8 Mellin Transforms and Their Applications
“One cannot understand ... the universality of laws of nature, the relationship of things, without an understanding of mathematics. There is no other way to do it.” Richard P. Feynman
“The research worker, in his efforts to express the fundamental laws of Nature in mathematical form, should strive mainly for mathe- matical beauty. He should take simplicity into consideration in a subordinate way to beauty. ... It often happens that the require- ments of simplicity and beauty are the same, but where they clash the latter must take precedence.” Paul Dirac
8.1 Introduction
This chapter deals with the theory and applications of the Mellin transform. We derive the Mellin transform and its inverse from the complex Fourier trans- form. This is followed by several examples and the basic operational properties of Mellin transforms. We discuss several applications of Mellin transforms to boundary value problems and to summation of infinite series. The Weyl trans- form and the Weyl fractional derivatives with examples are also included. Historically, Riemann (1876) first recognized the Mellin transform in his fa- mous memoir on prime numbers. Its explicit formulation was given by Cahen (1894). Almost simultaneously, Mellin (1896, 1902) gave an elaborate discus- sion of the Mellin transform and its inversion formula.
367 368 INTEGRAL TRANSFORMS and THEIR APPLICATIONS
8.2 Definition of the Mellin Transform and Examples We derive the Mellin transform and its inverse from the complex Fourier transform and its inverse, which are defined respectively by ∞ 1 F {g(ξ)} = G(k)=√ e−ikξg(ξ)dξ, (8.2.1) 2π −∞
∞ 1 F −1{G(k)} = g(ξ)=√ eikξG(k)dk. (8.2.2) 2π −∞ Making the changes of variables exp(ξ)=x and ik = c − p,wherec is a constant, in results (8.2.1) and (8.2.2) we obtain ∞ 1 G(ip − ic)=√ xp−c−1g(log x)dx, (8.2.3) 2π 0
c +i∞ 1 g(log x)=√ xc−pG(ip − ic)dp. (8.2.4) 2π c−i∞ 1 We now write √ x−cg(log x) ≡ f(x)andG(ip − ic) ≡ f˜(p) to define the 2π Mellin transform of f(x)andtheinverse Mellin transform as ∞ M {f(x)} = f˜(p)= xp−1f(x)dx, (8.2.5) 0 c +i∞ 1 M −1{f˜(p)} = f(x)= x−pf˜(p)dp, (8.2.6) 2πi c−i∞ where f(x) is a real valued function defined on (0, ∞) and the Mellin transform variable p is a complex number. Sometimes, the Mellin transform of f(x)is denoted explicitly by f˜(p)=M [f(x),p]. Obviously, M and M −1 are linear integral operators.
Example 8.2.1 (a) If f(x)=e−nx,wheren>0, then ∞ M {e−nx} = f˜(p)= xp−1e−nxdx, 0 Mellin Transforms and Their Applications 369 which is, by putting nx = t, ∞ 1 Γ(p) = tp−1e−tdt = . (8.2.7) np np 0 1 (b) If f(x)= ,then 1+x ! " ∞ 1 dx M = f˜(p)= xp−1 · , 1+x 1+x 0 t x which is, by substituting x = or t = , 1 − t 1+x 1 = tp−1(1 − t)(1−p)−1dt = B(p, 1 − p)=Γ(p)Γ(1 − p), 0 which is, by a well-known result for the gamma function,
= π cosec(pπ), 0 < Re(p) < 1. (8.2.8)
(c) If f(x)=(ex − 1)−1,then ! " ∞ 1 1 M = f˜(p)= xp−1 dx, ex − 1 ex − 1 0
∞ ∞ 1 1 which is, by using e−nx = and hence, e−nx = , 1 − e−x ex − 1 n=0 n=1
∞ ∞ ∞ Γ(p) = xp−1e−nxdx = =Γ(p)ζ(p), (8.2.9) np n=1 0 n=1
∞ 1 where ζ(p)= ,(Rep>1) is the famous Riemann zeta function. np n=1 2 (d) If f(x)= ,then e2x − 1
! " ∞ ∞ ∞ 2 dx M = f˜(p)=2 xp−1 =2 xp−1e−2nxdx e2x − 1 e2x − 1 0 n=1 0 ∞ ∞ Γ(p) 1 =2 =21−p Γ(p) =21−p Γ(p)ζ(p). (8.2.10) (2n)p np n=1 n=1 370 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 1 (e) If f(x)= ,then ex +1 ! " 1 M =(1− 21−p)Γ(p) ζ(p). (8.2.11) ex +1 This follows from the result 1 1 2 − = ex − 1 ex +1 e2x − 1 combined with (8.2.9) and (8.2.10). 1 (f) If f(x)= ,then (1 + x)n ! " ∞ 1 M = xp−1(1 + x)−ndx, (1 + x)n 0 t x which is, by putting x = or t = , 1 − t 1+x 1 = tp−1(1 − t)n−p−1dt 0 Γ(p)Γ(n − p) = B(p, n − p)= , (8.2.12) Γ(n) where B(p, q) is the standard beta function. Hence, Γ(n) M −1{Γ(p)Γ(n − p)} = . (1 + x)n (g) Find the Mellin transform of cos kx and sin kx. It follows from Example 8.2.1(a) that Γ(p) Γ(p) pπ pπ M [e−ikx]= = cos − i sin . (ik)p kp 2 2 Separating real and imaginary parts, we find πp M [cos kx]=k−p Γ(p)cos , (8.2.13) 2 πp M [sin kx]=k−p Γ(p)sin . (8.2.14) 2 These results can be used to calculate the Fourier cosine and Fourier sine transforms of xp−1. Result (8.2.13) can be written as ∞ Γ(p) πp xp−1 cos kx dx = cos . kp 2 0 Mellin Transforms and Their Applications 371
Or, equivalently, ! " π Γ(p) πp F xp−1 = cos . c 2 kp 2 Or, 2 Γ(p) πp F {xp−1} = cos . (8.2.15) c π kp 2 Similarly, 2 Γ(p) πp F {xp−1} = sin . (8.2.16) s π kp 2
8.3 Basic Operational Properties of Mellin Transforms If M {f(x)} = f˜(p), then the following operational properties hold: (a) (Scaling Property).
