# hurwitz zeta Integral representation

How can I prove

$$\zeta(s,z) = \frac{z^{-s}}{2}+\frac{z^{1-s}}{s-1}+2\int^\infty_0 \frac{\sin(s\arctan(x/z))}{(z^2+x^2)^{s/2}(e^{2\pi x}-1)}\,dx \ ?$$

My attempt

$$\frac{2}{e^{2\pi x}-1}\,dx = \frac {1}{\pi x} -1+\frac {2x}{\pi}\sum_{k=1}^\infty\frac {1}{x^2+k^2},$$

but the term $\sin(s\arctan(x/z))$ is difficult to deal with!

• It may be helpful to consider the imaginary part of the expression $(z +i\, x)^{-s}$. Aug 16, 2016 at 19:46
• en.wikipedia.org/wiki/Abel-Plana_formula Aug 16, 2016 at 21:15
• @RandomVariable, thanks RV. That's a very useful formula. Aug 17, 2016 at 20:36

This proof relies on this paper

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

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

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

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

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

Swapping the order of integration of $$(3)$$ we obtain

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

From this here we know that

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

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

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

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

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

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

Plugging $$(7)$$ in $$(2)$$

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

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

From this answer we know that

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

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

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

Plugging $$(10)$$ in $$(8)$$ we finally get

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

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

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

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 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

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