I am supposed to evaluate the following integral:
$$\int_{-\infty}^{\infty}\frac{\sin(5x)}{1+(x-\frac{\pi}{2})^2} dx$$
So does this guy have a pole at $\frac{\pi}{2}$? I wanna try and use the Residue theorem to solve it but my complex analysis is a bit rusty. Any hints much appreciated!
Best Answer
Whenever you want to use the residue theorem and you have trig functions, you'll generally want to be a bit clever and replace the trig functions with exponentials.
In this case, you're looking for the imaginary part of $\int\limits_{-\infty}^\infty \frac{e^{5ix}}{1 + (x - \pi / 2)^2} dx$.
We can consider the contour of a large semi-circle in the upper half-plane.
To determine where the function isn't holomorphic, we set its denominator $1 + (x - \pi/2)^2$ to zero. The function thus has simple poles at $\frac{\pi}{2} \pm i$. The only relevant pole here is at $\frac{\pi}{2} + i$, since the other pole is in the lower half-plane.
Computing the residue at this point gives us $\frac{e^{5i (\pi/2 + i)}}{2i} = \frac{i \cdot e^{-5}}{2i} = \frac{1}{2e^5}$.
So the integral with the complex exponential will be $2 \pi i \frac{1}{2e^5} = i \frac{\pi}{e^5}$. The imaginary part of this integral is $\frac{\pi}{e^5}$, which is the original integral.