HERMITE´S INTEGRAL REPRESENTATION HURWITZ ZETA FUNCTION

Today´s post we will proof Hermite´s integral representation for The Hurwitz´s zeta function. In general this integral is proved by means of Abel-Plana´s summation formula. Here we base our proof in Connon´s paper [1] providing all details.

The Hurwitz Zeta function is defined as following

$\zeta(s,a)=\sum_{k=0}^{\infty} \frac{1}{\left(k+a\right)^{s}}$

It admits the beautiful integral representation due to Hermite

$\zeta(s,u)=\frac{u^{-s}}{2}+\frac{u^{1-s}}{s-1}+ 2\int_{0}^{\infty}\frac{sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}\left( e^{2 \pi x}-1\right)}$

Proof:

Start with the following integral (proved below in the appendix)

$\frac{2}{\Gamma(s)}\int_0^{\infty}e^{-uy^2}y^{2s-1}\sin(xy^2)dy=\frac{sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}}$ (1)

Multiply both sides of (1) by $\frac{1}{e^{2 \pi x}-1}$ and integrate with respect to

$\frac{2}{\Gamma(s)}\int_0^{\infty}\frac{1}{e^{2 \pi x}-1}\int_0^{\infty}e^{-uy^2}y^{2s-1}\sin(xy^2)dy\,\,dx=\int_0^{\infty}\frac{sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}\left(e^{2 \pi x}-1\right)}dx$
(2)

Lets now focus on the double integral on the left hand side of (2), call it .

$J=\int_0^{\infty}\frac{1}{e^{2 \pi x}-1}\int_0^{\infty}e^{-uy^2}y^{2s-1}\sin(xy^2)dy\,\,dx$ (3)

Swapping the order of integration of (3) we obtain

$J=\int_0^{\infty}e^{-uy^2}y^{2s-1}\,\,\int_0^{\infty}\frac{\sin(xy^2)}{e^{2 \pi x}-1}\,dx\,dy$ (4)

From this post (equation (5))we know that

$2\int_0^{\infty}\frac{\sin(xs)}{e^{2 \pi x}-1}\,dx=\frac{1}{e^{s}-1}-\frac{1}{s}+\frac{1}{2}$ (5)

Letting $s=y^2 \,\text{in}\,\,(4)\,\,\text{we obtain}$

$2\int_0^{\infty}\frac{\sin(xy^2)}{e^{2 \pi x}-1}\,dx=\frac{1}{e^{y^2}-1}-\frac{1}{y^2}+\frac{1}{2}$ (6)

Plugging (6) back in (4) we get

$J=\frac{1}{2}\int_0^{\infty}e^{-uy^2}y^{2s-1}\left(\frac{1}{e^{y^2}-1}-\frac{1}{y^2}+\frac{1}{2} \right)dy$

Now, let $v=y^2\,\Rightarrow \, dy=\frac{dv}{2\sqrt{v}}$

$J=\frac{1}{2}\int_0^{\infty}e^{-uv}(v^{1/2})^{2s-1}\left(\frac{1}{e^{v}-1}-\frac{1}{v}+\frac{1}{2} \right)\frac{dv}{2\sqrt{v}}$

$J=\frac{1}{4}\int_0^{\infty}e^{-uv}v^{s-1}\left(\frac{1}{e^{v}-1}-\frac{1}{v}+\frac{1}{2} \right)dv$ (7)

Plugging (7) in (2)

$\frac{1}{2\Gamma(s)}\int_0^{\infty}\left(\frac{1}{e^{v}-1}-\frac{1}{v}+\frac{1}{2} \right)\frac{v^{s-1}}{e^{uv}}dv=\int_0^{\infty}\frac{sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}\left(e^{2 \pi x}-1\right)}dx$

$\frac{1}{\Gamma(s)}\int_0^{\infty}\left(\frac{1}{e^{v}-1}-\frac{1}{v}+\frac{1}{2} \right)\frac{v^{s-1}}{e^{uv}}dv=2\int_0^{\infty}\frac{sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}\left(e^{2 \pi x}-1\right)}dx$ (8)

From the appendix we know that

$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\left(\frac{1}{(e^x-1)}-\frac{1}{x}+\frac{1}{2}\right)\frac{x^{s-1}}{e^{ax}}dx=\zeta(s,a)-\frac{a^{-s}}{2}-\frac{a^{1-s}}{s-1}$ (9)

$\text{letting}\,\,x=v\,\,\text{and}\,\,a=u\,\,\text{in}\,(9)$

$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\left(\frac{1}{(e^v-1)}-\frac{1}{x}+\frac{1}{2}\right)\frac{v^{s-1}}{e^{av}}dv=\zeta(s,u)-\frac{u^{-s}}{2}-\frac{u^{1-s}}{s-1}$ (10)

