A Little Problem with a Simple Integral

[Alamino]

If we have

P(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2} x^2},

then

I=\int^{+\infty}_{-\infty}dx \, P(x) \theta(x)<br /> = \int^{+\infty}_0 dx \, P(x)=\frac{1}{2},

where \theta is the step function. Now, using its integral representation, we have

<br /> I=\int^{+\infty}_{-\infty}dx \, P(x) \frac{1}{2\pi i} \int^{+\infty}_{-\infty} \frac{dk}{k} \, e^{ikx} = \frac{1}{2\pi i} \int^{+\infty}_{-\infty} \frac{dk}{k} \int^{+\infty}_{-\infty}dx \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2+ikx}= \frac{1}{2\pi i}\int^{+\infty}_{-\infty} \frac{dk}{k} e^{-\frac{1}{2} k^2} .<br />

But the integral over k is calculated using residues and gives 2\pi i times the residue at k=0, what gives simply 2\pi i for the integral over k and the wrong result 1 for the integral I. I cannot see the catching. What is wrong with this calculation?

 

[StatusX]

Could you explain how you got from the second to last to the last expression? Also, I don't think you can close that last integral, because k is squared. So it will only go to 0 as R goes to infinity if -(x^2-y^2)<0, or |x|>|y|, where k=x+iy, and this is not true over the entire contour.

 

[Alamino]

The passage you're referring to is a gaussian integration. I'm not so sure (maybe all the problem is here), but I think that I can consider the integrated function as e^{-|k|^2/2}[\tex], what gives k squared for the reals and has not the problem you mentioned.

 

[StatusX]

I don't think you can do that. |k|^2 does not equal k^2 for complex numbers. In particular, the first is real while the second isn't.