Residue Theorem with real zero

[DCN]

Homework Statement

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

Homework Equations

The residue theorem

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.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

The residue theorem

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.

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.