# Integral with Bessel functions of the First Kind.

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.

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:

$\int_0^\infty \{bJ_0(at) J_1(bt) + aJ_0(bt) J_1(at) \}e^{-t} dt\ = J_0^2(0) - \int_0^\infty J_0(at) J_0(bt) e^{-t} dt$

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The formula of the series form of the product of two Bessel functions of the First Kind in people.math.sfu.ca/~cbm/aands/page_360.htm may be useful. – Harry Peter Aug 12 '13 at 15:01

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.

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Thanks very much, both to you and Bateman. :-) – bclifford Apr 11 '13 at 1:21

Unfortunately, I do not believe there is a simple closed form for this integral. One way forward, if you can call it that, is to express the integral as a series. For example, note that the Laplace Transform of $f(t) = J_0(a t)$ is

$$\hat{f}(s) = (s^2+a^2)^{-1/2}$$

Also note that Laplace transforms turn multiplication into differentiation as follows:

$$\int_0^{\infty} dt \: t^m \, f(t) e^{-s t} = (-1)^m \frac{\partial^m}{\partial s^m} \hat{f}(s)$$

Then using the Maclurin series for $J_1(b t)$:

$$J_1(b t) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!\,(k+1)!} \left ( \frac{b t}{2}\right )^{2 k+1}$$

we get the following expression for the integral:

$$\int_0^{\infty} dt \: J_0(a t) J_1(b t) e^{-t} = -\sum_{k=0}^{\infty} \frac{(-1)^k}{k!\,(k+1)!} \left ( \frac{b}{2}\right )^{2 k+1} \left [\frac{\partial^{2 k+1}}{\partial s^{2 k+1}} (s^2+a^2)^{-1/2} \right]_{s=1}$$

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Thanks, that's clever. – bclifford Apr 11 '13 at 1:19

$\int_0^\infty J_0(at)J_1(bt)e^{-t}~dt$

$=\int_0^\infty\left(\sum\limits_{m=0}^\infty\dfrac{(-1)^m\left(\dfrac{at}{2}\right)^{2m}}{(m!)^2}\right)\left(\sum\limits_{n=0}^\infty\dfrac{(-1)^n\left(\dfrac{bt}{2}\right)^{2n+1}}{n!(n+1)!}\right)e^{-t}~dt$

$=\int_0^\infty\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+n}a^{2m}b^{2n+1}t^{2m+2n+1}e^{-t}}{2^{2m+2n+1}(m!)^2n!(n+1)!}dt$

$=\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+n}a^{2m}b^{2n+1}(2m+2n+1)!}{2^{2m+2n+1}(m!)^2n!(n+1)!}$

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