M {f(ax)} = a−pf˜(p),a>0. (8.3.1)
PROOF By definition, we have, ∞ M {f(ax)} = xp−1f(ax)dx, 0 which is, by substituting ax = t, ∞ 1 f˜(p) = tp−1f(t)dt = . ap ap 0
(b) (Shifting Property).
M [xa f(x)] = f˜(p + a). (8.3.2)
Its proof follows from the definition. 1 p (c) M {f(xa)} = f˜ , (8.3.3) a a 372 INTEGRAL TRANSFORMS and THEIR APPLICATIONS ! " 1 1 M f = f˜(1 − p), (8.3.4) x x dn M {(log x)n f(x)} = f˜(p),n=1, 2, 3,.... (8.3.5) dpn The proofs of (8.3.3) and (8.3.4) are easy and hence, left to the reader. Result (8.3.5) can easily be proved by using the result d xp−1 = (log x)xp−1. (8.3.6) dp (d) (Mellin Transforms of Derivatives).
M [f (x)] = −(p − 1)f˜(p − 1), (8.3.7) provided [xp−1f(x)] vanishes as x → 0andasx →∞.
M [f (x)] = (p − 1)(p − 2)f˜(p − 2). (8.3.8)
More generally, Γ(p) M [f (n)(x)] = (−1)n f˜(p − n) Γ(p − n) Γ(p) =(−1)n M [f(x),p− n], (8.3.9) Γ(p − n) provided xp−r−1f (r)(x)=0 as x → 0forr =0, 1, 2,...,(n − 1).
PROOF We have, by definition, ∞ M [f (x)] = xp−1f (x) dx, 0 which is, integrating by parts, ∞ p−1 ∞ p−2 =[x f(x)]0 − (p − 1) x f(x) dx 0 = −(p − 1)f˜(p − 1).
The proofs of (8.3.8) and (8.3.9) are similar and left to the reader. (e) If M {f(x)} = f˜(p), then
M {xf (x)} = −pf˜(p), (8.3.10) Mellin Transforms and Their Applications 373 provided xpf(x) vanishes at x =0 andas x →∞.
M {x2f (x)} =(−1)2p(p +1)f˜(p). (8.3.11)
More generally,
Γ(p + n) M {xnf (n)(x)} =(−1)n f˜(p). (8.3.12) Γ(p)
PROOF We have, by definition, ∞ M {xf (x)} = xpf (x)dx, 0 which is, integrating by parts, ∞ p ∞ p−1 =[x f(x)]0 − p x f(x)dx = −pf˜(p). 0
Similar arguments can be used to prove results (8.3.11) and (8.3.12). (f) (Mellin Transforms of Differential Operators). If M {f(x)} = f˜(p), then 2 d M x f(x) = M [x2f (x)+xf (x)] = (−1)2 p2f˜(p), (8.3.13) dx and more generally, d n M x f(x) =(−1)npnf˜(p). (8.3.14) dx
PROOF We have, by definition, 2 d M x f(x) = M [x2f (x)+xf(x)] dx = M [x2f (x)] + M [xf(x)] = −pf˜(p)+p(p +1)f˜(p) by (8.3.10) and (8.3.11) =(−1)2 p2 f˜(p). 374 INTEGRAL TRANSFORMS and THEIR APPLICATIONS
Similar arguments can be used to prove the general result (8.3.14). (g) (Mellin Transforms of Integrals). ⎧ ⎫ ⎨ x ⎬ 1 M f(t)dt = − f˜(p +1). (8.3.15) ⎩ ⎭ p 0
In general, ⎧ ⎫ ⎨ x ⎬ n Γ(p) M {I f(x)} = M I −1f(t)dt =(−1) f˜(p + n), (8.3.16) n ⎩ n ⎭ Γ(p + n) 0 where In f(x)isthenth repeated integral of f(x) defined by
x
Inf(x)= In−1f(t)dt. (8.3.17) 0
PROOF We write x F (x)= f(t)dt 0 so that F (x)=f(x) with F (0) = 0. Application of (8.3.7) with F (x) as defined gives ⎧ ⎫ ⎨ x ⎬ M { } − − M − f(x)=F (x),p = (p 1) ⎩ f(t)dt, p 1⎭ , 0 which is, replacing p by p +1, ⎧ ⎫ ⎨ x ⎬ 1 1 M f(t) dt, p = − M {f(x),p+1} = − f˜(p +1). ⎩ ⎭ p p 0
An argument similar to this can be used to prove (8.3.16).
