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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. $$\int_{0}^{1/2}\frac{\sin(x)}{x^{3/2}\log(x)}\,dx.$$

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    the integral is disvergent – 2017-01-13
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    In a neighborhood of $0$, $\sin(x)\sim x$, so you only need to study the convergence of $$\int_0^{1/2}\frac{1}{x^{1/2}\log x}\,dx.$$ The convergence of this integral can be easily handled. – 2017-01-13

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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.