How to prove that $\int_{-\infty}^{\infty} \frac{1-\cos x}{x^2} dx$ equal to $\pi $?

Question

How to prove that $\int_{-\infty}^{\infty} \frac{1-\cos x}{x^2} dx$ equal to $\pi $?

Is there any simple approach that does not require knowledge in Fourier Analysis or Complex analysis?

Answer 1

Here is an outline. By integration by parts,

$$\int_{-\infty}^\infty \frac{1-\cos x}{x^2}\, dx = \int_{-\infty}^\infty \frac{\sin x}{x}\, dx = 2\int_0^\infty \frac{\sin x}{x}\, dx$$

To evaluate $\int_{0}^\infty \frac{\sin x}{x}\, dx$, consider the integral function

$$f(y) := \int_{0}^\infty e^{-xy}\frac{\sin x}{x}\, dx \quad (y > 0)$$

We have $f'(y) = \int_{0}^\infty -e^{-xy}\sin x\, dx = -\frac{1}{1 + y^2}$, and hence $f(y) = -\tan^{-1}(y) + C$, where $C$ is a constant. Since $\lim\limits_{y\to \infty} f(y) = 0$, we deduce $C = \pi/2$. The answer we seek is $2\,f(0) = \pi$.

To complete this argument, justify the differentiation under the integral sign and show that $\lim\limits_{y\to \infty} f(y) = 0$.

Answer 2

Assume $t>0$. One may observe that $$ \int_0^\infty te^{-xt}(1-\cos x)\:dx=\text{Re} \int_0^\infty te^{-xt}(1-e^{ix})\:dx=\frac1{1+t^2} $$ giving $$ \int_0^\infty\!\! \int_0^\infty te^{-xt}(1-\cos x)\:dx\:dt=\int_0^\infty \frac{dt}{1+t^2}=\left. \arctan t\frac{}{}\right|_0^\infty=\frac{\pi}2. \tag1 $$ On the other hand, since $$ \int_0^\infty te^{-xt}\:dt=\frac1{x^2},\qquad x>0, $$ one has $$ \int_0^\infty\!\! \int_0^\infty te^{-xt}(1-\cos x)\:dt\:dx= \int_0^\infty \frac{1-\cos x}{x^2}\:dx. \tag2 $$ Then by using Fubini's theorem, justified here since the integrands are positive and corresponding integrals are finite, from $(1)$ and $(2)$ one gets $$ \int_0^\infty \frac{1-\cos x}{x^2}\:dx=\frac{\pi}2 $$ or the integrand being even here,

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

as announced.

Answer 3

Note that $$I=\int_{-\infty}^{\infty}\frac{1-\cos\left(x\right)}{x^{2}}dx=2\int_{0}^{\infty}\frac{1-\cos\left(x\right)}{x^{2}}dx $$ $$\stackrel{IBP}{=}2\int_{0}^{\infty}\frac{\sin\left(x\right)}{x}dx=\frac{1}{i}\int_{0}^{\infty}\frac{e^{ix}-e^{-ix}}{x}dx=\frac{1}{i}\lim_{\epsilon\rightarrow0^{+}}\int_{0}^{\infty}\frac{e^{\epsilon+ix}-e^{\epsilon-ix}}{x}dx $$ and now using the Frullani's theorem with complex parameters we get $$I=\frac{1}{i}\lim_{\epsilon\rightarrow0^{+}}\log\left(\frac{\epsilon+i}{\epsilon-i}\right)=\color{red}{\pi}$$ as wanted.