A converging Integral

First, break up the integral into two parts : $\displaystyle\int_0^1 x^a\sin(x)dx + \displaystyle\int_1^{\infty} x^a\sin(x)dx$

By definition of convergence for improper integrals, the integral converges iff the two parts converges. Clearly if $a\geq 0$ the second one is divergent. If $a<0$ an integration by parts shows that the second part is convergent (because $a-1<-1$ and $|\cos(x)|<1$ ) :

$\displaystyle\int_1^{\infty} x^a\sin(x)dx = \bigg[-x^a\cos(x)\bigg]_1^{\infty} -\displaystyle\int_1^{\infty} ax^{a-1}\cos(x)dx$

Note: the following integral converges iff $\alpha>1$ $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{\alpha}}$

For the first one, what's important is the asymptotic behaviour at $0$ . We have $x^a\sin(x) \sim x^{a+1}$ therefore the first integral convergences iff $a+1>-1$ ie $a>-2$ .

Note: the following integral converges iff $\alpha<1$ $\displaystyle\int_0^1 \dfrac{dx}{x^{\alpha}}$

At the end, the two simultaneously converges iff $-2<a<0$ , which gives the answer $\boxed{-2}$ .