Plugging (10) in (8) we finally get

$\boxed{\zeta(s,u)=\frac{u^{-s}}{2}+\frac{u^{1-s}}{s-1}+2\int_0^{\infty}\frac{sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}\left(e^{2 \pi x}-1\right)}dx}$ (11)

If we let u=1 in (11) we obtain

$\boxed{\zeta(s)=\frac{1}{2}+\frac{1}{s-1}+2\int_0^{\infty}\frac{sin\left( s \arctan \left( x\right)\right)}{\left( 1+x^2\right)^{s/2}\left(e^{2 \pi x}-1\right)}dx}$ (12)

Appendix

$\int_0^{\infty}e^{-uy^2}y^{2s-1}\sin(xy^2)dy=\frac{\Gamma(s)}{2}\frac{sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}}$

Proof:

$I=\int_0^{\infty}e^{-uy^2}y^{2s-1}\sin(xy^2)dy$

let $y^2=w \Rightarrow 2ydy=dw\,\Rightarrow \,dy=\frac{dw}{2\sqrt{w}}$

$I=\int_0^{\infty}e^{-uw}(w^{1/2})^{2s-1}\sin(xw)\frac{dw}{2\sqrt{w}}$

$=\frac{1}{2}\int_0^{\infty}e^{-uw}w^{s-1}\sin(xw)dw$

From here (equation (10))we know that

$\int_0^{\infty}e^{-at}w^{x-1}\sin(bt)dt=\Gamma(x)\frac{sin\left( x \arctan \left( \frac{b}{a}\right)\right)}{\left( a^2+b^2\right)^{x/2}}$

Letting $a=w, \,x=s\,\text{and}\, b=x$ we obtain

$\int_0^{\infty}e^{-uw}w^{s-1}\sin(xw)dw=\Gamma(s)\frac{sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}}$

and we conclude that

$\boxed{\int_0^{\infty}e^{-uy^2}y^{2s-1}\sin(xy^2)dy=\frac{\Gamma(s)}{2}\frac{sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}}}$

First Integral Representation for the Hurwitz Zeta function

$\zeta(s,a)=\frac{1}{\Gamma(s)} \int_{0}^{\infty}\frac{t^{s-1}e^{-ta}}{1-e^{-t}}dt$

Proof:

The Hurwitz Zeta function is defined as following

$\zeta(s,a)=\sum_{k=0}^{\infty} \frac{1}{\left(k+a\right)^{s}}$

It admits a very simple integral representation that can easely be obtained as following: multiplying both sides of the above equation by $\Gamma(s)$

$\Gamma(s)\zeta(s,a)=\sum_{k=0}^{\infty} \frac{1}{\left(k+a\right)^{s}}\int_{0}^{\infty}t^{s-1}e^{-t}dt$

substituting $t\mapsto(k+a)t$

$\Gamma(s)\zeta(s,a)=\sum_{k=0}^{\infty} \int_{0}^{\infty}t^{s-1}e^{-t(k+a)}dt$

$\Gamma(s)\zeta(s,a)= \int_{0}^{\infty}t^{s-1}e^{-ta}\sum_{k=0}^{\infty}e^{-kt}dt$

Finally arraiving at:

$\boxed{\zeta(s,a)=\frac{1}{\Gamma(s)} \int_{0}^{\infty}\frac{t^{s-1}e^{-ta}}{1-e^{-t}}dt}$

Second Integral Representation for the Hurwitz Zeta function

$\zeta(s,a)=\frac{1}{2}a^{-s}+\frac{a^{1-s}}{s-1}+\frac{1}{\Gamma(s)} \int_{0}^{\infty}\left(\frac{1}{e^{t}-1}-\frac{1}{t}+\frac{1}{2}\right)\frac{t^{s-1}}{e^{at}}dt$

Proof:

Starting from

$\zeta(s,a)= \frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}e^{-ax}}{1-e^{-x}}dx$

rewriting as

$\zeta(s,a)= \frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}e^{-ax}}{e^{-x}(e^{x}-1)}dx$

$= \frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}e^{-ax}}{e^{-x}(e^{x}-1)}dx$

$=\frac{1}{\Gamma(s)}\int_{0}^{\infty}x^{s-1}e^{-ax}e^{x}(e^x-1)^{-1}dx\\&=\frac{1}{\Gamma(s)}\int_{0}^{\infty}x^{s-1}e^{-ax}e^{x}\left((e^x-1)^{-1}-\frac{1}{x}+\frac{1}{2}-\frac{1}{2}-\frac{1}{x}\right)dx$

