# A Little Problem with a Simple Integral

1. Dec 8, 2004

### Alamino

If we have

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

then

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

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

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

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?

2. Dec 8, 2004

### 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.

Last edited: Dec 8, 2004
3. Dec 9, 2004

### 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 [tex]e^{-|k|^2/2}[\tex], what gives k squared for the reals and has not the problem you mentioned.

4. Dec 9, 2004

### 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.