# Evaluation of $\int_0^\infty \frac{(x^2+y^2)^{-s/2}}{e^{2\pi y}-1}\cos(s \arctan(y/x))dy$

$$\mbox{Does the integral}\quad \int_{0}^{\infty}{\left(x^{2} + y^{2}\right)^{-s/2} \over {\rm e}^{2\pi y} - 1}\, \cos\left(s\arctan\left(y \over x\right)\right)\,{\rm d}y\quad \mbox{converge or diverge ?.}$$

Here $s$ is complex and $x$ is real.

This is similar to Hermites' integral formula for the Hurwitz zeta function, but uses $\large\cos$ in place of $\large\sin$.

The limit of the integrand tends to $\infty$ as $y \to 0^{+}$, but I know this does not necessarily imply divergence due to examples such as $\int_{0}^{1}x^{-1/2}\,{\rm d}x = 2 < +\infty$.

Suppose $x \neq 0$. We have, as $y$ tends to $0$, $$\frac{(x^2+y^2)^{-s/2}}{e^{2\pi y}-1}\cos(s \arctan(y/x)) \sim \frac{1}{2 \pi \: x^s \:y}$$ which gives a divergent integral.
Now if $x=0$, we have, as $y$ tends to $0$, $$\frac{(x^2+y^2)^{-s/2}}{e^{2\pi y}-1}\cos(s \arctan(y/x)) \sim \frac{\cos (\pi s/2)}{2 \pi \:y^{s+1}}$$ which gives a convergent integral for $\Re s <0$ or $\cos (\pi s/2)=0$.
• Why does the first part "give a divergent integral" ? I know the integrand behaves like $1/(2\pi y x^2)$ near $y=0$ which tends to $\infty$ as $y\to 0$, but how can we be sure that the integral of the integrand is not defined; c.f. $\int_0^1 x^{-1/2}dx = 2<+\infty$. – Pixel Aug 12 '14 at 16:26
• @pbs This a comparison test for improper integrals. Let's say $x>0$ (the integrand is even function of $x$). Near $y=0$, you may write $$f(y)= \frac{1}{2 \pi \: x^s \:y}+ \mathcal{O}(1)$$ giving $$\int_{\epsilon}^{b} f(y)\: \mathrm{d}y= \int_{\epsilon}^{b}\frac{1}{2 \pi \: x^s \:y} \mathrm{d}y+ \mathcal{O}(1), \qquad \epsilon>0,$$ and the integral on the right hand side is divergent since it is equal to $\frac{1}{2 \pi x^s} \log (b/\epsilon)$ which tends to $\infty$ as $\epsilon \rightarrow 0^+$. Thanks. – Olivier Oloa Aug 12 '14 at 17:45