Integral of exponential absolute functions

[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 separate 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?

 

[HallsofIvy]

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.)

 

[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?

 

[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.