Evaluating the integral: $\int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}~dx$ I am interested in evaluating the following integral:
$$
\int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}~dx
$$
Using Matlab, Numerically it seems that the integral is convergent, 
but I'm not sure about it. How can we prove that the integral is
convergent or not? 
Many Thanks in advance.
 A: The integral converges. There is potential trouble at $0$ and "at" infinity. So we split the integral into (i) the part from $0$ to $1$, and (ii) the part from $1$ to $\infty$.
(i) Look at the power series expansion of $2-2\cos x-x\sin x$. The first few terms are $2-2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}\right) -x\left(x-\frac{x^3}{31}\right)$.  There is cancellation, and the first non-zero term is the $x^4$ term. So our function behaves well as $x$ approaches $0$ from the right: It approaches $\frac{1}{12}$. If we define your integrand to be $\frac{1}{12}$, the resulting function is continuous on $[0,1]$, and hence integrable. 
(ii) For large $x$, the integrand is $\lt \frac{K}{x^3}$ for some constant $K$, and we know that $\int_1^\infty \frac{dx}{x^3}$ converges.  
A: let $I=\int_{0}^{\infty}\dfrac{2-2\cos{x}-x\sin{x}}{x^4}dx$
then using integration by parts we have
$$I=-\dfrac{1}{3}\dfrac{2-2\cos{x}-x\sin{x}}{x^3}|_{0}^{\infty}+\dfrac{1}{3}\int_{0}^{\infty}\dfrac{\sin{x}-x\cos{x}}{x^3}dx$$
then $$I=\dfrac{1}{3}\int_{0}^{\infty}\dfrac{\sin{x}-x\cos{x}}{x^3}dx$$
and application of integration by parts yields
$$I=-\dfrac{1}{6}\lim_{x\to 0^{+}}\dfrac{\sin{x}-x\cos{x}}{x^2}+\dfrac{1}{6}\int_{0}^{\infty}\dfrac{x\sin{x}}{x^2}dx=\dfrac{\pi}{12}$$
