Why $\displaystyle \int^{\infty}_{\pi} \frac{sinx}{lnx}dx$ converges?
You can rewrite the integral in the following way:
$\displaystyle \displaystyle \int_\pi^\infty \frac{\sin x}{\log x} \text{ d}x=\int_\pi^{2\pi} \frac{\sin x}{\log x} \text{ d}x+\int_{2\pi}^{3\pi} \frac{\sin x}{\log x} \text{ d}x+\int_{3\pi}^{4\pi} \frac{\sin x}{\log x} \text{ d}x+\ldots=\sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{\sin x}{\log x} \text{ d}x$.
Take note that the terms of the sum are alternating! We can write the sum as
$\displaystyle \displaystyle \sum_{k=1}^\infty (-1)^k \int_{k\pi}^{(k+1)\pi} \left|\frac{\sin x}{\log x}\right| \text{ d}x$
Also observe that as $\displaystyle k\to \infty$ the term $\displaystyle \displaystyle a_k=\int_{k\pi}^{(k+1)\pi} \left|\frac{\sin x}{\log x}\right| \text{ d}x\to 0$ monotonically. Therefore, the sum (and equivalently, the integral) converges via the alternating series test.
This is probably the cleanest way to do the problem, but you might be able to get away with an integration by parts, first. Let $\displaystyle u=\frac{1}{\log x}$ and $\displaystyle dv=\sin x\text{ d}x$. Then, $\displaystyle du=\frac{1}{x(\log x)^2} \text{ d}x$ and $\displaystyle v=-\cos x$. The integral becomes
$\displaystyle \displaystyle \int_\pi^\infty \frac{\sin x}{\log x}\text{ d}x=\left.\frac{-\cos x}{\log x} \right|_\pi^\infty+\int_\pi^\infty \frac{\cos x}{x (\log x)^2}\text{ d}x$
Can you see how the terms will converge?
Dirichlet's test for integrals!
Let,
$\displaystyle f(x)=sin(x)$
and
$\displaystyle g(x)=\frac{1}{ln(x)}$
$\displaystyle |\int^b_{\frac{\pi}{2}} sin(x)dx| \leq 2$ for all $\displaystyle b>\frac{\pi}{2}} $
and $\displaystyle g(x)$ is monotonic function and $\displaystyle lim_{x\to\infty}g(x)=0$
So, by the test of Dirichlet the integral converges!