How does one show the following integral (which was obtained in Maple)
$\int_0^{\infty } x \exp \left(-a x^2\right) Y_0(b x) \, dx=\frac{1}{2\pi a}\exp\left(-\frac{b^2}{4 a}\right) \text{Ei}\left(\frac{b^2}{4
a}\right)$,
where $Y_0$ is the Bessel function of the 2nd kind, and Ei is the exponential integral function?
Best Answer
This answer is based on the fact that $Y_\nu(2\sqrt{u})$ is a solution to the confluent hypergeometric limit equation $x y''(x) + (\nu+1)y'(x) + y(x) = 0$. Since this has only linear coefficients, it can be used to create a first-order differential equation for the integral.
Let $z = b^2/4a$. Then the integral is equivalent to $$ \int_0^\infty xe^{-ax^2}Y_0(bx)dx = \frac{2}{b^2}\int_0^\infty Y_0(2\sqrt{u})e^{-u/z}du = \frac{2}{b^2}I(z) $$ The integral $I(z)$ is now in the form of the Laplace transform of the function $f(u) = Y_0(2\sqrt{u})$ that satisfies the differential equation $$ (uf'(u))'+f(u) = 0. $$ Taking the Laplace transform of this equation gives $$ -s\frac{d}{ds}\left[sF(s)\right] - \lim_{u\rightarrow 0^+}uf'(u)+F(s) = 0 $$ The limit can be found from $uf'(u) = -\sqrt{u}Y_1(2\sqrt{u})$, and turns out to be $1/\pi$. So $F(s)$ satisfies $$ s^2F'(s) + (s-1)F(s) = -\frac{1}{\pi}, $$ and $I(z) = F(1/z)$ satisfies $$ zI'(z)+(z-1)I(z) = \frac{z}{\pi}\Longrightarrow \frac{d}{dz}\left[\frac{e^z}{z}i(z)\right] = \frac{e^z}{\pi z}\Longrightarrow i(z) = \frac{ze^{-z}}{\pi}\left[C + \mathrm{Ei}\left(z\right)\right] $$ Which gives for the original integral $$ \int_0^\infty xe^{-ax^2}Y_0(bx)dx = \frac{2}{b^2}I(z) = \frac{2ze^{-z}}{\pi b^2}\left[C + \mathrm{Ei}\left(z\right)\right] = \frac{e^{-b^2/(4a)}}{2\pi a}\left[C + \mathrm{Ei}\left(\frac{b^2}{4a}\right)\right]. $$
I'm not sure of an easy way to show $C = 0$ other than the asymptotic analysis ComplexButTrivial did.