Is this limit infinite?

Question

I'm trying to prove that $$ \frac {\sin x}{x^{3/2}} $$ is integrable (improper sense) on $(0,1)$. I am trying to find this limit :

$$ \lim_{a\to 0+} \int_a^1 \frac {\sin x}{x^{3/2}} dx $$

When I try integrating by parts, I end up with an infinite term $$ \frac {\cos (a)}{a^{1/2}} $$ plus a finite term and a finite integral. I think I'm doing something wrong, or does that simply mean that the integral is divergent and that the function isn't integrable in the improper sense on $(0,1)$? (I have some reasons to think it should be integrable... but could be wrong)

Answer 1

Note that the function is positive and that $\sin x<x$ for $x>0$; so you have $$ \frac{\sin x}{x^{3/2}}<\frac{x}{x^{3/2}}=\frac{1}{x^{1/2}} $$ Is $$ \int_0^1\frac{1}{x^{1/2}}\,dx $$ convergent?

Answer 2

There are two natural guesses for what your integration by parts looked like.

We can let $u=\sin x$ and $dv=x^{-3/2}\,dx$. Then $du=\cos x\,dx$ and we can take $v=-2x^{-1/2}$.

In this case, both $\left.uv\right|_a^1$ and $\int_a^1 v\,du$ behave nicely as $a$ approaches $0$ from the right. So our original integral converges.

Or else we can let $u=x^{-3/2}$ and $dv=\sin x\,dx$. Then both $\left.uv\right|_a^1$ and $\int_a^1 v\,du$ have an infinite limit as $a$ approaches $0$ from the right. (There happens to be cancellation, but the approach has to be viewed as unsuccessful.)