Investigate the absolute and conditional convergence of the integral $$\int_0^{+\infty}f(x)\mathrm{d}x = \int_0^{+\infty}\left(x + \frac{1}{x}\right)^{\alpha}\sin (x^3) \mathrm{d}x$$

for all values of $\alpha$.

As we know, the integral $\int_0^{+\infty}f(x)\mathrm{d}x$ convergence absolutely iff $\int_0^{+\infty}f(x)\mathrm{d}x$ converges and $\int_0^{+\infty}|f(x)|\mathrm{d}x$ converges.

1) Convergence of $\int_0^{+\infty}f(x)\mathrm{d}x$

I wrote

$$\int_0^{+\infty}f(x)\mathrm{d}x = \int_0^{1}f(x)\mathrm{d}x + \int_1^{+\infty}f(x)\mathrm{d}x$$

The first integral on the right converges for $\alpha <4$ and the second one for $\alpha <2$, so the original integral converges for $\alpha < 2$

2) Convergence of $\int_0^{+\infty}|f(x)|\mathrm{d}x$.

This is the part I'm having trouble with. This integral is

$$\int_0^{+\infty}|f(x)|\mathrm{d}x = \int_0^{+\infty}\left|\left(x + \frac{1}{x}\right)^{\alpha}\sin (x^3)\right| \mathrm{d}x$$

but I don't know how to deal with the integrand with absolute value. I would like to split the integral as I did above but whatever I do I get that the integral converges absolutely for $\alpha < 2$ (I get the same inequality as above), but the answer giving by my textbook is that:

  • For $\alpha< -1$ it converges absolutely
  • For $-1 \leq \alpha < 2$ converges conditionally

I was also thinking about using the inequality $\left|\int_a^bf(x)\mathrm{d}x\right| \leq \int_a^b|f(x)|\mathrm{d}x$, but then I realised that I need an integral bounding the integral I want to prove the convergence of and not the other way around.

  • $\begingroup$ is this $$\sin(x^3)$$ or $$\sin(x)^3$$? $\endgroup$ May 28, 2017 at 14:56
  • $\begingroup$ @Dr.SonnhardGraubner is $\sin(x^3)$ $\endgroup$
    – Joqsan
    May 28, 2017 at 15:00

1 Answer 1


Your proof of conditional convergence is correct: the integral converges for $\alpha<2$

Let's begin with a change of variable. Let $z=x^3$, or $x=z^{1/3}$: $$ \int _0^{\infty}\left(x+x^{-1}\right)^{\alpha}|\sin(x^3)|dx $$ $$ \Longrightarrow\frac{1}{3}\int_0^{\infty}\left(z^{1/3}+z^{-1/3}\right)^{\alpha}z^{-2/3}\cdot|\sin(z)|dz $$ $$ =\frac{1}{3}\int_0^{\infty}\left(1+z^{-2/3}\right)^{\alpha}z^{(\alpha-2)/3}\cdot|\sin(z)|dz $$Near $z=0^+$, the integrand is $O(z^{(1-\alpha)/3})$, in the sense that $$ \lim_{z\to0^+} \frac{\left(1+z^{-2/3}\right)^{\alpha}z^{(\alpha-2)/3}\cdot|\sin(z)|}{z^{(1-\alpha)/3}}=1 $$Since we are imposing $\alpha<2$, we have integrability near $z=0$, say on $[0,\pi]$. So it suffices to integrate over $[\pi,\infty)$. Also, for any $\alpha$ we have $(1+z^{-2/3})^{\alpha}\to 1$ as $z\to\infty$, so it will not affect convergence (strictly speaking, we could find a $M_{\alpha}$ such that for $x>M_{\alpha}$, $1/2 < (1+z^{-2/3})^{\alpha} <2$). We use the standard trick of splitting the integral into countably many intervals of length $\pi$: $$ \int_{\pi}^{\infty}z^{(\alpha-2)/3}|\sin(z)|dz $$A standard argument using the MVT or some such shows this integral diverges for $\alpha=-1$, and subsequently for $-1\le \alpha <2$ by direct comparison. However, for $\alpha<-1$, one can use direct comparison with the $p$-integral, since $|\sin(x)|\le 1$ for real $x$.

So in summary, the integral converges conditionally for $-1\le \alpha<2$ and absolutely for $\alpha<-1$.


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.