Mellin transforms and asymptotics: Harmonic sums

Phillipe Flajolet, Die Theorie der reziproken Xavier Gourdon, Funktionen und Integrale ist Philippe Dumas ein centrales Gebiet, welches manche anderen Gebiete der Analysis miteinander verbindet.

Reporter : Hjalmar Mellin Ilya Posov Saint-Petersburg State University Faculty of Mathematics and Mechanics

1 Joint Advanced Student School 2004 Contents

• Mellin transform and its basic properties • Direct mapping • Converse mapping • Harmonic sums

2 Joint Advanced Student School 2004 Robert Hjalmar Mellin

Born: 1854 in Liminka, Northern Ostrobothnia, Finland Died: 1933

3 Joint Advanced Student School 2004 Some terminology

2 • Analytic function fs()= c01+−c()s s0+c2()s−s0+L

• Meromorphic function

• Open strip 〈ab, 〉

4 Joint Advanced Student School 2004 Mellin transform ∞ M[(fx);s]==f*1(s) ∫ f(x)xs− dx 0 fx()=→O(xu ) x 0 fx()=→O(xv ) x +∞

fundamental strip f * (su) exists in the strip 〈−−, v〉

5 Joint Advanced Student School 2004 Mellin transform example

1 fx()= fundamental strip : 1+ x uv= 01=− 1 =→O(1) x 0 〈−uv,0− 〉 = 〈 ,1〉 1+ x 1 =→O(xx−1) +∞ 1+ x +∞ xs−1 π fs* ()==∫ dx 0 1s+ xsinπ

6 Joint Advanced Student School 2004 Gamma function

fx()= e−x ex−x =→O(1) 0 ex−−xb=∀O( ) b>0 x→+∞ +∞ fs*1()==∫ e−−xsxdxΓ()s sΓ()s=Γ(s+1) 0

fundamental strip : uv==0 −∞ 〈−uv,0− 〉 = 〈 ,+∞〉

7 Joint Advanced Student School 2004 Transform of step function

⎪⎧0, x ∈[0,1] H(x) = ⎨ ⎩⎪1, x ∈+()1, ∞ H(xx) =→O(1) 0 H(xx) =∀O( −b ) b>0 x→+∞ +∞ 1 1 H(*1xH)==∫∫(x)xss−−dxx1dx=, s∈〈0,+∞〉 00s * 1 H(xx) =−1 H( ), H (x) =− , s∈〈−∞,0〉 s

8 Joint Advanced Student School 2004 Basic properties (1/2)

fx() f* (s) 〈〉αβ, xfν ()x f* (s+〈να) −ν,β−ν〉

ρ 1 * ⎛⎞s fx() f⎜⎟ 〈ρα,ρβ 〉>ρ 0 ρρ⎝⎠ fx(1 / ) −−fs* ( ) 〈−βα, −〉 1 fx()µαf* (s) 〈 ,β〉>µ0 µ s λµfx() λµ−s f* (s) ∑∑kkkk ()k

9 Joint Advanced Student School 2004 Basic properties (2/2)

fx() f* (s) 〈〉αβ, d fx()logx f* (s) 〈〉αβ, ds d Θ−f ()xsf* (s) 〈αβ′′, 〉Θ=x dx d fx() −−(s 1)f* (s−1) 〈αβ′′+1, +1〉 dx x 1 ft()dt −+f* (s 1) ∫0 s

10 Joint Advanced Student School 2004 Zeta function

Mf⎡⎤λµ()x;s= λµ−s f* (s) ⎣⎦∑∑kkkk ( k) e−x gx( ) ==e−−xx+e23+e−x+... 1− e−x −x λµkk==1, kf, ( x) =e ⎛⎞111 gs* ()=+++M⎡⎤e−x ;s=ζ (s)Γ(s) ⎜⎟sssL ⎣⎦ ⎝⎠123 s ∈〈1, +∞〉

11 Joint Advanced Student School 2004 Some Mellin tranforms

es−x Γ〈() 0,+∞〉 es−x −Γ1()〈−1,0〉

2 −x 11 es22Γ〈() 0, +∞〉 e−x ζ ()ssΓ〈() 1,+∞〉 1− e−x 1 π 〈〉0,1 1s+ xsinπ π log(1+〈x) −1,0〉 sssinπ 1 H(x) ≡〈1 0,+∞〉 01<

13 Joint Advanced Student School 2004 Laurent expansion

+∞ k φ()sc=−∑ k (ss0 ) kr≥−

c−r ≠ 0 Pole of order r if r>0

Analytic in sr0 if ≤ 0 Example : 11 =+23+(ss+1)+( =−1) ss2 (1++)s1 L 0 111 =−+1(+ s =0) ss22(1+ )s s L 0

