Hello salla2,
I have rewritten your problem as:
I=d e^{-c} \cdot \int_{0}^{\infty}e^{-x}J_1(ax)J_1(bx)dx
Leaving out the factor before the integral it comes down to:
I=\int_{0}^{\infty}e^{-x}J_1(ax)J_1(bx)dx
This is not easy. I tried several ways.
*) The online mathematica integrator: NOK
*) xmaxima: NOK
*) The following integral book:
"Integraltafel, zweiter teil bestimmte integrale" by grobner - hofreiter
gives on page 203 the following formula:
\int_{0}^{\infty}e^{-ax}J_{v}(bx)J_{v}(cx)dx=\frac{1}{\pi\sqrt{bc}}Q_{v-\frac{1}{2}}<br />
\left( \frac{a^2+b^2+c^2}{2bc}\right)
Q_{v-\frac{1}{2}}(z)=\frac{\sqrt{2\pi}\Gamma\left(v+\frac{1}{2}\right)}<br />
{2^{v+1}\Gamma\left(v+1\right)} \cdot z^{-v-\frac{1}{2}}<br />
F\left(\frac{2v+1}{4},\frac{2v+3}{4},v+1;\frac{1}{z^2}\right)
Which is a "Legendresche Function" F is the hypergeometric series.
*) The following book:
"Table of Laplace Transforms" by Roberts and Kaufman
gives on page 57, the following function for the Laplace transform of J_{v}(at)J_{v}(bt)
which can be used considering the definition of the Laplace transform, s is then afterwards set to 1.
\frac{1}{\pi\sqrt{ab}}Q_{v-\frac{1}{2}}<br />
\left( \frac{s^2+a^2+b^2}{2ab}\right)
The definition of Q_{v-\frac{1}{2}} was given as:
Q_v^{\mu}(x)=\frac{e^{\mu \pi i}\pi^{1/2}\Gamma\left(\mu+v+1\right)(x^2-1)^{\mu/2}}<br />
{2^{v+1}\Gamma\left(v+\frac{3}{2}\right)x^{\mu +v+1}} \cdot {\left.}_2F_1{\right.}<br />
\left[\frac{\mu+v+1}{2},\frac{\mu+v+2}{2};v+\frac{3}{2};\frac{1}{x^2}\right]
In which \mu=0
I can't help you any further on this, hope it helps a bit. It is a difficult question...
coomast