(h) (Convolution Type Theorems). If M {f(x)} = f˜(p)andM {g(x)} =˜g(p), then ⎡ ⎤ ∞ x dξ M [f(x) ∗ g(x)] = M ⎣ f(ξ) g ⎦ = f˜(p)˜g(p), (8.3.18) ξ ξ ⎡0 ⎤ ∞ M [f(x) ◦ g(x)] = M ⎣ f(xξ) g(ξ)dξ⎦ = f˜(p)˜g(1 − p). (8.3.19) 0 Mellin Transforms and Their Applications 375
PROOF We have, by definition, ⎡ ⎤ ∞ x dξ M [f(x) ∗ g(x)] = M ⎣ f(ξ) g ⎦ ξ ξ 0 ∞ ∞ x dξ = xp−1dx f(ξ) g ξ ξ 0 0 ∞ ∞ dξ x x = f(ξ) xp−1g dx, = η , ξ ξ ξ 0 0 ∞ ∞ dξ = f(ξ) (ξη)p−1g(η) ξdη ξ 0 0 ∞ ∞ = ξp−1f(ξ)dξ ηp−1g(η)dη = f˜(p)˜g(p). 0 0 Similarly, we have ⎡ ⎤ ∞ M [f(x) ◦ g(x)] = M ⎣ f(xξ) g(ξ)dξ⎦ 0 ∞ ∞ = xp−1dx f(xξ) g(ξ)dξ, (xξ = η), 0 0 ∞ ∞ dη = g(ξ)dξ ηp−1ξ1−pf(η) ξ 0 0 ∞ ∞ = ξ1−p−1g(ξ)dξ ηp−1f(η)dη =˜g(1 − p)f˜(p). 0 0
Note that, in this case, the operation ◦ is not commutative. Clearly, putting x = s, ∞ M −1{f˜(1 − p)˜g(p)} = g(st)f(t)dt. 0 Putting g(t)=e −t andg ˜(p)=Γ(p), we obtain the Laplace transform of f(t) ∞ M −1 {f˜(1 − p)Γ(p)} = e−stf(t)dt = L {f(t)} = f¯(s). (8.3.20) 0 376 INTEGRAL TRANSFORMS and THEIR APPLICATIONS
(i) (Parseval’s Type Property). If M {f(x)} = f˜(p)andM {g(x)} =˜g(p), then
c +i∞ 1 M [f(x)g(x)] = f˜(s)˜g(p − s)ds. (8.3.21) 2πi c−i∞
Or, equivalently,
∞ c +i∞ 1 xp−1f(x)g(x)dx = f˜(s)˜g(p − s)ds. (8.3.22) 2πi 0 c−i∞
In particular, when p = 1, we obtain the Parseval formula for the Mellin trans- form,
∞ c +i∞ 1 f(x)g(x)dx = f˜(s)˜g(1 − s)ds. (8.3.23) 2πi 0 c−i∞
PROOF By definition, we have
∞ M [f(x)g(x)] = xp−1f(x)g(x)dx 0 ∞ c +i∞ 1 = xp−1g(x)dx x−sf˜(s)ds 2πi 0 c−i∞ c +i∞ ∞ 1 = f˜(s)ds xp−s−1g(x)dx 2πi c−i∞ 0 c +i∞ 1 = f˜(s)˜g(p − s)ds. 2πi c−i∞
When p = 1, the above result becomes (8.3.23). Mellin Transforms and Their Applications 377
8.4 Applications of Mellin Transforms
Example 8.4.1 Obtain the solution of the boundary value problem
2 x uxx + xux + uyy =0, 0 ≤ x<∞, 0
We apply the Mellin transform of u(x, y) with respect to x defined by ∞ u˜(p, y)= xp−1u(x, y) dx 0 to reduce the given system into the form
2 u˜yy + p u˜ =0, 0 The solution of the transformed problem is A sin py u˜(p, y)= , 0 < Re p<1. p sin p The inverse Mellin transform gives c +i∞ A x−p sin py u(x, y)= dp, (8.4.3) 2πi p sin p c−i∞ whereu ˜(p, y) is analytic in the vertical strip 0 < Re (p)=c<π. The integrand of (8.4.3) has simple poles at p = nπ, n =1, 2, 3,... which lie inside a semi- circular contour in the right half plane. Evaluating (8.4.3) by theory of residues gives the solution for x>1as ∞ A 1 u(x, y)= (−1)n x−nπ sin nπy. (8.4.4) π n n=1 378 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 8.4.2 (Potential in an Infinite Wedge). Find the potential φ(r, θ) that satisfies the Laplace equation 2 r φrr + rφr + φθθ = 0 (8.4.5) in an infinite wedge 0 φ(r, α)=f(r),φ(r, −α)=g(r)0≤ r<∞, (8.4.6ab) φ(r, θ) → 0asr →∞ for all θ in − α<θ<α. (8.4.7) y = 0 x - =- Figure 8.1 An infinite wedge. We apply the Mellin transform of the potential φ(r, θ) defined by ∞ M [φ(r, θ)] = φ˜(p, θ)= rp−1φ(r, θ) dr 0 to the differential system (8.4.5)–(8.4.7) to obtain d2φ˜ + p2φ˜ =0, (8.4.8) dθ2 φ˜(p, α)=f˜(p), φ˜(p, −α)=˜g(p). (8.4.9ab) The general solution of the transformed equation is φ˜(p, θ)=A cos pθ + B sin pθ, (8.4.10) Mellin Transforms and Their Applications 379 where A and B are functions of p and α. The boundary conditions (8.4.9ab) determine A and B, which satisfy A cos pα + B sin pα = f˜(p), A cos pα − B sin pα =˜g(p). f˜(p)+˜g(p) f˜(p) − g˜(p) These give A = ,B= . 