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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.

https://brainmass.com/math/complex-analysis/computing-integral-sin-minus-infinity-to-infinity-634199

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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 part.

However, it's also possible to write sin(z) = [exp(iz) - exp(-iz)]/(2i) and then to split up the integral in two parts and close the contours for the two parts differently. At the point of the splitting a singularity will arise at z = 0, so this must be avoided, but that can be done at the start. So, let's try now from the start and see if we can get to a computation of the integral.

I = Integral from minus infinity to plus infinity of sin(x)/x dx

Let's now use Cauchy's theorem that states that the integral of an analytic function along a closed contour is zero, to rewrite this as:

I = Integral along C of sin(z)/z dz

where C is an open contour that from minus infinity on the real axis to plus infinity that avoids the origin by moving above it. Unlike in the traditional approach we don't need to choose a small half circle of radius epsilon and let epsilon tend to zero at the end, the deformation doesn't need to be infinitesimal. The next step is to write:

I = Integral along C of [exp(iz) - exp(-iz)]/(2 i z) dz

And there is no problem with splitting this in two parts:

I = I1 - I2

where

I1 = Integral along C of exp(i z)/(2 i z) dz

and

I2 = Integral along C of exp(-i z)/(2 i z) dz

Then to make progress, we need to close the contour in I1 in the upper half plane, and since there are no poles inside that contour, we have that I1 = 0, as the contribution from the half circle used to close the contour vanishes in the limit where the radius is taken to infinity. The contour in I2 has to be closed in the lower half plane, which makes that a clockwise contour. The residue theorem and the fact that the contribution of the half circle vanishes in the limit if the radius to infinity, then implies says that I2= -2 pi i / (2 i) = -pi.

This means that I = I1 - I2 = pi.

So, while superficially similar to the traditional approach, we see that this method has a number of advantages. At the start we do need to modify the contour from minus infinity to plus infinity in anticipation of what is to come next, but this modification is simpler than the traditional principal value. Unlike in case of the principal value there is now no limit procedure attached to it, the contour is deformed in some arbitrary way that makes it move in the upper half plane and we don't need to revisit this later. And unlike the traditional approach there is no hole that was plugged that needs to be treated separately at the end, the result is a sum of two contour integrals that are valuated in the usual straightforward way.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!