14 Joint Advanced Student School 2004 Singular element Definition 1

A singular element of φ(ss) at 0 is an initial sum of Laurent expansion truncated at terms of O(1) or smaller : 111 1 φ()ss=−22==+1−LL+2+3( +1)+ ss(1++)ss→→01s s −s1 ⎡⎤11⎡11⎤ s.e. at s0 =0 : 22−−, +1 ,K ⎣⎦⎢⎥ss⎣⎢ss⎦⎥ ⎡⎤11⎡ ⎤⎡1 ⎤ s.e. at ss0 = −+1 : , 2 , +2 +3( +1) ,K ⎢⎥⎣⎦ss++11⎢⎣ ⎥⎦⎢⎣s+1 ⎦⎥

15 Joint Advanced Student School 2004 Singular expansion Definition 2 Let φ(s) be meromorfic in Ω with ℘ including all the poles of φφ(ss) in Ω. A singular expansion of ( ) in Ω is a formal sum of singular elements of φ(s) at all points of ℘ Notation : φ(sE) ^ (s∈Ω) 11⎡⎤ ⎡11⎤⎡1⎤ ^ + −+ s ∈〈−2, 2〉 2 ⎢⎥ ⎢2 ⎥⎢⎥ ss(1+ )⎣⎦ss+12ss=−10⎣ s⎦= ⎣⎦s=1

16 Joint Advanced Student School 2004 Singular expansion of gamma function ssΓ=() Γ(s+1) Γ+(1sm+) Γ=()s ss(1++)(s 2)L(s+m) Thus Γ=(ss) has poles at the points −m with m∈ (1− )m 1 Γ()s ms! + m +∞ (1− )k 1 Γ∈()ss^ ∑ k =0 ks! + k poles of gamma function

17 Joint Advanced Student School 2004 Direct mapping (1/3) Theorem 2 Let fx( ) have a transform f* (s) with nonempty fundamental strip 〈,αβ〉. ξγk + fx()=+∑ cξ ,k x(logx) O(x) x→0 ()ξ ,∈kA k ∈− , γξ<−≤α Then fs* ( ) is continuable to a meromorphic function in the strip 〈−γβ, 〉 where it admits the singular expansion (1− )k k! fs* () c (s ) ^ ∑ ξ ,k k +1 ∈〈−γβ, 〉 ()ξ ,∈kA ()s +ξ

18 Joint Advanced Student School 2004 Direct mapping (2/3)

fx() f* (s) Order at 0: O(xs−α ) Leftmost boundary of f.s. at ℜ=( ) α Order at +∞ℜ: O(xs−β ) Rightmost boundary of f.s. at ( ) =β Expansion till O(xsγ ) at 0 Meromorphic continuation till ℜ( ) ≥−γ Expansion till O(xδ ) at +∞ Meromorphic continuation till ℜ≤(s) −δ

19 Joint Advanced Student School 2004 Direct mapping (3/3) fx() f* (s) (1− )k k! Term xxak(log ) at 0 Pole with s.e. ()sa+ k+1 (1− )k k! Term xxak(log ) at +∞−Pole with s.e. ()sa+ k +1 1 Term xa at 0 Pole with s.e. sa+ 1 Term xxa log at 0 Pole with s.e. − ()sa+ 2

20 Joint Advanced Student School 2004 Example 0

23 M j −+xjxx (1− ) M1 fx()==e 1−x+−+L =∑ x+O()x 2! 3! x→0 j=0 j! fs* ()=Γ()s, s∈〈0,+∞〉 fs* ( ) is meromorphically continuable to 〈−−M1,+∞〉 M (1− )j 1 fs* ()^ ∑ s∈〈−M−1,+∞〉 j=0 js! + j finally +∞ (1− )j 1 Γ∈()ss^ ∑ j=0 js! + j

poles of gamma function 21 Joint Advanced Student School 2004 Example 1 e− x fx()==, f* (s) Γ(s)ζ (s) s∈〈1,+∞〉 1− e−x ex−xj+∞ 1 ==BB,1,B=−, −x ∑ j +101L 1(−+ejx→0 j =−1 1)!2 +∞ Bj +1 1 Γ∈()ssζ ()^ ∑ s j =−1 (1js++)! j +∞ (1− )j 1 Γ∈()ss^ ∑ j =0 js! + j

11Bm+1 ζζ()ss^ ∈ , (0)=− , ζ(−m)=− sm−+121

result: ζ (s) is meromorphic in with the only pole at s0 = 1

22 Joint Advanced Student School 2004 Example 2 1 +∞ fx()==∑(−1)nnx 1+ x x→0+ n=0 1/ x +∞ =−∑(1)nn−−1 x 11+ /x x→+∞ n=1 π fs*()=∈, s〈0,1〉 sinπ x

