Residue Theorem with real zero

1. Jul 11, 2016

DCN

1. The problem statement, all variables and given/known data

Find $$\int_{0}^{\infty} \frac{\cos(\pi x)}{1-4x^2} dx$$

2. Relevant equations

The residue theorem

3. The attempt at a solution

The residue of this function at $$x=\pm\frac{1}{2}$$ is zero. Therefore shouldn't the integral be zero, if you take a closed path as a hemisphere in the upper half of the complex plane? Yet the integral evaluates to $$\pi/4$$

I am completely lost.

Last edited: Jul 11, 2016
2. Jul 11, 2016

Ray Vickson

You cannot complete the integral along a semicircle in the upper half-plane because the function $\cos(\pi x)$ grows exponentially large when you let the imaginary part of $x$ go to infinity. However, you can instead write the integral as
$$I = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos(\pi x)}{1-4x^2} \, dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{e^{i \pi x}}{1-4x^2} \, dx$$
because the imaginary part of the second integrand is an odd function, which integrates to zero.

That last form can be completed in the upper half-plane, and when you do that the residues do not vanish.

Last edited: Jul 12, 2016
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