2cospα 2sinpα Thus, solution (8.4.10) becomes sin p(α + θ) sin p(α − θ) φ˜(p, θ)=f˜(p). +˜g(p) sin(2 pα) sin(2 pα) = f˜(p)h˜(p, α + θ)+˜g(p)h˜(p, α − θ), (8.4.11) where sin pθ h˜(p, θ)= . sin(2 pα) Or, equivalently, ! " sin pθ 1 rn sin nθ h(r, θ)=M −1 = , (8.4.12) sin 2 pα 2 α (1+2rn cos nθ + r2n) where π π n = or, 2α = . 2α n Application of the inverse Mellin transform to (8.4.11) gives φ(r, θ)=M −1 f˜(p)h˜(p, α + θ) + M −1 g˜(p)h˜(p, α − θ) , which is, by the convolution property (8.3.18), ⎡ ∞ rn cos nθ ξn−1f(ξ)dξ φ(r, θ)= ⎣ 2α ξ2n − 2(rξ)n sin nθ + r2n 0 ⎤ ∞ ξn−1g(ξ)dξ π + ⎦ , |α| < . (8.4.13) ξ2n +2(rξ)n sin nθ + r2n 2n 0 This is the formal solution of the problem. In particular, when f(r)=g(r), solution (8.4.11) becomes cos pθ φ˜(p, θ)=f˜(p) = f˜(p)h˜(p, θ), (8.4.14) cos pα where cos pθ h˜(p, θ)= = M {h(r, θ)}. cos pα 380 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Application of the inverse Mellin transform to (8.4.14) combined with the convolution property (8.3.18) yields the solution ∞ r dξ φ(r, θ)= f(ξ) h ,θ , (8.4.15) ξ ξ 0 where ! " cos pθ rn (1 + r2n)cos(nθ) h(r, θ)=M −1 = , (8.4.16) cos pα α (1+2r2n cos 2nθ + r2n) π and n = . 2α Some applications of the Mellin transform to boundary value problems are given by Sneddon (1951) and Tranter (1966). Example 8.4.3 Solve the integral equation ∞ f(ξ) k(xξ)dξ = g(x),x>0. (8.4.17) 0 Application of the Mellin transform with respect to x to equation (8.4.17) combined with (8.3.19) gives f˜(1 − p)k˜(p)=˜g(p), which gives, replacing p by 1 − p, f˜(p)=˜g(1 − p)h˜(p), where 1 h˜(p)= . k˜(1 − p) The inverse Mellin transform combined with (8.3.19) leads to the solution ∞ f(x)=M −1 g˜(1 − p)h˜(p) = g(ξ)h(xξ)dξ, (8.4.18) 0 provided h(x)=M −1 h˜(p) exists. Thus, the problem is formally solved. If, in particular, h˜(p)=k˜(p), then the solution of (8.4.18) becomes ∞ f(x)= g(ξ) k(xξ)dξ, (8.4.19) 0 Mellin Transforms and Their Applications 381 provided k˜(p)k˜(1 − p)=1. Example 8.4.4 Solve the integral equation ∞ x dξ f(ξ) g = h(x), (8.4.20) ξ ξ 0 where f(x) is unknown and g(x)andh(x) are given functions. Applications of the Mellin transform with respect to x gives 1 f˜(p)=h˜(p)k˜(p), k˜(p)= . g˜(p) Inversion, by the convolution property (8.3.18), gives the solution ∞ x dξ f(x)=M −1 h˜(p)k˜(p) = h(ξ) k . (8.4.21) ξ ξ 0 8.5 Mellin Transforms of the Weyl Fractional Integral and the Weyl Fractional Derivative DEFINITION 8.5.1 The Mellin transform of the Weyl fractional integral of f(x) is defined by ∞ 1 W −α[f(x)] = (t − x)α−1f(t)dt, 0 < Re α<1,x>0. (8.5.1) Γ(α) x −α −α Often xW∞ is used instead of W to indicate the limits to integration. Result (8.5.1) can be interpreted as the Weyl transform of f(t), defined by ∞ 1 W −α[f(t)] = F (x, α)= (t − x)α−1f(t)dt. (8.5.2) Γ(α) x We first give some simple examples of the Weyl transform. If f(t)=exp(−at), Re a>0, then the Weyl transform of f(t)isgivenby ∞ 1 W −α[exp(−at)] = (t − x)α−1 exp(−at)dt, Γ(α) x 382 INTEGRAL TRANSFORMS and THEIR APPLICATIONS which is, by the change of variable t − x = y, ∞ e−ax = yα−1 exp(−ay)dy Γ(α) 0 which is, by letting ay = t, ∞ e−ax 1 e−ax W −α[f(t)] = tα−1e−tdt = . (8.5.3) aα Γ(α) aα 0 Similarly, it can be shown that Γ(μ − α) W −α[t−μ]= xα−μ, 0 < Re α W −α[W −βf(x)] = W −(β+α)[f(x)] = W −β[W −αf(x)]. (8.5.7) Invoking a change of variable t − x = y in (8.5.1), we obtain ∞ 1 W −α[f(x)] = yα−1f(x + y)dy. (8.5.8) Γ(α) 0 d We next differentiate (8.5.8) to obtain, D = , dx ∞ 1 ∂ D[W −αf(x)] = tα−1 f(x + t)dt Γ(α) ∂x 0 ∞ 1 = tα−1Df(x + t)dt Γ(α) 0 = W −α[Df(x)]. (8.5.9) A similar argument leads to a more