How Can Contour Integration Solve This Integral?

[JohnSimpson]

Homework Statement

Use contour integration to obtain the result int(sin(x)^2/x^2, x=-Inf..Inf) = Pi

Homework Equations

The Attempt at a Solution

I defined a contour that encircles the pole at z=0. It looks like a bridge over the pole. The outer integral is easily shown to be zero, and the two along the x-axis converge to the integral of interest. Every time I do the small integral around the pole, the integrand blows up as a i tend the radius to zero.

Note that the lemma typically used to evaluate integrals such as sinx/x is not valid because z=0 is not a simple pole.

Any help would be appreciated

 

[mighty2000]

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Hello! Thank you for your question. Contour integration is a powerful tool in solving complex integrals, and in this case, it can indeed be used to evaluate the given integral. Here is a possible solution:

First, let's define the contour C as a semicircle in the upper half-plane, centered at the origin with radius R. The contour C encircles the pole at z=0, as you have correctly identified. Now, let's consider the integral along this contour:

∮C sin(z)^2/z^2 dz = ∫_0^π sin(Re^(iθ))^2/(Re^(iθ))^2 iRe^(iθ) dθ

= ∫_0^π sin^2(θ)/R dθ

As R→∞, this integral goes to 0 since sin^2(θ) is bounded and 1/R goes to 0. This means that the integral along the upper half of the contour, from R to -R, is equal to the integral along the lower half, from -R to R. This is because the only difference between the two is the sign of R in the denominator, which goes to 0 as R→∞.

Now, let's focus on the integral along the real axis, from -R to R. This is the integral we want to evaluate. We can write this integral as the sum of two integrals: one from -R to 0 and another from 0 to R.

∫_-R^R sin(x)^2/x^2 dx = ∫_-R^0 sin(x)^2/x^2 dx + ∫_0^R sin(x)^2/x^2 dx

Now, let's evaluate each of these integrals separately. For the first one, we can use the substitution x=-u, and the integral becomes:

∫_R^0 sin(-u)^2/(-u)^2 (-du) = ∫_0^R sin(u)^2/u^2 du

And for the second one, we can use the substitution x=u, and the integral becomes:

∫_0^R sin(u)^2/u^2 du

Now, let's add these two together:

∫_-R^R sin(x)^2/x^2 dx = 2∫_0^R sin(x)^2/x^2 dx

Now, as