)Km−1(by) −m+1 m m −m+1 = ay K−m+1(ay)y Km(by) − by Km(ay)y K−m+1(by).

We diﬀerentiate using the recurrence relations. We obtain

2 −m+1 m −m+1 m a y K−m(ay)y Km(by) + aby K−m+1(ay)y Km−1(by) m −m+1 2 m −m+1 −aby Km−1(ay)y K−m+1(by) − b y Km(ay)y K−m(by) 2 2 = (a − b )yKm(ay)Km(by).

56 Theorem 8.2 We have the following deﬁnite integrals:

Z ∞ π(amb−m − a−mbm) xKm(ax)Km(bx)dx = 2 2 , 0 2 sin mπ(a − b ) m 6= 0, |Rem| < 1, Re(a + b) > 0; Z ∞ ln a − ln b xK0(ax)K0(bx)dx = 2 2 , Re(a + b) > 0; 0 a − b Z ∞ 2 πm xKm(ax) dx = 2 , 0 2 sin mπa m 6= 0, |Rem| < 1, Rea > 0; Z ∞ 2 1 xK0(ax) dx = 2 , Rea > 0. 0 2a Proof. Assume that 0 < Rem < 1. Then for small z π π z −m K (z) ≈ I (z) ≈ , m 2 sin πm −m 2 sin πmΓ(1 − m) 2 π π z m−1 K (z) ≈ − I (z) ≈ . m−1 2 sin π(m − 1) m−1 2 sin πmΓ(m) 2 Therefore, for small y, y aK (ay)K (by) − bK (ay)K (by) a2 − b2 m−1 m m m−1 πy ay m−1by −m ay −mby m−1 ≈ a − b (a2 − b2)(2 sin πm)2Γ(m)Γ(1 − m) 2 2 2 2 π(amb−m − a−mbm) = . 2 sin mπ(a2 − b2)

8.2 Barnes integrals Theorem 8.3 For Rez > 0, c > 0, c + Rem > 0, 1 Z ∞ is is x−2c−is−m Km(z) = Γ c + Γ c + + m ds. (8.365) 8π −∞ 2 2 2 Proof. 1 Z is is z −2c−is−m ids Γ c + Γ c + + m 2πi γ 2 2 2 2 ∞ ∞ X (−1)n z 2n−m X (−1)n z 2n+m = Γ(m − n) + Γ(−m − n) n! 2 n! 2 n=0 n=0 ∞ ∞ X π z 2n−m X π z 2n+m = + n!Γ(1 − m + n) sin πm 2 n!Γ(1 + m + n) sin πm 2 n=0 n=0 π = I (z) − I (z) = 2K (z). sin πm −m m m

57 The integral is convergent because of the estimates

is is 2c+Rem − π s Γ c + Γ c + + m ≤ chsi e 2 . (8.366) 2 2

−2c−is−m z −2c−Rem s arg z ≤ |z| e . (8.367) 2 Of course, we have a version for the Hankel functions. The following representation holds only for real x

1 Theorem 8.4 For 0 < c < 2 Rem,

∞ it 1 1 Z Γ(c + ) xm+ 2 −2c−it J (x) = √ 2 dt. (8.368) m 4 π it 2 −∞ Γ(m + 1 − c − 2 )

8.3 Integral of Sonine and Schafheitlin ([1] Exercise 4.14, p. 236): Z ∞ Jm(xξ)Jk(yξ)dξ λ 0 ξ ykΓ 1+m+k−λ 1 − m + k − λ 1 + m + k − λ y2 = 2 F , ; k + 1; , y < x; λ 1+k−λ 1+m−k−λ 2 2 x2 2 x Γ(k + 1)Γ 2 xmΓ 1+m+k−λ 1 + m − k − λ 1 + m + k − λ x2 = 2 F , ; m + 1; , x < y. λ 1+m−λ 1−m+k−λ 2 2 y2 2 y Γ(m + 1)Γ 2

8.4 Another integral The following integrals essentially Watson, 13.31 (1): Z ∞ 2 exp(−qy )Jm(ay)Jm(by)ydy (8.369) 0 1 a2 + b2 ab π = exp − I , Rem > −1, | arg q| < . (8.370) 2q 4q m 2q 2 Here is its another version: Z ∞ 2 exp(−qy )Im(αy)Im(βy)ydy (8.371) 0 1 α2 + β2 αβ π = exp I , Rem > −1, | arg q| < . (8.372) 2q 4q m 2q 2

58 9 Klein-Gordon equation in 1 + 1 dimensions 9.1 Hyperbolic coordinates We have the coordinates 1 1 x := (t + y), x := (t − y). (9.373) + 2 − 2 The space R1,1 is divided into 4 sectors:

