How do I find the set of convergence of improper parametric integral?

Question

I have the integral:

$$\int_1^{+\infty} \frac{x^{\alpha}\cdot \sin x}{1+x^3}\, dx$$

And I have to find the set of convergence? How should I do act? And how should I treat the parametric variable $\alpha$? It is the main problem here, cause if it were "simple" (non-parametric integral), it would be easy to solve

Answer 1

The short answer is that the sine function makes this integral equivalent to an alternating series, and by the alternating series test it will converge provided the integrand goes to zero at infinity. In other words, $\alpha < 3$.

Going into more detail, the integral can be expanded into an infinite sum like this $$ \int_1^\infty \frac{x^\alpha \sin x}{1+x^3}dx = \int_1^\pi \frac{x^\alpha \sin x}{1+x^3}dx + \sum_{n=1}^\infty \int_{n\pi}^{(n+1)\pi} \frac{x^\alpha \sin x}{1+x^3}dx. $$ Calling the individual terms of the sum $I_n$ for convenience, we see \begin{multline} I_n = \int_{n\pi}^{(n+1)\pi} \frac{x^\alpha \sin x}{1+x^3}dx = \int_0^\pi\frac{(u+n\pi)^\alpha \sin(u+n\pi)}{1+(u+n\pi)^3}du \\= (-1)^n\int_0^\pi \frac{(u+n\pi)^\alpha \sin u}{1+(u+n\pi)^3}du \end{multline} The integrand in that last integral is nonnegative on $[0,\pi]$, so the terms do indeed alternate sign. As for their magnitude, $$ |I_n| = \int_0^\pi \frac{(u+n\pi)^\alpha \sin u}{1+(u+n\pi)^3}du = (n\pi)^{\alpha-3}\int_0^\pi \frac{(1 + \frac{u}{n\pi})^\alpha \sin u}{(1+\frac{u}{n\pi})^3 + (n\pi)^{-3}}du $$ In the limit as $n\rightarrow \infty$, the integral goes to the constant value $\int_0^\pi \sin u\,du = 2$, so $\lim_{n\rightarrow\infty}|I_n| = 0$ only if $\alpha < 3$. By the alternating series test, that sum (and thus the integral) converges only if $\alpha < 3$.

Answer 2

By integration by parts, one has \begin{eqnarray*} &&\int_1^{+\infty} \frac{x^{\alpha}\cdot \sin x}{1+x^3}\, dx\\ &=&-\int_1^{+\infty} \frac{x^{\alpha}}{1+x^3}\, d\cos x\\ &=&-\frac{x^{\alpha}\cos x}{1+x^3}\bigg|_0^{\infty}+\int_1^{+\infty} \cos xd\bigg(\frac{x^{\alpha}}{1+x^3}\bigg)\\ &=&-\frac{x^{\alpha}\cos x}{1+x^3}\bigg|_0^{\infty}-\int_1^{+\infty}\frac{x^{\alpha-1}[(3-\alpha)x^3-\alpha]\cos x}{(1+x^3)^2}\;dx\\ \end{eqnarray*} If $\alpha<3$, one has $$ \lim_{x\to\infty} \frac{x^{\alpha}\cos x}{1+x^3}=0$$ and then for big $x$, $$ \bigg|\frac{x^{\alpha-1}[(3-\alpha)x^3-\alpha]\cos x}{(1+x^3)^2}\bigg|\le \frac{C}{x^2} $$ and hence the integral converges. If $\alpha\ge 3$, it is easy to see the integral diverges.