# Absolute and conditional convergence of $\int_0^{+\infty}\left(x + \frac{1}{x}\right)^{\alpha}\sin (x^3) \mathrm{d}x$

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.

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

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 $$\alpha<-1$$.