J++ = {(t, y): t > |y|} = {x+ > 0, x− > 0},

J−− = {(t, y): t < −|y|} = {x+ < 0, x− < 0},

J+− = {(t, y): y > |t|} = {x+ > 0, x− < 0},

J−+ = {(t, y): y < −|t|} = {x+ < 0, x− > 0}. There are 4 hyperbolic coordinate systems:

J++ : t = r cosh φ, y = r sinh φ, p 1 1 r = t2 − y2, φ = log(t + y) − log(t − y), 2 2 1 1 x = reφ, x = re−φ; + 2 − 2 J−− : t = −r cosh φ, y = −r sinh φ, p 1 1 r = t2 − y2, φ = log(−t − y) − log(−t + y), 2 2 1 1 x = − reφ, x = − re−φ; + 2 − 2 J+− : t = r sinh φ, y = r cosh φ, p 1 1 r = y2 − t2, φ = log(y − t) − log(y + t), 2 2 1 1 x = reφ, x = − re−φ; + 2 − 2 J−+ : t = −r sinh φ, y = −r cosh φ, p 1 1 r = y2 − t2, φ = log(−y + t) − log(−y − t), 2 2 1 1 x = − reφ, x = re−φ; + 2 − 2

The d’Alembertian in all sectors is

2 2 2 −1 −2 2 2 = −∂t + ∂y = −∂x+ ∂x− = ∂r + r ∂r − r ∂φ. (9.374)

9.2 Plane waves We look for solutions of the Klein-Gordon equation (−2 + 1)F = 0

59 of the form 1 ZZ F (t, y) = F (ξ, η)ei(−tτ+yη)dξdη. 2π We obtain (−τ 2 + η2 + 1)F (ξ, η) = 0.

Positive frequency plane waves are parametrized by ψ ∈ R and given by i(−x cosh ψ+y sinh ψ) fψ(x, y) := e . They are solutions of the Klein-Gordon equation. Here is the plane wave in hyperbolic coordinates:

−ir cosh(φ−ψ) J++ : fψ(r, φ) = e , ir cosh(φ−ψ) J−− : fψ(r, φ) = e , −ir sinh(φ−ψ) J+− : fψ(r, φ) = e , ir sinh(φ−ψ) J−+ : fψ(r, φ) = e . Positive frequency solutions are given by Z g(x, y) = fψ(x, y)g(ψ)dψ, (9.375) where g is a distribution on R. We will denote by H the Hilbert space of functions on R2 of the form (9.375) with g ∈ L2(R). We will treat them as “nice” solutions of the Klein-Gordon equation.

9.3 Hyperbolic waves

Let us introduce for µ ∈ R 1 Z f := f eiµψdψ. (9.376) µ 2π ψ Proposition 9.1 Here are the expressions of hyperbolic waves in hyperbolic coordinates: 1 J : f (r, φ) = K (ir)eiµφ, (9.377) ++ µ π iµ 1 J : f (r, φ) = K (−ir)eiµφ, (9.378) −− µ π iµ ± π m e 2 J : f (r, φ) = K (±r)eiµφ, (9.379) +− µ π iµ ∓ π m e 2 J : f (r, φ) = K (±r)eiµφ. (9.380) −+ µ π iµ Proposition 9.2 We can expand plane waves in hyperbolic waves: Z −iµψ fψ = fµe dµ. (9.381)

60 9.4 Wave equation in 1 + 1 dimension We will use polar coordinates. Using the expressions (4.141) we obtain

2 2 2 −1 −2 2 2 := ∂x − ∂t = cos(2φ)∂r − 2 sin(2φ)r ∂r∂φ − cos(2φ)r ∂φ −2 −1 +2r sin(2φ)∂φ − r cos(2φ)∂r. (9.382)

Thus

2 2 2 2 r 2 = cos(2φ) r ∂r − r∂r − 2 sin(2φ) r∂r − 1 ∂φ − cos(2φ)∂φ

On functions of the form rλf(φ) we obtain an operator

2 2 Λλ = cos(2φ)(λ − 2λ) − 2 sin(2φ)(λ − 1)∂φ − cos(2φ)∂φ.

Let us substitute w = sin(2φ). One can distinguish two regions: √ p 2 2 ∂φ = 2 1 − w ∂w, cos(2φ) = 1 − w , cos(2φ) > 0, √ p 2 2 ∂φ = −2 1 − w ∂w, cos(2φ) = − 1 − w , cos(2φ) < 0.

