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Computing the integral of sin(x)/x, minus infinity to infinity

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The standard textbook way to compute the integral of sin(x)/x from minus infinity to plus infinity is to replace this integral by the principal value which then allows one to replace sin(x) by Im[exp(ix)], and we can take the imaginary part operator out of the integral. One then proceeds by completing the contour in the upper half plane and by plugging the hole that arises when taking the principal part using a small half circle in either the upper or lower half plane that makes the contour avoid the singularity at the origin. The contribution from the half circle must then be subtracted from the entire contour integral, and taking the appropriate limits will then yield the final result.

One problem with this approach is the complexity of attacking the problem this way. Many of the steps taken here are not obvious to beginning students, they are needed to avoid problems later due to choices one needs to make later, and when these avoidance steps come first this makes the entire approach quite dense. One can then ask if there exists a more direct, straightforward, way to compute the integral using contour integration.

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Solution Summary

A straightforward way to compute the integral of sin/x from minus infinity to infinity using contour integration is worked out in detail.

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To get to a more straightforward approach to compute the integral of sin(x)/x from minus to plus infinity, let's try to deconstruct the traditional method outlined in the problem description. The reason why we immediately take the step to replace the integral by its principal part is because we want to replace Sin(x) by exp(ix). And the rationale for that replacement is because we need to close the contour and you can either chose the upper or lower half plane, the former choice makes it necessary to have exp(iz) in the integrand to make sure the integral along the half circle vanishes in the limit of the radius to infinity, while the latter choice would make it necessary to have exp(-i z) in the integrand. Both choices are possible but we need to make a choice and whatever we choose will lead to a singularity at z = 0 unless we take the principal ...

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