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Consider the following integral: $$ I(x) \ = \ \int_{0}^{\infty} dp \int_{-\infty}^{\infty} dr \int_{-\infty}^{\infty} ds\ \sinh(\pi p) \left[ K_{ip} \left( \sqrt{ r^{2} + s^{2} } \ x \right) \right]^{2} $$

Where $K_{ip}$ is the modified Bessel function of the second kind, or imaginary order $\nu = i p$.

I have a strong feeling that this integral should diverge, but I am unsure how to show this explicitly. Is there a way to see this?

I feel like maybe I could use the identity $$ K_{\nu}(x) K_{\nu}(y) \ = \ \frac{1}{2} \int_{0}^{\infty} \frac{1}{2a} \exp\left( - \frac{a}{2} - \frac{x^{2} + y^{2}}{2a}\right)\ K_{\nu}\left( \frac{xy}{a} \right) \ dp $$

Should I make a variable change $(r,s) \mapsto$ to some kind of polar coordinates? in this case, the "angle" wouldn't appear in the integrand. I'm unsure what is the best way to approach such a problem.

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$$I(x)= \int_{0}^{+\infty}\int_{0}^{+\infty}2\pi r \sinh(\pi p)K_{ip}^2(rx)\,dr\,dp=\frac{2\pi}{x^2}\int_{0}^{+\infty}\int_{0}^{+\infty}r K_{ip}^2(r)\sinh(\pi p)\,dr\,dp$$ and since $\int_{0}^{+\infty}r K_a^2(r)\,dr = \frac{\pi a}{2\sin(\pi a)}$ for any $a$ such that $\text{Re}(a)\in(-1,1)$, the given integral equals $$ \frac{\pi^2}{x^2}\int_{0}^{+\infty}p\,dp = +\infty.$$

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  • $\begingroup$ @Jack_D'Aurizio Where did you find the integral $\int_{0}^{\infty} r K_{a}^{2}(r) dr = \frac{\pi a}{2 \sin(\pi a)}$? $\endgroup$ Commented Apr 19, 2017 at 4:58
  • $\begingroup$ @Greg.Paul: classical tables of integrals, it is related with a Hankel transform. $\endgroup$ Commented Apr 19, 2017 at 5:11

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