We obtain

 λ λ λ 2 2  1 − + 2 − 1 w∂w + (1 − w )∂w , p 2 2 2 2 Λλ = −4 1 − w λ λ 2 2  2 2 − 1 − λw∂w + (1 − w )∂w .

This corresponds to the Jacobi equation

2 2 (1 − w )∂w − 2(m + 1)w∂w + n(n + 2m + 1) (9.383) with λ n = −m = , (9.384) 2 λ −n = m = . (9.385) 2 10 Elements of partial diﬀerential equations 10.1 General formalism Let X α P (k) = Pαk (10.386) α be a polynomial in d variables k = (k1, . . . , kd). We set 1 D := ∂ . (10.387) i i i

61 We consider the diﬀerential operator

X α P (D) = PαD . (10.388) α One can consider two problems: ﬁnd solutions of the homogeneous problem

P (D)ζ = 0, (10.389) and, given f, ﬁnd solutions of the inhomogeneous problem.

P (D)ζ = f. (10.390)

To solve the inhomogeneous problem, it is useful to introduce a Green’s function or a fundamental solution of P (D), which is a distribution G satisfying

PG(x) = δ(x). (10.391)

Note that if we know Green’s function, then Z Gf(x) = G(x − y)f(y)dy (10.392) solves the inhomogeneous equation. Green’s function is not uniquely deﬁned. In fact, if G is Green’s function and ζ solves the homogeneous problem, then G + ζ is also Green’s function. We will see, however, that often we will have distinguished Green’s functions. Sometimes we will also have distinguished solutions. We can look for Green’s functions using the Fourier transformation. In fact, suppose that G ∈ S(Rd). We can write Z dk G(x) = G(k)eikx , (10.393) (2π)d Z dk δ(x) = eikx . (10.394) (2π)d

Equation (10.391) becomes

Z dk Z dk P (k)G(k)eikx = eikx . (10.395) (2π)d (2π)d

Thus formally, Z 1 dk G(x) = eikx . (10.396) P (k) (2π)d If (10.396) is well deﬁned, then it provides a distinguished Green’s function for 1 P (D). Unfortunately, often, especially if P has zeros, P (k) is not a well deﬁned distribution and (10.396) is problematic.

62 10.2 Laplace equation in d = 1 2 2 Consider P (D) = −∂x, so that P (k) = k . Space of solutions is a + bx. (10.397) Examples of Green’s functions:

G+(x) = −θ(x)x, (10.398) G−(x) = −θ(−x)|x|, (10.399) 1 G0(x) = − |x|. (10.400) 2

1 k2 is not a distribution, but it can be regularized.

10.3 Helmholtz equation in d = 1 2 2 2 2 Consider P (D) = −∂x + m , so that P (k) = k + m . Space of solutions is mx −mx a+e + a−e (10.401) Examples of Green’s functions:

sinh(mx) G+(x) = −θ(x) , (10.402) m | sinh(mx)| G−(x) = −θ(−x) , (10.403) m e−m|x| G(x) = . (10.404) 2m

1 k2+m2 is a distribution. By the method of residues we can compute:

1 Z eikx e−m|x| dk = , (10.405) 2π k2 + m2 2m which reproduces (10.404).

10.4 Laplace equation in d = 2 2 2 2 2 Consider P (D) = −∂x − ∂y , so that P (k) = kx + ky. Introduce complex coordinates 1 1 z = (x + iy), z = (x − iy). (10.406) 2 2 Then ∂z = ∂x − i∂y, ∂z = ∂x + i∂y. (10.407) We have 2 2 −∂x − ∂y = ∂z∂z. (10.408)

63 Therefore, solutions are sums of a holomorphic and antiholomorphic function:

ζ1(z) + ζ2(z). (10.409)

In polar coordinates x = r cos φ, y = r sin φ, the Laplacian is 1 1 −∂2 − ∂2 = − ∂ r∂ − ∂2. (10.410) x y r r r r2 φ We claim that rotationally symmetric Green’s functions have the form 1 1 1 a − log r = a − log(z) − log(z) + log 4. (10.411) 2π 4π 4π We easily check that a + b log(r) solves the Laplace equation outside of the origin, either in the polar coordinates, or noticing the decomposition into a holomorphic and an antiholomorphic function. It is more diﬃcult to determine the coeﬃcient in front of log r. ∞ 2 Note that PG = δ means that for any test function φ ∈ Cc (R ) Z Z P (D)φ(x, y)G(x, y)dxdy = φ(x, y)P (D)G(x, y)dxdy = φ(0, 0). (10.412)

