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 $$\sin(x)^3$$? $\endgroup$ – 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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