general result Dn[W −αf(x)] = W −α[Dnf(x)], (8.5.10) Mellin Transforms and Their Applications 383 where n is a positive integer. Or, symbolically, Dn W −α = W −αDn. (8.5.11) We now calculate the Mellin transform of the Weyl fractional integral by x % &α−1 % & x putting h(t)=tαf(t)andg = 1 1 − x H 1 − x ,whereH 1 − t Γ(α) t t t is the Heaviside unit step function so that (8.5.1) becomes ∞ x dt F (x, α)= h(t) g , (8.5.12) t t 0 which is, by the convolution property (8.3.18), F˜(p, α)=h˜(p)˜g(p), where h˜(p)=M {xαf(x)} = f˜(p + α), and ! " 1 g˜(p)=M (1 − x)α−1H(1 − x) Γ(α) 1 1 B(p, α) Γ(p) = xp−1(1 − x)α−1dx = = . Γ(α) Γ(α) Γ(p + α) 0 Consequently, Γ(p) F˜(p, α)=M [W −αf(x),p]= f˜(p + α). (8.5.13) Γ(p + α) It is important to note that this result is an obvious extension of result 7(b) in Exercise 8.8 DEFINITION 8.5.2 If β is a positive number and n is the smallest integer greater than β such that n − β = α>0, the Weyl fractional derivative of a function f(x) is defined by W β[f(x)] = En W −(n−β)[f(x)] ∞ (−1)n dn = (t − x)n−β−1f(t)dt, (8.5.14) Γ(n − β) dxn x where E = −D. Or, symbolically, W β = EnW −α = EnW −(n−β). (8.5.15) 384 INTEGRAL TRANSFORMS and THEIR APPLICATIONS It can be shown that, for any β, W −βW β = I = W βW −β. (8.5.16) And, for any β and γ, the Weyl fractional derivative satisfies the laws of exponents W β[W γ f(x)] = W β+γ[f(x)] = W γ[W βf(x)]. (8.5.17) We now calculate the Weyl fractional derivative of some elementary functions. If f(x)=exp(−ax), a>0, then the definition (8.5.14) gives W βe−ax = En[W −(n−β)e−ax]. (8.5.18) Writing n − β = α>0 and using (8.5.3) yields W βe−ax = En[W −αe−ax]=En[a−αe−ax] = a−α(ane−ax)=aβe−ax. (8.5.19) Replacing β by −α in (8.5.19) leads to result (8.5.3) as expected. Similarly, we obtain Γ(β + μ) W βx−μ = x−(β+μ). (8.5.20) Γ(μ) It is easy to see that W β(cos ax)=E[W −(1−β) cos ax], which is, by (8.5.6), 1 = aβ cos ax − πβ . (8.5.21) 2 Similarly, 1 W β(sin ax)=aβ sin ax − πβ , (8.5.22) 2 provided α and β lie between 0 and 1. If β is replaced by −α, results (8.5.20)–(8.5.22) reduce to (8.5.4)–(8.5.6), respectively. Finally, we calculate the Mellin transform of the Weyl fractional derivative with the help of (8.3.9) and find M [W βf(x)] = M [EnW −(n−β)f(x)] = (−1)nM [DnW −(n−β)f(x)] Γ(p) = M [W −(n−β)f(x),p− n], Γ(p − n) Mellin Transforms and Their Applications 385 which is, by result (8.5.13), Γ(p) Γ(p − n) = · f˜(p − β) Γ(p − n) Γ(p − β) Γ(p) = M [f(x),p− β] Γ(p − β) Γ(p) = f˜(p − β). (8.5.23) Γ(p − β) Example 8.5.1 (The Fourier Transform of the Weyl Fractional Integral). πiα F{W −αf(x)} =exp − k−αF{f(x)}. (8.5.24) 2 We have, by definition, ∞ ∞ 1 1 F{W −αf(x)} = √ e−ikxdx (t − x)α−1f(t)dt 2π Γ(α) −∞ x ∞ t 1 1 = √ f(t)dt · exp(−ikx)(t − x)α−1dx. 2π Γ(α) −∞ −∞ Thus, ∞ ∞ 1 1 F{W −αf(x)} = √ e−iktf(t)dt · eikτ τ α−1dτ, (t − x = τ) 2π Γ(α) −∞ 0 1 = F{f(x)} M {eikτ } Γ( α) πiα =exp − k−αF{f(x)}. 2 In the limit as α → 0 lim F{W −αf(x)} = F{f(x)}. α→0 This implies that W 0{f(x)} = f(x). We conclude this section by proving a general property of the Riemann- Liouville fractional integral operator D−α, and the Weyl fractional integral operator W −α. It follows from the definition (6.2.1) that D−αf(t)canbe expressed as the convolution −α D f(x)=gα(t) ∗ f(t), (8.5.25) 386 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where tα−1 g (t)= ,t>0. α Γ(α) Similarly, W −αf(x) can also be written in terms of the convolution −α W f(x)=gα(−x)∗f(x). (8.5.26) Then, under suitable conditions, Γ(1 − α − p) M [D−αf(x)] = f˜(p + α), (8.5.27) Γ(1 − p) Γ(p) M [W −αf(x)] = f˜(p + α). (8.5.28) Γ(α + p) Finally, a formal computation gives ∞ ∞ x 1 {D−αf(x)}g(x)dx = g(x)dx (x − t)α−1f(t)dt Γ(α) 0 0 0 ∞ ∞ 1 = f(t)dt · (x − t)α−1g(x)dx Γ(α) 0 t ∞ = f(t)[W −αg(t)] dt, 0 which is, using the inner product notation, D−αf, g = f, W−αg. (8.5.29) This show that D−α and W −α behave like adjoint operators. Obviously, this result can be used to define fractional integrals of distributions. This result is taken from Debnath and Grum (1988). 8.6 Application of Mellin Transforms to Summation of Series In this section we discuss a method of summation of series that is particularly associated with the work of Macfarlane (1949). THEOREM 8.6.1 If M {f(x)} = f˜(p), then ∞ c +i∞ 1 f(n + a)= f˜(p) ξ(p, a)dp, (8.6.1) =0 2πi n c−i∞ Mellin Transforms and Their Applications 387 where ξ(p, a)istheHurwitz zeta function defined by ∞ 1 ξ(p, a)= , 0 ≤ a ≤ 1, Re(p) > 1. (8.6.2) (n + a)p n=0 PROOF If follows from the inverse Mellin transform that c +i∞ 1 f(n + a)= f˜(p)(n + a)−p dp. (8.6.3) 2πi c−i∞ Summing this over all n gives ∞ c +i∞ 1 f(n + a)= f˜(p) ξ(p, a) dp. =0 2πi n c−i∞ This completes the proof. Similarly, the scaling property (8.3.1) gives c +i∞ 1 f(nx)=M −1{n−p f˜(p)} = x−p n−pf˜(p)dp. 2πi c−i∞ Thus, ∞ c +i∞ 1 f(nx)= x−pf˜(p) ζ(p)dp = M −1{f˜(p) ζ(p)}, (8.6.4) =1 2πi n c−i∞ ∞ where ζ(p)= n−p is the Riemann zeta function. n=1 When x = 1, result (8.6.4) reduces to ∞ c +i∞ 1 f(n)= f˜(p) ζ(p)dp. (8.6.5) =1 2πi n c−i∞ This can be obtained from (8.6.1) when a =0. Example 8.6.1 Show that ∞ (−1)n−1n−p =(1− 21−p) ζ(p). (8.6.6) n=1 388 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Using Example 8.2.1(a), we can write the left-hand side of (8.6.6) multiplied by tn as ∞ ∞ ∞ 1 (−1)n−1n−ptn = (−1)n−1tn · xp−1e−nxdx Γ(p) n=1 n=1 0 ∞ ∞ 1 = xp−1dx (−1)n−1tnxe−nx Γ(p) 0 n=1 ∞ 1 te−x = xp−1 · · dx Γ(p) 1+te−x 0 ∞ 1 t = xp−1 · dx. Γ(p) ex + t 0 In the limit as t → 1, the above result gives ∞ ∞ 1 1 (−1)n−1n−p = xp−1 dx Γ(p) ex +1 n=1 0 ! " 1 1 = M =(1− 21−p) ζ(p), Γ(p) ex +1 in which result (8.2.11) is used. Example 8.6.2 Show that ∞ sin an 1 = (π − a), 0 c +i∞ ∞ − sin an − 1 Γ(p 1) πp = p−1 ζ(p)cos dp. (8.6.8) =1 n 2πi a 2 n c−i∞ Mellin Transforms and Their Applications 389 We next use the well-known functional equation for the zeta function πp (2π)p ζ(1 − p)=2Γ(p) ζ(p)cos (8.6.9) 2 in the integrand of (8.6.8) to obtain c +i∞ ∞ p − sin an −a · 1 2π ζ(1 p) = − dp. =1 n 2 2πi a p 1 n c−i∞ The integral has two simple poles at p =0andp = 1 with residues 1 and −π/a, respectively, and the complex integral is evaluated by calculating the residues at these poles. Thus, the sum of the series is ∞ sin an 1 = (π − a). n 2 n=1 8.7 Generalized Mellin Transforms In order to extend the applicability of the classical Mellin transform, Naylor (1963) generalized the method of Mellin integral transforms. This generalized Mellin transform is useful for finding solutions of boundary value problems in regions bounded by the natural coordinate surfaces of a spherical or cylindrical coordinate system. They can be used to solve boundary value problems in finite regions or in infinite regions bounded internally. The generalized Mellin transform of a function f(r) defined in a Obviously, this generalized transform seems to be very useful for finding the solution of boundary value problems in which f(r) is prescribed on the internal boundary at r = a. On the other hand, if