Integrating a Modified Bessel function of the second kind with a singularity

Question

Does someone know how to handle the integral

$$\int_{-\infty}^{\infty} \frac{K_0\!\left(\lvert \tau \rvert \sqrt{q^2 \alpha ^2}\right)}{q^2}\cos (q x)\,\mathrm{d}q $$

$\alpha$ is a real number and $\tau$ is imaginary time. We have two problems at $q=0$, namely: the singularity originating from $\frac{1}{q^2}$ and the Modified Bessel function of the second kind blows up in the other direction. The latter behaviour is dominant. I was considering making the case that the argument of the Bessel function is small, thus claiming $K_0\!\left(\lvert \tau \rvert \sqrt{q^2 \alpha ^2}\right) \approx \log\left(\lvert \tau \rvert \sqrt{q^2 \alpha ^2}\right)$. But in the end, I'm left with the same problems as before. Does anyone have an idea how to approach this?

Kind regards, Roeland

Answer 1

Mathematica gives the conditional result

$$\int_{-\infty}^{\infty} \frac{\cos(q x) K_0\left(|\tau| \sqrt{q^2 \alpha^2}\right)}{q^2} \, dq=\pi \left(\sqrt{\alpha^2 \tau \tau^*+x^2}-x \sinh^{-1}\left(\frac{x}{|\alpha \tau| }\right)\right)\tag{1}$$

which is valid for $\tau \neq 0\land |\alpha \tau| \geq |\Im(x)|$.

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Note the integrand is an even function of $q$ and Mathematica gives the same result for

$$2 \int\limits_0^{\infty} \frac{\cos(q x) K_0\left(|\tau|\, \sqrt{q^2 \alpha^2}\right)}{q^2} \, dq=\pi \left(\sqrt{\alpha^2 \tau \tau^*+x^2}-x \sinh^{-1}\left(\frac{x}{|\alpha \tau|}\right)\right)\tag{2}$$

which is again valid for $\tau \neq 0\land | \alpha \tau| \geq |\Im(x)|$.

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I believe formula (2) above is equivalent to

$$2\, \mathcal{M}_q\left[K_0\left(|\tau|\, \sqrt{q^2 \alpha^2}\right) \cos(q x)\right](-1)=\pi x \left(\sqrt{\frac{\alpha^2}{x^2}} \sqrt{\frac{x^2 \left(\frac{\alpha^2 |\tau|^2}{x^2}+1\right)}{\alpha^2}}-\sinh^{-1}\left(\frac{1}{|\tau| \sqrt{\frac{\alpha^2}{x^2}}}\right)\right)\tag{3}$$

where

$$\mathcal{M}_q[f(q)](s)=\int_0^{\infty } f(q)\, q^{s-1} \, dq\tag{4}$$

is the Mellin transform of $f(q)$.

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See WolframAlpha evaluation of formula (3) above.

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I believe formula (2) above is also equivalent to the Mathematica evaluation

$$2 \sqrt{\frac{\pi}{2}}\, \text{FourierCosTransform}\left[\frac{K_0\left(|\tau| \sqrt{q^2 \alpha^2}\right)}{q^2},q,x\right]=\pi \left(\sqrt{\alpha^2 \tau \tau^*+x^2}-x \sinh^{-1}\left(\frac{x}{|\alpha \tau|}\right)\right),\quad\tau\neq 0\tag{5}$$

where

$$\text{FourierCosTransform}[f(q),q,x]=\sqrt{\frac{2}{\pi}} \int\limits_{-\infty}^{\infty} f(q)\, \cos(x q) \, dq\tag{6}$$