$\int_{0}^{\infty} \frac{J_1 (bx)}{x+a} = \pi \frac{0}{0}$?

Question

I need to solve

$I=\int_{0}^{\infty} \frac{J_1 (bx)}{x+a}$.

According to "Table of integrals, series, and products"(page 678) by I.S.Gradshteĭn & I.M.Ryzhik,

$\int_{0}^{\infty} \frac{J_{\nu}(bx)}{x+a}= \frac{\pi}{sin(\pi \nu)}(\mathbf{J}_{\nu}(ab)-J_{\nu}(ab)), \;\;\; [b>0, \;\; |arg(a)|< \pi, \;\; Re(\nu)> -1]$,

where $\mathbf{J}_{\nu}(x)$ is the Anger function and $J_{\nu} (x)$ a Bessel function of the first kind. Now, for $\nu$ integer, both the Anger and the Bessel function are the same, in which case, I am left with:

$\int_{0}^{\infty} \frac{J_1 (bx)}{x+a} = \pi \frac{0}{0}$

Does anybody know an alternative way to get the solution for $I$ ? Is there even a solution?

Thank you.

Answer 1

Use $$ \frac{1}{z}=\int_0^\infty ds\ e^{-s z} $$ to write your integral as $$ I=\int_0^\infty ds\ e^{-a s}\int_0^\infty dx\ J_1(bx)e^{-s x}= \int_0^\infty ds\ e^{-a s}\left[\frac{1}{b}-\frac{1}{b \sqrt{\frac{b^2}{s^2}+1}}\right] $$ $$ =\frac{1}{ab}-\frac{1}{b}b\int_0^\infty ds\ e^{-ab s}\frac{s}{\sqrt{1+s^2}}=\boxed{\frac{1}{ab}+\frac{1}{2} \pi H_{-1}(a b)+\frac{1}{2} \pi Y_1(a b)}\ , $$ where $H$ is a Struve function and $Y$ a Bessel function (for $a,b>0$).