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
Best Answer
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 $\tau \neq 0\land |\alpha \tau| \geq |\Im(x)|$.
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 $\tau \neq 0\land | \alpha \tau| \geq |\Im(x)|$.
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 $f(q)$.
See WolframAlpha evaluation of formula (3) above.
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
$$\text{FourierCosTransform}[f(q),q,x]=\sqrt{\frac{2}{\pi}} \int\limits_{-\infty}^{\infty} f(q)\, \cos(x q) \, dq\tag{6}$$