Integral : $\int_0^{\infty} \frac{1-\cos(ax)}{x^2} dx. (a>0) $

Question

I have to calculate $\displaystyle\int_0^{\infty} \dfrac{1-\cos(ax)}{x^2} dx. (a>0) $

My attempt:

Let $f(z):= \dfrac{1-e^{iaz}}{z^2}.$

$C_1 : z=t, t:0 \to R$

$C_R : z=Re^{i\theta}, \theta : 0 \to 2\pi $

$C_2 : z=te^{2\pi i}, t:R \to 0$

$C_r : z=re^{i\theta}, \theta : 2\pi \to 0$

$(R>r)$

$f(z)$ is a regular analytic function, so I can use Cauchy's Integral Theorem.

$\displaystyle\int_{C_1} f(z)dz +\displaystyle\int_{C_R} f(z)dz +\displaystyle\int_{C_2} f(z)dz +\displaystyle\int_{C_r} f(z)dz =0. $

But this didn't work because $\lim_{r\to 0} \displaystyle\int_{C_r} f(z)dz \neq 0$.

How should I set $f(z)$ and Integral route?

Answer 1

First note that $$ I(a) = \int_0^{ + \infty } {\frac{{1 - \cos (ax)}}{x^2}dx} = a\int_0^{ + \infty } {\frac{{1 - \cos t}}{t^2}dt} = aI(1). $$ Now $$ I(1) = I'(a) = \int_0^{ + \infty } {\frac{{\sin (ax)}}{x}dx} = \int_0^{ + \infty } {\frac{{\sin t}}{t}dt} = \frac{\pi }{2}. $$ Hence, $I(a)=\frac{\pi }{2} a$.