TITLE: How to evaluate $\int \frac{\cos(x) - 1}{x^2}\mathrm dx$?
QUESTION [3 upvotes]: would like a hint with the integral $$\int \frac{\cos(x) - 1}{x^2}\mathrm dx$$Thanks

REPLY [4 votes]: I was perusing some past unanswereds, and this came up. So let's give an answer.
So we look to solve $\displaystyle \int \frac{\cos(x) - 1}{x^2}\mathrm dx$.
Then we will integrate by parts, as suggested in the comments. Let $u = \cos (x) - 1, \mathrm dv = 1/{x^2}$. So $\int u \mathrm dv = uv - \int v \mathrm du \to$ 
$$ \int \frac{ \cos (x) - 1}{x^2} \mathrm dx = - \frac{\cos (x) - 1}{x} - \int \frac{\sin (x)}{ x} \mathrm dx$$   
The integral $\int \sin(x) / x \mathrm dx$ is commonly denoted as the sine integral, $si(x)$
This integral looks to be very interesting if we evaluate it from $0$ to $\infty$, and this demands that we evaluate the limit as $x \to 0$ of $\frac{\cos (x) - 1}{x}$. Of course, this is trivial, and we get $- \pi /2$ overall. This leads to the interesting statement that
$$ \int_0^\infty \frac {\sin (x)}{x} \mathrm dx + \int_0^\infty \frac {\cos (x) - 1}{x^2} \mathrm dx = 0$$
This interesting line is why I ended up writing this answer.
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