No you're all right, that's not the same problem at all, which is:
(sorry but I don't believe because that won't converge.)
You can split it into two bits: one for x positive, one for x negative. Let's do
Put and that gives you an easy integral in .
Now do the same for
and add them together.
If I'm wrong about it NOT being then it could be the answer is supposed to be zero because one half equals minus the other half ...?
e^(-x^2) is always positive
Int e^(-x^2) from -inf to pos inf is sqrt(pi)-- this is one of those well known results - the Gaussian Integral See Gaussian integral - Wikipedia, the free encyclopedia
however i read the original pblm here incorrectly --didn't see the abs|x| out front so for your problem it works out to 1