How to evaluate $\int \frac{\cos(x) - 1}{x^2}\mathrm dx$?

would like a hint with the integral $$\int \frac{\cos(x) - 1}{x^2}\mathrm dx$$Thanks

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Do you mean $\frac{\cos(x)-1}{x^2}$ or ${\cos(x)}-\frac{1}{x^2}$ ? – lhf Apr 27 '11 at 16:19
I will assume that you're actually considering the integral $\int_{0}^{\infty} \frac{\cos x - 1}{x^2} \, dx$. If you are cognizant of $\int_{0}^{\infty} \frac{\sin x}{x} \, dx = \frac{\pi}{2}$, you may simply apply integration by parts. – Sangchul Lee Apr 27 '11 at 16:23
If you meant the first of lhf's choices, you'll require the (nonelementary) sine integral. – J. M. Apr 27 '11 at 16:25
If you mean $\frac{\cos x - 1}{x^2}$, Wolfram Alpha returns something containing the sine integral, so it's probably not solvable with more elementary functions. – joriki Apr 27 '11 at 16:25
sos440's hint of using integration by parts also applies if you're after the indefinite integral; your expression will then involve $\int\frac{\sin\;x}{x}\mathrm dx$, and this is where the sine integral comes in. – J. M. Apr 27 '11 at 16:34

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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