2
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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: $$ I(\alpha)= \color{red}{\frac{\pi\,(2-\alpha)}{2\,\Gamma(\alpha)\,\sin\left(\frac{\pi\alpha}{2}\right)}}.\tag{4}$$

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

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}

$\endgroup$
  • $\begingroup$ How to obtain these formulae? $\endgroup$ – alans Jun 29 '16 at 14:20
  • 1
    $\begingroup$ @alans: Did you notice the Laplace transform in the other answer? $\endgroup$ – Jack D'Aurizio Jun 29 '16 at 14:21
  • $\begingroup$ Double integration reversal $\endgroup$ – Behrouz Maleki Jun 29 '16 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.