How can i study the convergence of the following improper integral? I have tried to narrow $\sin(x)$ by substituting to $\int_{0}^{1/2}\frac{dx}{x^{3/2}\log(x)}$, but anyway, I have not obtained anything so far.
$$\int_{0}^{1/2}\frac{\sin(x)}{x^{3/2}\log(x)}\,dx.$$

For technical purposes let's say the integrand is $- \frac{\sin x}{x^{3/2} \log x}$. This is nonnegative over $(0,1/2]$. We have:
$$-\frac{\sin x}{x^{3/2} \log x} \sim - \frac{x}{x^{3/2} \log x} = - \frac{1}{\sqrt x \log x}$$
Now:
$$I := \int_0^{1/2} -\frac1{\sqrt x \log x} dx = \int_{\log 2}^{\infty} \frac{e^{-u/2}}{u}du$$
We have:
$$\lim_{u\to \infty} u^{2} \frac{e^{-u/2}}{u} = 0$$
so there exists $M > \log 2$ such that for all $u \ge M$, $\frac{e^{-u/2}}{u} < \frac1{u^2}$
Thus $I$ is convergent and so is the original integral.
