Integral of exponential absolute functions

1. Sep 20, 2010

Wishe Deom

Hello,

I am having difficulty solving the following integral:

$$\int^{\infty}_{-\infty}e{-(a|x|+ikx)}dx$$

I have tried to use an explicit form of the absolute, eg.

$$-(a|x|+ikx) = \left\{\stackrel{-(ik+a)x\ x>0} {-(ik-a)\ x<0}$$

Does this allow me to seperate the integral into a sum of two integrals?

$$\int^{0}_{-\infty}e{-(ik+a)x}dx+\int^{\infty}_{0}e{-(ik+a)x}dx$$

This was my best guess, but the result I got did not converge, so either I did the integral improperly, or else this is not a legal method.

Would someone be so kind as to share their knowledge?

Last edited: Sep 20, 2010
2. Sep 20, 2010

HallsofIvy

Staff Emeritus
Yes, you can but I think a better reduction is
$$e^{-a|x|+ ikx}dx= e^{-a|x|} e^{ikx}= e^{-a|x|}(cos(kx)+ i sin(kx))$$
Then make use of the symmetries involved.
(I am assuming you mean that to be an exponent. You missed the "^" in your LaTex.)

3. Sep 20, 2010

Wishe Deom

Thank you.

So, by symmetry, the sin term dissappears, and the rest of the integral can be taken as twice the integral from zero to infinity?

4. Sep 20, 2010

mathman

Yes. In your original separation into two integrals, the exponent in the first term is wrong. It should be (a-ik)x. You had it right on the previous line.