Will integral be $\frac{\pi}{2}$?

Question

Show that $\int \frac{1-cosx}{x^2}\ dx=\frac{\pi}{2}$.

I used Taylor's series for cosx to find integral but I don't see intergal becoming equal to $\frac{\pi}{2}$ without any limits of integration.

I have written the question as it is given. But I think question is wrong as integral can't be equal to a constant value.

left hand side is a funtion deperndin on $x$ and right hand side is a constant. ?? — mesel, Dec 18 '15 at 22:24
Yes, $\int_0^{\infty} \frac{1 - \cos x}{x^2},dx = \frac{\pi}{2}$. — Daniel Fischer, Dec 18 '15 at 22:34
http://www.wolframalpha.com/input/?i=%E2%88%AB_0%5E%7B%5Cinfty%7D+%5Cfrac%7B1-%5Ccos+x%7D%7Bx%5E2%7D%5C%2Cdx — janmarqz, Dec 18 '15 at 22:35
That depends on what you can work with. I guess complex analysis is not something you can use? — Daniel Fischer, Dec 18 '15 at 22:39
@DanielFischer No I can't use complex analysis. Is there any simpler way? — Mathematics, Dec 18 '15 at 22:40
An integration by parts to get $;\displaystyle\int_0^\infty \frac {\sin(x)}x,dx$ for example... — Raymond Manzoni, Dec 18 '15 at 22:41
Well, for me, complex analysis (the residue theorem) is the simplest way ;) But of course one can do it with real methods too. What can you use? Do you know that $\int_0^{\infty} \frac{\sin x}{x},dx = \frac{\pi}{2}$? Can you use Fourier theory? — Daniel Fischer, Dec 18 '15 at 22:43
@DanielFischer Of course, invoking Parseval's Theorem comes in handy here as the integral of the square of the rectangular function (hmm, square of a rectangle) is trivial. — Mark Viola, Dec 18 '15 at 23:59

score 4 · Accepted Answer · answered Dec 18 '15 at 23:02

You can do, as Raymond Manzoni said, an integration by parts and write $$I=\int_0^{+\infty}\frac{1-\cos x}{x^2}\mathrm{d}x=\underset{A}{\underbrace{\Big[-\frac{1-\cos x}{x}\Big]_0^{+\infty}}}+\underset{B}{\underbrace{\int_0^{+\infty}\frac{\sin x}{x}\mathrm{d}x}}.$$

For the first term $A$, there is no problem to find the limit in $+\infty,$ and in $0$ write Taylor-Young development : $\cos x=1-\frac{x^2}{2}+o(x^2)$ to get finally $A=0.$
For the second term $B,$ you can use Daniel Fischer's indication that $\int_0^{+\infty}\frac{\sin x}{x}\mathrm{d}x=\frac{\pi}{2}$ to get your result. This last equality comes from Fourier transformation and Fourier inverse formula, by seeing that $\widehat{\mathbb{1}_{[-1,1]}}(\xi)=\frac{2\sin\xi}{\xi}$ if $\xi\neq 0$ and $0$ if $\xi=0,$ then $$I=\underset{\text{integral on $\mathbb{R^+}$}}{\underbrace{\frac{1}{2}}}\cdot\underset{\text{to compense the $2$}}{\underbrace{\frac{1}{2}}}\cdot\widehat{\widehat{\mathbb{1}_{[-1,1]}}}(0)\underset{\text{inverse formula}}{\underbrace{=}}\frac{\pi}{2}.$$ Other ways can be found on this Wikipedia page.

Mark Viola · Answer 2 · 2015-12-19T00:07:42.917

Another approach is to use Parseval's Theorem. First we write

$$\frac{1-\cos x}{x^2}=\frac{2\sin^2(x/2)}{x^2}=2\left(\frac{\sin (x/2)}{x}\right)^2$$

Then, we have

$$\int_0^\infty \frac{1-\cos x}{x^2}\,dx=\int_{-\infty}^\infty\left(\frac{\sin (x/2)}{x}\right)^2\,dx$$

Recall that the Fourier Transform of $\frac{\sin (x/2)}{x}$ is $\int_{-\infty}^\infty \frac{\sin(x/2)}{x}\,e^{ikx}\,dx= \pi\text{rect}(x)$, where $\text{rect}(k)$ is the Rectangular Function. Then, Parseval's Theorem yields

$$\int_{-\infty}^\infty\left(\frac{\sin (x/2)}{x}\right)^2\,dx=\frac1{2\pi}\int_{-\infty}^\infty \left(\pi \text{rect}(k)\right)^2\,dk=\frac{\pi}2$$

Will integral be $\frac{\pi}{2}$?

2 Answers2