the derivative of f(r) is prescribed at r = a,itis convenient to define the associated integral transform by ∞ 2p p−1 a M +[f(r)] = F +(p)= r + f(r) dr, |Re(p)| In this case, we can show by integration by parts that 2 2 ∂ f ∂f 2 p+1 M + r + r = p F +(p) − 2 a f (a), (8.7.7) ∂r2 ∂r where f (r)existsatr = a. THEOREM 8.7.1 (Convolution). If M +{f(r)} = F +(p), and M +{g(r)} = G +(p), then 1 M +{f(r) g(r)} = F +(ξ)G +(p − ξ) dξ. (8.7.8) 2πi L Or, equivalently, ⎡ ⎤ −1 ⎣ 1 ⎦ f(r)g(r)=M F +(ξ) G +(p − ξ)dξ . (8.7.9) + 2πi L PROOF We assume that F +(p)andG +(p) are analytic in some strip Mellin Transforms and Their Applications 391 |Re(p)| <γ.Then ∞ 2p p−1 a M +{f(r) g(r)} = r + f(r)g(r)dr rp+1 a ∞ ∞ a2p = rp−1f(r)g(r)dr + f(r)g(r)dr (8.7.10) rp+1 a a ∞ 1 p−ξ−1 = F +(ξ)dξ r g(r)dr 2πi L a ∞ 2p 1 a −ξ + g(r)dr r F +(ξ) dξ. (8.7.11) 2π rp+1 a L 2ξ Replacing ξ by −ξ in the first integral term and using F +(ξ)=a F +(−ξ), which follows from the definition (8.7.5), we obtain −ξ ξ −2ξ r F +(ξ)dξ = r a F +(ξ)dξ. (8.7.12) L L The path of integration L,Re(ξ)=c, becomes Re(ξ)=−c, but these paths can be reconciled if F (ξ) tends to zero for large Im(ξ). In view of (8.7.11), we have rewritten ∞ ∞ a2p 1 a2p−2ξ f(r) g(r)dr = F +(ξ)dξ g(r) dr. (8.7.13) rp+1 2πi rp−ξ+1 a L a This result is used to rewrite (8.7.10) as ∞ 2p p−1 a M +{f(r)g(r)} = r + f(r)g(r)dr rp+1 a ∞ ∞ a2p = rp−1f(r) g(r) dr + f(r) g(r) dr rp+1 a a ∞ 1 p−ξ−1 = F +(ξ) dξ r g(r) dr 2πi L a ∞ 1 a2p−2ξ + F +(ξ)dξ g(r) dr 2πi rp−ξ+1 L a 1 = F +(ξ) G +(p − ξ) dξ. 2πi L 392 INTEGRAL TRANSFORMS and THEIR APPLICATIONS This completes the proof. If the range of integration is finite, then we define the generalized finite Mellin transform by a a2p M a{f(r)} = F a (p)= rp−1 − f(r)dr, (8.7.14) − − rp+1 0 where Re p<γ. The corresponding inverse transform is given by 1 r p f(r)=− F a (p)dp, 0 8.8 Exercises 1. Find the Mellin transform of each of the following functions: (a) f(x)=H(a − x), a>0, (b) f(x)=xme−nx, m, n > 0, 1 2 (c) f(x)= 1+x2 , (d) f(x)=J0 (x), z − (e) f(x)=x H(x x0), z (f) f(x)=[H(x − x0) − H(x)]x , (g) f(x)=Ei(x), % & (h) f(x)=exEi(x), (i) f(x)=exp −ax2 ,a>0, (k) f(x)=Ci(x), (j) f(x)=erfc(x), − (m) f(x)=(1+x)−1, (l) f(x)=(1+xa) b, where the exponential integral is defined by ∞ ∞ Ei(x)= t−1 e−t dt = ξ−1 e−ξxdξ. x 1 2. Derive the Mellin transform-pairs from the bilateral Laplace transform and its inverse given by ∞ c +i∞ 1 g¯(p)= e−ptg(t)dt, g(t)= ept g¯(p)dp. 2πi −∞ c−i∞ 3. Show that 1 M =Γ(p) L(p), ex + e−x 1 1 1 where L(p)= − + −··· is the Dirichlet L-function. 1p 3p 5p 4. Show that ! " 1 Γ(p)Γ(n − p) M = . (1 + ax)n ap Γ(n) 5. Show that p n−p Γ M { −n } 1 a 2 −1 x Jn(ax) = p ,a>0,n> . 2 2 Γ n − +1 2 2 6. Show that 394 INTEGRAL TRANSFORMS and THEIR APPLICATIONS ! " πp π (a) M −1 cos Γ(p) f˜(1 − p) = F f(x) , 2 c 2 ! " πp π (b) M −1 sin Γ(p) f˜(1 − p) = F f(x) . 2 s 2 ∞ 7. If In f(x) denotes the nth repeated integral of f(x) defined by ∞ ∞ ∞ In f(x)= In−1f(t)dt, x show that ⎡ ⎤ ∞ 1 (a) M ⎣ f(t)dt, p⎦ = f˜(p +1), p x Γ(p) (b) M [I∞ f(x)] = f˜(p + n). n Γ(p + n) 8. Show that the integral equation ∞ f(x)=h(x)+ g(xξ) f(ξ) dξ 0 has the formal solution c +i∞ 1 h˜(p)+˜g(p) h˜(1 − p) f(x)= x−p dp. 2πi 1 − g˜(p)˜g(1 − p) c−i∞ 9. Find the solution of the Laplace integral equation ∞ 1 e−xξ f(ξ) dξ = . (1 + x)n 0 10. Show that the integral equation ∞ x dξ f(x)=h(x)+ f(ξ) g ξ ξ 0 has the formal solution c +i∞ 1 x−p h˜(p) f(x)= dp. 2πi 1 − g˜(p) c−i∞ Mellin Transforms and Their Applications 395 11. Show that the solution of the integral equation ∞ x dξ f(x)=e−ax + exp − f(ξ) ξ ξ 0 is c +i∞ ! " 1 Γ(p) f(x)= (ax)−p dp. 2πi 1 − Γ(p) c−i∞ 12. Assuming (see Harrington, 1967) ∞ , - M f(reiθ ) = rp−1f(reiθ ) dr, p is real, 0 and putting reiθ = ξ, M {f(ξ)} = F (p) show that (a) M [f(reiθ ); r → p]=exp(−ipθ) F (p). Hence, deduce (b) M −1 {F (p)cos pθ} =Re[f(reiθ )], (c) M −1 {F (p)sin pθ} = −Im[f(reiθ)]. 