I have seen in my complex analysis class that limϵ→0∫|x|≥ϵ1−eixx2dx=π.
From here, by taking the real part, we concluded that ∞∫−∞1−cos(x)x2dx=π.
I thought that now by taking to imaginary part, one should get ∞∫−∞sin(x)x2dx=0.
However, it does seem to me that it even converges, since near 0 it looks like 1/x from both sides.
On the other hand, it is an odd function, so maybe it does exist and is 0?
You can see the same argument in the following notes: Example 2
Thanks!
Answer
Through complex Analysis you may only show that
PV∫+∞−∞sinxx2dx=0
since we are not allowed to integrate through a pole, and ∫+∞−∞sinxx2dx is not convergent in the usual sense (as an improper Riemann integral). On the other hand (1) is trivial since sinxx2 is an odd function.
No comments:
Post a Comment