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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.

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  • $\begingroup$ is this $$\sin(x^3)$$ or $$\sin(x)^3$$? $\endgroup$ – Dr. Sonnhard Graubner May 28 '17 at 14:56
  • $\begingroup$ @Dr.SonnhardGraubner is $\sin(x^3)$ $\endgroup$ – Jazz May 28 '17 at 15:00
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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 $\alpha<-1$.

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