$=\frac{1}{\Gamma(s)}\int_{0}^{\infty}x^{s-1}(1+e^{-x}-e^{-x})e^{-ax}e^{x}\left((e^x-1)^{-1}-\frac{1}{x}+\frac{1}{2}-\frac{1}{2}-\frac{1}{x}\right)dx$

$=\frac{1}{\Gamma(s)}\int_{0}^{\infty}(e^{x}+1-1)\left((e^x-1)^{-1}-\frac{1}{x}+\frac{1}{2}-\frac{1}{2}-\frac{1}{x}\right)\frac{x^{s-1}}{e^{ax}}dx$

Now I´ll break down the last integral into three integrals

$\zeta(s,a)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}e^{x}\left((e^x-1)^{-1}-\frac{1}{x}+\frac{1}{2}-\frac{1}{2}-\frac{1}{x}\right)\frac{x^{s-1}}{e^{ax}}dx+\frac{1}{\Gamma(s)}\int_{0}^{\infty}\left((e^x-1)^{-1}-\frac{1}{x}+\frac{1}{2}-\frac{1}{2}-\frac{1}{x}\right)\frac{x^{s-1}}{e^{ax}}dx-\frac{1}{\Gamma(s)}\int_{0}^{\infty}\left((e^x-1)^{-1}-\frac{1}{x}+\frac{1}{2}-\frac{1}{2}-\frac{1}{x}\right)\frac{x^{s-1}}{e^{ax}}dx$

which is equal to

$\zeta(s,a)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}e^{-ax}}{1-e^{-x}}dx+\frac{1}{\Gamma(s)}\int_{0}^{\infty}\left((e^x-1)^{-1}-\frac{1}{x}+\frac{1}{2}-\frac{1}{2}-\frac{1}{x}\right)\frac{x^{s-1}}{e^{ax}}dx-\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}e^{-(1+a)x}}{1-e^{-x}}dx$

We can recognize the first and the last integrals as Hurwitz zeta functions of $\zeta(s,a)\,\,\text{and}\,\,\zeta(s,a+1)$ respectively and using the functional function of the Hurwitz Zeta Function

$\zeta(s,a+1)=\zeta(s,a)-a^{-s}$

we get

$\zeta(s,a)=\zeta(s,a)-\zeta(s,a)+a^{-s}+ \frac{1}{\Gamma(s)}\int_{0}^{\infty}\left((e^x-1)^{-1}-\frac{1}{x}+\frac{1}{2}-\frac{1}{2}-\frac{1}{x}\right)\frac{x^{s-1}}{e^{ax}}dx$

$\zeta(s,a)=a^{-s}+ \frac{1}{\Gamma(s)}\int_{0}^{\infty}\left((e^x-1)^{-1}-\frac{1}{x}+\frac{1}{2}-\frac{1}{2}-\frac{1}{x}\right)\frac{x^{s-1}}{e^{ax}}dx$

Breaking down the remaining integral into three integrals we get

$\zeta(s,a)=a^{-s}-\frac{1}{2}\frac{1}{\Gamma(s)}\int_{0}^{\infty}x^{s-1}e^{-ax}dx+\frac{1}{\Gamma(s)}\int_{0}^{\infty}x^{s-2}e^{-ax}dx+ \frac{1}{\Gamma(s)}\int_{0}^{\infty}\left(\frac{1}{(e^x-1)}-\frac{1}{x}+\frac{1}{2}\right)\frac{x^{s-1}}{e^{ax}}dx$

The first and second integrals can be rewritten as Gamma functions and the third one is the integral we are looking for, therefore we get

$\zeta(s,a)=a^{-s}-\frac{1}{2}\frac{1}{\Gamma(s)}\frac{\Gamma(s)}{a^s}+\frac{1}{\Gamma(s)}\frac{\Gamma(s)a^{1-s}}{s-1}+ \frac{1}{\Gamma(s)}\int_{0}^{\infty}\left(\frac{1}{(e^x-1)}-\frac{1}{x}+\frac{1}{2}\right)\frac{x^{s-1}}{e^{ax}}dx$

simplifying we finally get the result

$\boxed{\zeta(s,a)=\frac{a^{-s}}{2}+\frac{a^{1-s}}{s-1}+ \frac{1}{\Gamma(s)}\int_{0}^{\infty}\left(\frac{1}{(e^x-1)}-\frac{1}{x}+\frac{1}{2}\right)\frac{x^{s-1}}{e^{ax}}dx}$

[1] A new derivation of Hermite´s integral for the Hurwitz zeta function. pdf available here

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