13. (a) If M [exp(−r)] = Γ(p), show that , - M exp(−reiθ ) =Γ(p) e−ipθ, π (b) If M [log(1 + r)] = , then show that p sin πp , - π cos pθ M Re log (1 + reiθ) = . p sin πp ! " π 1 14. Use M −1 = = f(x), and Exercises 12(b) and 12(c), re- sin pπ 1+x spectively, to show that ! " π cos pθ 1+r cos θ (a) M −1 ; p → r = , sin pπ 1+2r cos θ + r2 ! " π sin pθ r sin θ (b) M −1 ; p → r = . sin pπ 1+2r cos θ + r2 15. Find the inverse Mellin transforms of π π (a) Γ(p)cospθ, where − <θ< , (b) Γ(p)sinpθ. 2 2 396 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 16. Obtain the solution of Example 8.4.2 with the boundary data (a) φ(r, α)=φ(r, −α)=H(a − r). (b) Solve equation (8.4.5) in 0 17. Show that ∞ ∞ cos kn k2 πk π2 1 π2 (a) = − + , and (b) = . n2 4 2 6 n2 6 n=1 n=1 ∞ −nx 18. If f(x)= ane , show that n=1 M {f(x)} = f˜(p)=Γ(p) g(p), ∞ −p where g(p)= an n is the Dirichlet series. n=1 If an =1 forall n, derive f˜(p)=Γ(p) ζ(p). Show that ! " exp(−ax) M =Γ(p) ξ(p, a). 1 − e−x 19. Show that ∞ − n−1 ( 1) − 1−p (a) p =(1 2 ) ζ(p). =1 n n ' ( ∞ (b) M (−1)n−1f(nx) =(1− 21−p) ζ(p)f˜(p). n=1 Hence, deduce ∞ ∞ (−1)n−1 π2 (−1)n−1 7 π4 (c) = , (d) = . n2 12 n4 8 90 n=1 n=1 20. Find the sum of the following series ∞ ∞ (−1)n−1 (−1)n−1 (a) cos kn, (b) sin kn. n2 n n=1 n=1 Mellin Transforms and Their Applications 397 21. Show that the solution of the boundary value problem 2 r φrr + rφr + φθθ =0, 0 φ(r, 0) = φ(r, π)=f(r), is + ∞ π c i f˜(p)cos p θ − dp 1 −p 2 φ(r, θ)= r πp . 2πi cos c−i∞ 2 22. Evaluate ∞ cos an 1 = (a3 − 3πa2 +2π2a). n3 12 n=1 23. Prove the following results: ⎡ ⎤ ∞ (a) M ⎣ ξnf(xξ) g(ξ)dξ⎦ = f˜(p)˜g(1 + n − p), ⎡0 ⎤ ∞ x (b) M ⎣ ξnf g(ξ)dξ⎦ = f˜(p)˜g(p + n +1). ξ 0 24. Show that (a) W −α[e−x]=e−x,α>0, √ % & 1 1 √ K1( x) (b) W 2 √ exp − x = √ ,x>0, x πx where K1(x) is the modified Bessel function of the second kind and order one. 25. (a) Show that the integral (Wong, 1989, pp. 186–187) π/2 1 I(x)= J 2(x cos θ) dθ, ν > − , ν 2 0 can be written as a Mellin convolution ∞ I(x)= f(xξ) g(ξ) dξ, 0 where ' ( 2 − 1 − 2 2 (1 ξ ) , 0 <ξ<1 f(ξ)=Jν (ξ)andg(ξ)= . 0,ξ≥ 1 398 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b) Prove that the integration contour in the Parseval identity c +i∞ 1 I(x)= x−p f˜(p)˜g(1 − p) dp, −2ν cannot be shifted to the right beyond the vertical line Re p =2. ∞ 2 2 sin t 26. If f(x)= exp(−x t ) · J1(t)dt, show that t2 0 3 1 − p Γ p + Γ 2 2 M {f(x)} = . p Γ(p +3) 27. Prove the following relations to the Laplace and the Fourier transforms: (a) M [f(x),p]=L [f(e −t),p], (b) M [f(x); a + iω]=F[f(e −t)e −at; ω], where L is the two-sided Laplace transform and F is the Fourier trans- − 1 form without the factor (2π) 2 . 28. Prove the following properties of convolution: (a) f ∗ g = g ∗ f, (b) (f ∗ g) ∗ h = f ∗ (g ∗ h), x ∗ − (d) δ(x − a) ∗ f(x)=a −1f , (c) f(x) δ(x 1) = f(x), a d n (e) δ n(n − 1) ∗ f(x)= (xnf(x)), dx d n d n d n (f) x (f ∗ g)= x f ∗ g = f ∗ x g . dx dx dx M { } ˜ ∇2 1 1 29. If f(r, θ) = f(p, θ)and f(r, θ)=frr + r fr + r2 fθθ, show that * + d2 M ∇2f(r, θ) = +(p − 2)2 f˜(p − 2,θ). dθ2