Thursday, June 22, 2017

integration - intlimitsinftyinftyfracsin(x)x2




I have seen in my complex analysis class that limϵ0|x|ϵ1eixx2dx=π.



From here, by taking the real part, we concluded that 1cos(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.


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