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

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 $\int_{0}^{1}x^{-1/2}\,{\rm d}x = 2 < +\infty$.

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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 $\cos (\pi s/2)=0$.

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  • $\begingroup$ 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. $\int_0^1 x^{-1/2}dx = 2<+\infty$. $\endgroup$ – Pixel Aug 12 '14 at 16:26
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    $\begingroup$ @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. $\endgroup$ – Olivier Oloa Aug 12 '14 at 17:45
  • $\begingroup$ @ OlivierOloa thank you that's very helpful. $\endgroup$ – Pixel Aug 14 '14 at 9:43

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