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


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