Mellin transform and Bessel function

Question

In this article, the following identities are stated without proof: $$\int_0^\infty x^{s - 1}K_0(4\pi x^{1/2}) dx = \frac{1}{2}(2\pi)^{-s}\Gamma(s)^2$$$$\int_0^\infty x^{s - 1}Y_0(4\pi x^{1/2}) dx = -\frac{1}{\pi}(2\pi)^{-s}\cos\pi s\Gamma(s)^2.$$ I assume these identities are standard, though I have been unable to locate a reference and don't know of any way to prove this myself. How might one prove this, and is there an analogue of this for the $J$-Bessel function?

Answer 1

Using the differential equation or equivalent definitions it is not difficult to check that the Laplace transform of $Y_0(4\pi t)$ is $-\frac{1}{2\pi^2}\cdot\frac{\text{arcsinh}\left(\frac{s}{4\pi}\right)}{\sqrt{1+\left(\frac{s}{4\pi}\right)^2}}$. Since the inverse Laplace transform of $2 t^{2n-1}$ is $\frac{2}{s^{2n}\Gamma(1-2n)}$, by the self-adjointness of the Laplace transform we have

$$\begin{eqnarray*} \int_{0}^{+\infty}x^{n-1}Y_0(4\pi\sqrt{x})\,dx &=& \int_{0}^{+\infty}2t^{2n-1}Y_0(4\pi t)\,dt\\&=&-\frac{1}{\pi^2\Gamma(1-2n)}\int_{0}^{+\infty}\frac{\text{arcsinh}\left(\frac{s}{4\pi}\right)}{s^{2n}\sqrt{1+\left(\frac{s}{4\pi}\right)^2}}\,ds\end{eqnarray*} $$ or, via the substitution $s=4\pi\sinh(u)$, $$-\frac{1}{\pi^2\Gamma(1-2n)(4\pi)^{2n-1}}\int_{0}^{+\infty}\frac{u}{\left(\sinh u\right)^{2n}}\,ds $$ which boils down to a derivative of the Beta function via the substitution $u=-\log v$.
The argument for $K_0$ is analogous, as it is the one for $J_0$: $$ \mathcal{L}(J_0(4\pi\sqrt{x})) = \frac{1}{s} e^{-\frac{4\pi^2}{s}},$$ $$\begin{eqnarray*} \int_{0}^{+\infty}x^{n-1}J_0(4\pi\sqrt{x})\,dx &=& \int_{0}^{+\infty}\frac{s^{-n}}{\Gamma(1-n)}\cdot \frac{1}{s} e^{-\frac{4\pi^2}{s}}\,ds\\&=&\frac{\Gamma(n)}{(2\pi)^{2n}\Gamma(1-n)}=\frac{\Gamma(n)^2 \sin(\pi n)}{\pi(2\pi)^{2n}}.\end{eqnarray*} $$