# Convergence of an improper integral $I=\int_0^\infty \frac{\sin{x}-x\cos{x}}{x^\alpha}dx$

Problem: Analyze convergence of an improper integral $I=\int_0^\infty \frac{\sin{x}-x\cos{x}}{x^\alpha}dx$.

My work: Problematic points are $0$ and $\infty$. Therefore, we will write integral as a sum of two integrals : $$I=\int_0^1 \frac{\sin{x}-x\cos{x}}{x^\alpha}dx+\int_1^\infty \frac{\sin{x}-x\cos{x}}{x^\alpha}dx=I_1+I_2.$$ Second integral converges absolutely for $\alpha>3$. First integral converges for $\alpha<2$ by comparison test ($I_1\leq \int_0^1 \frac{dx}{x^{\alpha-1}}$).

Now I'm stuck at this point of problem. Any hint or help is welcome. Thanks in advance.

We have that $\sin(x)-x\cos(x)$ behaves like $x^3$ in a right neighbourhood of the origin, hence integrability over there is ensured by $\color{red}{\alpha < 4}$. By Dirichlet's test, $$\int_{1}^{+\infty}\frac{\sin x}{x^\beta}\,dx,\qquad \int_{1}^{+\infty}\frac{\cos x}{x^\beta}\,dx$$ are convergent as soon as $\color{red}{\beta>0}$, hence the original integral is convergent as soon as $\color{red}{\alpha\in(1,4)}$.

If you are allowed to use Laplace transforms, you may also compute the value of the integral, since $$\mathcal{L}\left(\sin x-x\cos x\right) = \frac{2}{(1+s^2)^2}, \tag{1}$$ $$\mathcal{L}^{-1}\!\!\left(x^{-\alpha}\right) = \frac{s^{\alpha-1}}{\Gamma(\alpha)} \tag{2}$$ and the equivalent integral $$I(\alpha)=\frac{1}{\Gamma(\alpha)}\int_{0}^{+\infty}\frac{s^{\alpha-1}}{(1+s^2)^2}\,ds \tag{3}$$ is convergent iff ${\alpha >0}$ (that is needed to grant integrability in a right neighbourhood of the origin) and ${\alpha < 4}$ (that is needed to grant integrability in a left neighbourhood of $+\infty$). In such a case, through the substitution $\frac{1}{1+s^2}=u$, Euler's beta function and the $\Gamma$ reflection formula we have: $$I(\alpha)= \color{red}{\frac{\pi\,(2-\alpha)}{2\,\Gamma(\alpha)\,\sin\left(\frac{\pi\alpha}{2}\right)}}.\tag{4}$$

• I like it +1 :) – Behrouz Maleki Jun 29 '16 at 14:29
• @FelixMarin: no, you're right, I missed the factor $2$ in $(1)$ before performing my computations. Now fixed, thank you. – Jack D'Aurizio Jun 29 '16 at 21:32

May be useful \begin{align} & \int_{0}^{\infty }{\frac{\sin x}{{{x}^{\alpha }}}}dx=\frac{\pi }{2\Gamma (\alpha )\sin \left( \frac{\alpha \pi }{2} \right)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,\,0<\alpha <2 \\ \\ & \int_{0}^{\infty }{\frac{\cos x}{{{x}^{\alpha -1}}}}dx=\frac{\pi }{2\Gamma (\alpha-1 )\sin \left( \frac{\alpha\pi }{2} \right)}\,\,\,\,,\,\,\,\,1<\alpha <2 \\ \end{align}

• How to obtain these formulae? – alans Jun 29 '16 at 14:20
• @alans: Did you notice the Laplace transform in the other answer? – Jack D'Aurizio Jun 29 '16 at 14:21
• Double integration reversal – Behrouz Maleki Jun 29 '16 at 14:21