Assume that φ is rotationally symmetric and G(r) = a + b log(r). We have Z P (D)φ(x, y)G(x, y)dxdy (10.413)

x2+y2>2 Z ∞ = 2π (−∂rr∂r)φ(r)G(r)dr (10.414) Z ∞ = 2π φ(r)(−∂rr∂r)G(r)dr (10.415) +2π(φ0()G() − φ()G0()) (10.416) → −2πbφ(0). (10.417)

1 Hence, b = − 2π . 1 k θ(k) is not a distribution. We can regularize it as in (??). Then we obtain a Green’s function with a rather strange looking constant:

ZZ ixkx+iyky e − 1 dkxdky G(x, y) = 2 2 2 (10.418) (kx + ky) (2π) 2 2 kx+ky <1

ZZ ixkx+iyky e dkxdky + 2 2 2 (10.419) (kx + ky) (2π) 2 2 kx+ky >1 1 r γ = − log − . (10.420) 2π 2 2π

64 10.5 Helmholtz equation in d = 2 2 2 2 2 2 2 Consider P (D) = −∂x − ∂y + m , so that P (k) = kx + ky + m . The method of Fourier transformation gives a distinguished Green’s function

ZZ ixkx+iyky e dkxdky G(x, y) = 2 2 2 2 (10.421) (m + kx + ky) (2π) 1 ZZ ei|k|r cos φ = |k|d|k|dφ (10.422) (2π)2 (m2 + |k|2) 1 = K (mr), (10.423) 2π 0 where K0(z) is the 0th MacDonald function. Note the asymptotics around zero: z K (z) ' − log − γ. (10.424) 0 2 Thus, in order to obtain a zero-mass Green’s function, we need to renormalize. Writing Gm(r) for the Green’s function with mass m, as deﬁned in (10.420) and (10.423), we obtain the massless Green’s functions by the following limit:

1 lim Gm + log m = G0. (10.425) m→0 2π

10.6 Wave equation in d = 1 + 1 2 2 Consider 2 = ∂t − ∂y . Introducing coordinates 1 1 u := (t + y), u := (t − y), (10.426) + 2 − 2 we have

2 = ∂u+ ∂u− . (10.427) Therefore, the general solution is

χ+(t + y) + χ−(t − y). (10.428)

We have the retarded Green’s function, the advanced Green’s function, and the Pauli-Jordan solution:

D+(t, y) = θ(t − |x|) = θ(t + x)θ(t − x), (10.429) D−(t, y) = θ(−t − |x|) = θ(−t − x)θ(−t + x), (10.430) DPJ(t, y) = θ(t − |x|) − θ(−t − |x|) (10.431) = θ(t − x) − θ(−t − x) = θ(t + x) − θ(−t + x). (10.432)

Let us compute the retaqrded Green’s function by the Fourier transform method. We introduce E, p, the dual variables to t, y. Besides, p+ := E + p and

65 1 p− := E − p. We have dEdp = 2 dp+dp−.

Z ei(−Et+px)dEdp D+(t, y) = (10.433) (−E2 + p2 − i0sgnE)(2π)2 1 Z e−i(u+p−+u−p+)dp dp = − + (10.434) 2 2 − p+p− − i0sgn(p+ + p−) (2π) 1 Z e−iu−p+ dp Z e−iu+p− dp = − + − (10.435) 2 (p+ + i0)2π (p− + i0)2π

= θ(u+)θ(u−). (10.436)

The Feynman propagator is obtained by the Wick rotation from the Eu- clidean propagator (Green’s function of the Laplacian). More precisely, we set E = ikx, t = ix:

1 ZZ e−iEt+ipydEdp DF(t, y) = (10.437) (2π)2 −E2 + p2 − i0 i ZZ e−ikxx+ipydEdp = = iDE(i−1t, y). (10.438) (2π)2 −E2 + p2 − i0

Setting 1 x2 + y2 γ DE(x, y) = − log − , (10.439) 4π 4 2π we obtain i −t2 + y2 + i0 iγ DF(t, y) = − log − 4π 4 2π i | − t2 + y2| 1 iγ = − log + θ|t| − |y| − , 4π 4 2 2π i | − t2 + y2| 1 1 iγ D(±)(t, x) = ∓ log + θt − |y| − θ − t − |y| ∓ . 4π 4 2 2 2π 11 Miscellanea 11.1 ... Identity

(1 − w2) m + κ (2m + κ) ∂ + (2m + κ)w∂ = + ∂ mw + (1 − w2)∂ r w r r r w m + − ∂ − (m + κ)w + (1 − w2)∂ . r r w