+∞ (1− )n fs*( )^ ∑ , s∈〈−∞,1〉 (continuation to the left of the f.s.) n=0 sn+ +∞ (1− )n−1 fs*( )^ −∈∑ , s〈0,+∞〉 (continuation to the right of the f.s.) n=1 sn− n * π (1− ) fs()≡∈^ ∑ (s ) sinπ xsn∈ + n

23 Joint Advanced Student School 2004 Example 3 (nonexplicit transform)

1 1 7 139 5473 fx()==1−x24+x−x6+x8+O(x10) x→0 cosh x 4 96 5760 645120 fs* ( ) has a fundamental strip 〈+0, ∞〉 111 7 1 139 1 5473 1 fs* ()^ −+ − + s∈〈−10,+∞〉 ss4 ++2 96 s4 5760 s+6 645120 s+8 111 7 1 139 1 5473 1 fs* ()^ −+ − + +∈s ss4 ++2 96 s4 5760 s+6 645120 s + 8 L

24 Joint Advanced Student School 2004 Converse mapping (1/3) Theorem 3 Let fx( )∈+C(0, ∞) have a transform f*(s) with nonempty fundamental strip 〈,αβ〉. Let fs*( ) be meromorphically continuable to 〈−γβ, 〉 with a finite number of poles there, and be analytic on ℜ=(s) −γ.

* −r Let fs( ) =>O()sr with 1 when s→+∞ in 〈−γβ,〉. L

25 Joint Advanced Student School 2004 Converse mapping (2/3) Theorem 3 (continue) L 1 If fx* ( ) d s∈〈−γβ, 〉 ^ ∑ ξ ,k k ()ξ ,∈kA ()s − ξ k ∈− , γξ<−≤β Then an asymptotic expansion of fx( ) at 0 is

k −1 ⎛⎞(1− ) −−ξ k 1 γ f ()xd=+∑ ξ ,k ⎜⎟x(log x)O(x) ()ξ ,∈kA ⎝⎠(1k − )!

26 Joint Advanced Student School 2004 Converse mapping (3/3) fs* () f(x) Pole at ξ Term in asymptotic expansion ≈ x−ξ left of f.s. expansion at 0 right of f.s expansion at +∞ Simple pole 1 left: x−ξ at 0 s − ξ 1 right: −∞x−ξ at + s − ξ Multiple pole Logarifmic factor 1(−1)k left: xx−ξ (log )k at 0 ()sk− ξ k +1 ! 1(−1)k right: −∞xx−ξ (log )k at + ()sk− ξ k +1 !

27 Joint Advanced Student School 2004 Example 4 ΓΓ()ss(ν −) φν(=s) , analytic in strip 〈0, 〉 Γ(ν ) +∞ (1−Γ)j (ν +j −1) 1 φν(∈ss),^ ∑ 〈−∞〉 j=0 js!(Γ+ν ) j 1 ν /2+∞i fx( ) = ∫ φ(s)x−sds - original function 2iπ ν /2−∞i M (1−Γ)j (ν +j −1) fx()=+∑ xjMO(x+12) x→0 j=0 j!(Γ ν ) fx()=+(1 x)−ν has the same expansion at 0 fx( ) =+(1 x)−ν +ϖϖ(x), where (x) =O(xM ) ∀M>0

28 Joint Advanced Student School 2004 Example 5 π φ()ss=Γ(1− ) s∈〈0,1〉 sinπ s +∞ 1 φ()sn^ ∑(−∈1)n ! s〈−∞,1〉 n=0 sn+ +∞ nn fx( ) ∑(−1) n!x - the expansion is only asymptotic! n=0 M ⎛⎞nn M+12 ⎜⎟fx()=−∑(1)n!x+O(x ) x→0 ⎝⎠n=0 ⎛⎞+∞ e−t ⎜⎟In fact fx( ) = ∫ dt ⎝⎠0 1+ xt

29 Joint Advanced Student School 2004 Harmonic sums Definition 3

Gx( ) = ∑λµjkg( x) harmonic sum j

λ j amplitudes

µ j frequencies

−s Λ(s) = ∑λµjj The Dirichlet series j Mg[]()µµx;s= −s g* (s) (separation property) Gs**()=Λ()sg()s

30 Joint Advanced Student School 2004 Harmonic sum formula

The Mellin transform of the harmonic sum

Gx( ) = ∑λµkkg( x) is defined in the intersection k of the the fundamental strip of g(x) and the domain of absolute convergence of the Dirichlet −s series Λ=(s) ∑λµkk (which is of the form k

ℜ>(s) σσaa for some real ) Gs**()=Λ(s)g(s)

