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Nov3-11, 09:52 PM
P: 13
Hello, thanks for the help a few days ago, I found a way to solve this problem and I came back to give some closure. If we use sin(x)=(e^ix-e^-ix)/2i, and multiply that by itself four times (to the power 4), we will get some positive and some negative exponentials. There is a pole at z=0. We need to consider 2 contours, one for the positive exponentials and one for the negative exponentials.

For the positive exponentials, the contour is above the real axis and goes around the pole at z=0. Since there is no residue in here, the integral is 0.

For the negative exponentials, the contour is below the real axis and has the a residue inside. Therefore, to compute the integral of sinc(x)^4 between negative infinity and infinity, we just need to find the value of the residue in the lower contour. Using the formula for a residue of a pole of order 4:

Residue=-2ipi * lim(z->0) of (1/((4-1)!))*(dł/dzł)(((z-0)^4)*(negative exponentials))

We find that the integral is equal to 2pi/3.