66 11.2 We set z = cos φ: 1 1 ∂2 + ∂2 + ( − m2) x y 4 y2

2 1 1 2 1 2 1 = ∂ + ∂r + ∂ + ( − m ) r r r2 φ 4 sin2 φ 1 1 1 1 = ∂2 + ∂ + (1 − z2)∂2 − z∂ + ( − m2) r r r r2 z z 4 1 − z2 We use

2 − 1 (m+ 1 ) 2 1 (m+ 1 ) 1 z (1 − z ) 2 2 ∂ (1 − z ) 2 2 = ∂ − (m + ) , z z 2 1 − z2 obtaining

2 − 1 (m+ 1 ) 2 2 1 2 1 2 1 (m+ 1 ) (1 − z ) 2 2 (1 − z )∂ − z∂ + ( − m ) (1 − z ) 2 2 z z 4 1 − z2 1 2 = (1 − z2)∂2 − (2m + 2)z∂ − m + . z z 2 11.3 √ 2 1−t 2 1+t 2 We set t = cos 2φ, so that sin φ = 2 , cos φ = 2 , ∂φ = −2 1 − t ∂t. 1 1 1 1 ∂2 + ( − α2) + ∂2 + ( − β2) + ∂2 x 4 x2 y 4 y2 z

2 1 = ∂2 + ∂ + (1 − w2)∂2 − 2w∂ + r r r r2 w w ! 1 1 2 1 2 4(1 − t2)∂2 − 4t∂ + ( − α2) + ( − β2) . 1 − w2 t t 4 1 + t 4 1 − t

We use

− 1 (α+ 1 ) − 1 (β+ 1 ) 1 (α+ 1 ) 1 (β+ 1 ) (1 + t) 2 2 (1 − t) 2 2 ∂t(1 + t) 2 2 (1 − t) 2 2 1 1 1 1 1 1 = ∂ + (α + ) − (β + ) . t 2 2 1 + t 2 2 1 − t We obtain

− 1 (α+ 1 ) − 1 (β+ 1 ) (1 + t) 2 2 (1 − t) 2 2 1 2 1 2 × 4(1 − t2)∂2 − 4t∂ + ( − α2) + ( − β2) t t 4 1 + t 4 1 − t 1 (α+ 1 ) 1 (β+ 1 ) ×(1 + t) 2 2 (1 − t) 2 2 ! α − β α + β + 2 (α + β + 2)(α + β) = 4 (1 − t2)∂2 + − t ∂ + . t 2 2 t 4

67 .1 The digamma function In our paper we use the digamma function: ∂ Γ(z) ψ(z) := z . (.440) Γ(z) Here are its properties:

1 ψ(1 + z) = ψ(z) + , (.441) z ψ(z) − ψ(1 − z) = −π cot(πz), (.442) 1 1 ψ + z − ψ − z = π tan(πz), (.443) 2 2 1 2 log 2 + ψ(z) + ψ z + = 2ψ(2z), (.444) 2 ψ(1) = −γ, (.445) 1 ψ = −γ − 2 log 2. (.446) 2 The inverse of the Gamma function is an analytic function with the derivative 1 ψ(z) ∂ = − , (.447) z Γ(z) Γ(z)

1 n ∂z = (−1) n!, n = 0, 1, 2, ... (.448) Γ(z) z=−n It is also useful to introduce 1 1 the shifted kth harmonic number H (z) := + ··· + , (.449) k z z + k − 1 1 1 the kth harmonic number H := + ··· + = H (1), (.450) k 1 k k Γ(z + k) the Pochhammer symbol (z) := (.451) k Γ(z) ( (z)(z + 1) ··· (z + k − 1), k ≥ 0, = 1 (z+k)(z+k+1)···(z−1) , k ≤ 0. Some of their properties are collected below:

Hk+n(z) = Hn(z) + Hk(z + n), (.452)

Hk(z) = −Hk(1 − z − k), (.453)

ψ(z + k) = ψ(z) + Hk(z), (.454)

ψ(1 + k) = −γ + Hk, (.455) k (z)k = (−1) (1 − k − z)k, (.456)

∂z(z)n = Hn(z)(z)n. (.457)

68 References

[1] Andrews, G.E., Askey, R., Roy, R.: Special functions, Cambridge Univer- sity Press, 1999 [2] Everitt, W.N. and Kalf, H.: The Bessel diﬀerential equation and the Han- kel transform, to appear in J. Comput. Appl. Math. [3] Watson, G.N. A treatise on the theory of Bessel functions, Cambridge University Press, 2nd ed. 1948 [4] Titchmarsh, E.C.: Eigenfunction expansions I, Oxford University Press, 2nd ed. 1962