31 Joint Advanced Student School 2004 Example 6 +∞ ⎡⎤11 +∞ 1xk/ 1 1 x hx()=−∑∑⎢⎥= , λµkk=, =, g(x)= kk==11⎣⎦kk++x k1/xk k k 1+x 11 1 hn()==Hn 1+++L + 23 n +∞ +∞ −−ss1+ Λ=()sk∑∑λµkk = =ζ(1−s) kk==11 π hs**()=Λ()sg()s=−ζ (1−s) s∈〈−1,0〉 sinπ s 11 ζγ()ss=++L, ζ(1−)=−+γ ss−1 ⎡⎤1 γ +∞ ζ (−1)k hs*()^ − −−(1)k s ∈〈−1,+∞〉 ⎢⎥2 ∑ ⎣⎦ssk =1 sk− (1− )k B 1 1 1 1 Hnlog ++γγk log n++ − + + n ∑ k 4 L k ≥1 kn 21n2n120n

32 Joint Advanced Student School 2004 Example 7 +∞ ⎡⎤xx⎛⎞ lx()=Γlog (x+1)−γ x=∑ ⎢⎥−log⎜⎟1+ s∈<−2,+∞> n=1 ⎣⎦nn⎝⎠ 1 λµ==1, , gx( ) =x−log(1+x) nnn +∞ +∞ −ss Λ(sn))= ∑∑λµkk = = ζ(−s nn==11 π Simple poles in positive integers ls**()=Λ(s)g(s)=−ζ (−s) , s∈〈−2,−1〉 sssinπ Double poles in 0 and -1 ⎡⎤11− γπ⎡⎤1log2+∞ (−(1) n−1ζ −n) ls* () ++− + ^ ⎢⎥22⎢⎥∑ ⎣⎦(1ss+() +1) ⎣⎦2ssn=1 ns()+ n 11+∞ B lx( ) =−log(x!) γγx=xlog x−(1−)x−log x+log 2π+ 2n ∑ 21n− 22n=1 nn(2−1)x +∞ B 1 log(xx!) =+log xxe− 2π x 2n ()∑ 21n− n=1 2(nn2 −1)x

33 Joint Advanced Student School 2004 Example 8 +∞ −nx e +∞ Lx( ) == (polylogarithm Li(z) znwn− ) w=k∈ ww∑∑w n=1 n=1 n 1 λµ==,,ng(x)=e− x nknw * Λ(ss)(=+ζζk),Lk (ss)=+(k)Γ(s)s∈〈1,+∞〉 +∞ ζ (−kn)1 (−1)k −1 ⎡ 1 H ⎤ Ls* () (−+1)n +k −1 k ^ ∑ ⎢ 2 ⎥ n=0 n!(s+ n k−+1)!⎣(sk−1)sk+−1⎦ nk≠−1 s ∈〈−∞,+∞〉 k −1 +∞ (1−() kn−1 ζ kn−) n Lxkk()=−x [ logx+H−1]+∑ (−1) x (1kn− )! n=0 ! nk≠−1

+∞ 1 π nnζ ()2 − n Lx12()=+∑(−1) x xnn=0 ! 34 Joint Advanced Student School 2004 Example 9 modified theta function +∞ −nx22 e 1 2 Θ=()xn, λµ=, =, g(x)=e−x wn∑ wwn n=1 nn

* 1 ⎛⎞s Θ=1 ()ssΓ⎜⎟ζ (+1) s∈〈0,+∞〉 22⎝⎠ double pole at ss==0 and simple poles at −2m

* ⎡⎤11γ 111 Θ+1 ()ss^ 2 +++L∈ ⎣⎦⎢⎥ss2+12 s2 240 s+4 γ 11 Θ=()x−logx++xxx24+ + →0 1 2 12 240 L

* 1 ⎛⎞s 11π M Θ=0 ()s Γ(⎜⎟ζ sx)OΘ0 = −+()x→0 22⎝⎠ 2x 2 35 Joint Advanced Student School 2004 Example 10

+∞ −−kx x Dx()==∑ d(k)e , λµkkd(k), =k, g(x)=e k =1 +∞ +∞ dk() Λ(ss)(= λµ−2s = = ζ ) ∑∑kk −s kk==11k Ds* ()=Γ(s)ζ 2 (s) ⎡⎤11γζ⎡⎤+∞ 2 (−−2k 1)1 Ds* ()^ ++− ⎢⎥2 ⎢⎥∑ ⎣⎦(1s −−) ss1 ⎣⎦4 k =0 ()2k+1!s+2k+1 +∞ 2 11ζ (−−2k 1)21k + Dx() (−+logx γ )+−∑ x xk42k =0 ()+1!

36 Joint Advanced Student School 2004 The end

37 Joint Advanced Student School 2004