How to compute $\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx$?

Question

I want to compute $$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx$$

My approach is $$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx=\int_{-\infty}^{\infty} \frac{1}{(x -\pi)^2}dx+\int_{-\infty}^{\infty} \frac{\cos(x)}{(x -\pi)^2}dx.$$

However, the first term on the RHS does not exist since there is a singularity $\pi$.

How to overcome this problem?

Now follow the hint, I get $$\int_{-\infty}^{\infty} \frac{\sin x}{x}dx$$

BTW, if I want to use residue calculus to do this problem, how to do it? I am confused the pole at origin. Thanks!

Answer 1

HINT: With the substitution $x\to x+\pi$, we get

$$\int_{-\infty}^\infty \frac{1-\cos x}{x^2}dx$$

then integrate by parts with $u=1-\cos x$ and $dv=dx/x^2$. You should end up with a familiar integral...

Answer 2

I used @Franklin Pezutti Dyer's hint:

$-\int_{-\infty}^{+\infty}(1-\cos{x})d\left(\frac{1}{x}\right)=-\int_{-\infty}^{\infty}\frac{\sin{x}}{x}dx=-\pi$

Answer 3

$u=x-\pi,du=dx$ $$\int_{-\infty}^\infty\frac{1+\cos(x)}{(x-\pi)^2}dx=\int_{-\infty}^\infty\frac{1+\cos(u+\pi)}{u^2}du=\int_{-\infty}^\infty\frac{1-\cos(u)}{u^2}du$$ and now if we look at this for $u\to\infty$: $$\frac{1-\cos(u)}{u^2}\approx\frac 1{u^2}$$ and we can also calculate that: $$\lim_{u\to 0}\frac{1-\cos(u)}{u^2}=\frac12$$ so we can determine that the function is finite for the domain its in and is convergent due to the comparison test.

You can also use the fact that: $$1-\cos(x)=2\sin^2\left(\frac x2\right)$$