Weber-type integral

Question

In connection with this answer, I came across the following integral:

$$\int_{0}^{\infty} \frac{du}{u} \: \,e^{-t u^2} \frac{J_0(u) Y_0(r u)-J_0(r u) Y_0(u)}{J_0^2(u)+Y_0^2(u)}$$

where $r \gt 1$. I know this looks like an inverse Weber transform, but beyond the fact that the integrand has a name, I could not see how this could help. Some context:

1) It turns out that

$$\int_{0}^{\infty} \frac{du}{u} \: \frac{J_0(u) Y_0(r u)-J_0(r u) Y_0(u)}{J_0^2(u)+Y_0^2(u)} = \frac{\pi}{2}$$

I know this is correct from the boundary conditions of the problem I solved. Gradsteyn & Rhyzhik agree. Of course, I did not derive this directly, so I have no hard proof.

2) Asymptotically, the integrand behaves as

$$\frac{\sin{(r-1) u}}{\sqrt{r} u} e^{-t u^2}$$

This is very interesting, and somewhat expected. While this is a solution in a 2D, radially-symmetric geometry, the solution of the analogous problem in 1D has an integrand of the form $(\sin{(x u)}/u)\, e^{-t u^2}$.

3) The integral produces a function resembling an error function.

Anyway, any analytical evaluation of the integral that results in a faster evaluation of the result than my using numerical integration will get my appreciation.

Answer 1

We want to re-express $$F(k)=\int_0^{\infty}\frac{dx}{x}e^{-kx^2}\frac{J_0(rx)Y_0(x)-J_0(x)Y_0(rx)}{J_0^2(x)+Y_0^2(x)}.$$ Start with $$\int_0^{\infty}\frac{x}{b+x^2}\frac{J_0(rx)Y_0(x)-J_0(x)Y_0(rx)}{J_0^2(x)+Y_0^2(x)}dx=-\frac{\pi}{2}\frac{K_0(r\sqrt{b})}{K_0(\sqrt{b})}$$ which can be written as $$ \int_0^{\infty}dt e^{-bt}\int_0^{\infty}e^{-tx^2}x\frac{J_0(rx)Y_0(x)-J_0(x)Y_0(rx)}{J_0^2(x)+Y_0^2(x)}dx. $$ $$=\int_0^{\infty}dt e^{-tb}\frac{\partial}{\partial t}F(t).$$ Now, take the inverse Laplace transform with respect to $b$ to get $$\frac{\partial}{\partial t}F(t)=-\frac{\pi}{2}\int_{c-i\infty}^{c+i\infty}\frac{db}{2\pi i}e^{bt}\frac{K_0(r\sqrt{b}}{K_0(\sqrt{b})}$$ or $$F(k)=-\frac{\pi}{2}\int_{c-i\infty}^{c+i\infty}\frac{ds}{2\pi is}e^{ks}\frac{K_0(r\sqrt{s})}{K_0(\sqrt{s})}.$$ The analysis of this Inverse Laplace transfer is carried out in Carslaw and Jaeger, e.g.

Answer 2

This is not an answer, but rather a derivation of one of your results that you possibly don’t have a proof for (at least not in the forum anyway). Firstly consider the Hankel functions of the first and second kind; $H_{v}^{1}\left( z \right)={{J}_{v}}\left( z \right)+i{{Y}_{v}}\left( z \right)$ and $H_{v}^{2}\left( z \right)={{J}_{v}}\left( z \right)-i{{Y}_{v}}\left( z \right)$. Note that $H_{v}^{1}\left( z \right)$ has zeros in the lower half plane, but not in the upper half plane or on the real axis (the zeros are distributed asymptotically to $-i\log \left( 2 \right)/2n$)

Note

$$\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)}\sim \frac{{{a}^{v}}}{{{b}^{v}}} \> as \> \left| u \right|\to 0$$ and $$\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)}={{e}^{i\left( b-a \right)u}}\sqrt{\frac{a}{b}}\left( 1+O\left( {{\left| u \right|}^{-1}} \right) \right) \> as \> \left| u \right|\to \infty $$ Therefore consider a contour consisting of the real line, indented above z=0 with a small semi-circle of radius $\delta $, and a large semi-circle of radius R in the upper half plane. Therefore $$\int\limits_{C}^{{}}{\frac{H_{v}^{1}\left( bz \right)}{H_{v}^{1}\left( az \right)z}dz}=\int\limits_{R}^{{}}{{}}+\int\limits_{-\infty }^{\infty }{{}}+\int\limits_{\delta }^{{}}{{}}=0$$ and hence $$\int\limits_{-\infty }^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}=\frac{\pi i{{a}^{v}}}{{{b}^{v}}}$$ Here the integral is understood to be a principal value. Note the following properties $${{Y}_{v}}\left( z{{e}^{m\pi i}} \right)={{e}^{-vm\pi i}}{{Y}_{v}}\left( z \right)+2i\sin \left( vm\pi \right)\cot \left( v\pi \right){{J}_{v}}\left( z \right)$$ $${{J}_{v}}\left( z{{e}^{m\pi i}} \right)={{e}^{-vm\pi i}}{{J}_{v}}\left( z \right)$$ From which it is noted $$H_{v}^{1}\left( -z \right)={{J}_{v}}\left( -z \right)+i{{Y}_{v}}\left( -z \right)=-{{\left( -1 \right)}^{v}}{{J}_{v}}\left( z \right)+{{\left( -1 \right)}^{v}}\left( i{{Y}_{v}}\left( z \right) \right)={{\left( -1 \right)}^{v-1}}\left( {{J}_{v}}\left( z \right)-i{{Y}_{v}}\left( z \right) \right)={{\left( -1 \right)}^{v-1}}H_{v}^{2}\left( z \right)$$ From this we break the integral in two, i.e. $$\int\limits_{-\infty }^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}=-\int\limits_{0}^{\infty }{\frac{H_{v}^{1}\left( -bu \right)}{H_{v}^{1}\left( -au \right)u}du}+\int\limits_{0}^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}$$ Or using the results above $$\int\limits_{-\infty }^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}=\int\limits_{0}^{\infty }{\frac{H_{v}^{1}\left( bu \right)H_{v}^{2}\left( au \right)-H_{v}^{2}\left( bu \right)H_{v}^{1}\left( au \right)}{H_{v}^{1}\left( au \right)H_{v}^{2}\left( au \right)u}du}$$ Simplifying this one obtains $$\int\limits_{-\infty }^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}=2i\int\limits_{0}^{\infty }{\frac{{{J}_{v}}\left( au \right){{Y}_{v}}\left( bu \right)-{{J}_{v}}\left( bu \right){{Y}_{v}}\left( au \right)}{\left( J_{v}^{2}\left( au \right)+Y_{v}^{2}\left( au \right) \right)u}du}$$ Finally $$\int\limits_{0}^{\infty }{\frac{{{J}_{v}}\left( au \right){{Y}_{v}}\left( bu \right)-{{J}_{v}}\left( bu \right){{Y}_{v}}\left( au \right)}{\left( J_{v}^{2}\left( au \right)+Y_{v}^{2}\left( au \right) \right)u}du}=\frac{\pi {{a}^{v}}}{2{{b}^{v}}}$$

This result is listed in Bateman. As for your integral, I have several results, but nothing I am happy with as yet.