I'd like to solve the following integral:
$I = \int_0^\infty J_0(at) J_1(bt) e^{-t} dt\ $
where $J_n$ is an $n^{th}$ order Bessel Function of the First Kind and $a$ and $b$ are both positive real constants.
Any information about this integral would be useful.
Thanks!
I've included two relations specific to $0^{th}$ and $1^{st}$ order Bessel function that may be useful:
$ J_0(t)' = -J_1(t) $
$ J_1(t) = -J_{-1}(t) $
The following recursion relation may also be useful:
$ J_n(t)' = \frac{1}{2}\{J_{n-1}(t) – J_{n+1}(t) \} $
I've also found the following relation via integration by parts:
$ \int_0^\infty \{bJ_0(at) J_1(bt) + aJ_0(bt) J_1(at) \}e^{-t} dt\ =
J_0^2(0) – \int_0^\infty J_0(at) J_0(bt) e^{-t} dt $
Best Answer
The integral does not admit a closed form, I am afraid. It is discussed in the vol. 2 of Bateman's "Higher Transcendental functions", section 7.7.3, formula 15, which gives $$ 2^{\mu + \nu} \alpha^{-\mu} \beta^{-\nu} \gamma^{\lambda + \mu+\nu} \Gamma\left(\nu+1\right) \int_0^\infty J_{\mu}(\alpha t) J_\nu(\beta t) \mathrm{e}^{-\gamma t} t^{\lambda -1} \mathrm{d} t = \sum_{m=0}^\infty \frac{\Gamma\left(2m+\lambda+\mu+\nu\right)}{m! \Gamma(m+\mu+1)} \cdot {}_2F_1\left(-m,-m-\mu; \nu+1; \frac{\beta^2}{\alpha^2}\right) \left( -\frac{\alpha^2}{4 \gamma^2} \right)^m $$ In your case, $\lambda = 1$, $\mu=0$, $\nu=1$, $\gamma=1$: $$ \frac{2}{\beta} \int_0^\infty J_0(\alpha t) J_1(\beta t) \mathrm{e}^{-t} \mathrm{d}t = \sum_{m=0}^\infty \frac{\Gamma\left(2m+2\right)}{m! \Gamma(m+1)} \cdot {}_2F_1\left(-m,-m; 2; \frac{\beta^2}{\alpha^2}\right) \left( -\frac{\alpha^2}{4} \right)^m $$
By expanding the Bessel function $J_1(b t)$ in its defining series and integrating term-wise we can find other series representations: $$ \int_0^\infty J_0(a t) J_1(b t) \mathrm{e}^{-t} \mathrm{d}t = \frac{b}{2} \sum_{m=0}^\infty \binom{2m+1}{m} \frac{\left(-\frac{b^2}{4}\right)^m}{(1+a^2)^{2m+3/2}} \cdot {}_2F_1\left(-m, -m-\frac{1}{2}; 1; -a^2\right) $$ where, additionally, the Euler's transformation of the Gauss' hypergeometric function had been used.