loloPF said:
Tom Mattson: No this is not a high school problem, any recommendation on where to post it?
If it's a homework assignment, then it is most likely at the College Level. (In the USA, College=University).
I had another attempt that didn't work, but maybe you can play with it to get something. The trick is called "parameterizing the integral".
I started by generalizing your function to include a second variable.
F(\gamma,a)=\int_{-\infty}^{\infty} e^{-\sqrt{ax^2+\gamma^2}}dx
Then I differentiated with respect to a:
\frac{\partial F(\gamma,a)}{\partial a}=\int_{-\infty}^{\infty}\frac{\partial}{\partial a} e^{-\sqrt{ax^2+\gamma^2}}dx
\frac{\partial F(\gamma,a)}{\partial a}=\int_{-\infty}^{\infty}\frac{-a}{\sqrt{ax^2+\gamma^2}} e^{-\sqrt{ax^2+\gamma^2}}dx
Then I tried to do a u-substitution:
u=\sqrt{ax^2+\gamma^2}
du=\frac{ax}{\sqrt{ax^2+\gamma^2}}
The idea is to integrate the new integral with respect to x, then integrate
that with respect to a, and then recognize that I(\gamma)=F(\gamma,1).
Of course, I couldn't get it to work because of that x in my du. You can solve for x in terms of u, but it gets ugly again. But maybe you can find a parameterization that works.
And then again, there's always numerical integration.
shame on me!) but nothing good so far.
"... nevermind.
