<s>[INST] Does the following integral converge? $ \int\limits_0^\pi\frac{\sin x}{\sqrt{x}}\ dx $ Does the following integral converge?
$$
\int\limits_0^\pi\frac{\sin x}{\sqrt{x}}\ dx
$$
I haven't solved such problems for a while. So, I would really appreciate it if someone gave me a hint.
Or maybe my solution is correct?
$$
\sin x\sim x\Rightarrow\frac{\sin x}{\sqrt{x}}\sim\sqrt{x}
$$
$$
\int\limits_0^\pi\sqrt{x}\ dx\ \ \text{is convergent}
$$
Therefore, the initial integral is convergent as well.
 [/INST] Just for the fun of it !
The problem of the convergence being solved, there are analytical solution for this kind of integrals (and antiderivatives; have a look here.
Since @Von Neumann wrote an answer where complex numbers do appear, I wondered what would give the $1,400$ years old approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.
$$\int\frac{\sin (x)}{\sqrt{x}}\, dx \sim \int \frac{16 (\pi -x) \sqrt{x}}{5 \pi ^2-4 (\pi -x) x} \,dx=$$ and then the integral is$$-8 \sqrt{\pi }+2 i \sqrt{(-2-4 i) \pi } \cot
   ^{-1}\left(\sqrt{-\frac{1}{2}-i}\right)-(4+3 i) \sqrt{\left(-\frac{2}{5}+\frac{4
   i}{5}\right) \pi } \cot ^{-1}\left(\sqrt{-\frac{1}{2}-i}\right)-2 i \sqrt{(-2+4 i)
   \pi } \cot ^{-1}\left(\sqrt{-\frac{1}{2}+i}\right)-(4-3 i)
   \sqrt{\left(-\frac{2}{5}-\frac{4 i}{5}\right) \pi } \cot
   ^{-1}\left(\sqrt{-\frac{1}{2}+i}\right)$$ which is $\approx 1.78995$ while the "exact" value is $1.78966$.
Edit
Another amazing approximation is
$$\sin(x)=\pi \sum_{n=1}^\infty a_n \Big[\left(1-\frac x \pi\right)\frac x \pi\Big]^n$$ where coefficients $a_n$ make the sequence
$$\left\{1,1,2-\frac{\pi ^2}{6},5-\frac{\pi ^2}{2},14-\frac{3 \pi ^2}{2}+\frac{\pi
   ^4}{120},42-\frac{14 \pi ^2}{3}+\frac{\pi ^4}{24},132-15 \pi ^2+\frac{\pi
   ^4}{6}-\frac{\pi ^6}{5040}\right\} $$
This makes the integration very easy
$$\int\limits_0^\pi\frac{\sin (x)}{\sqrt{x}}\, dx=\pi ^2\sum_{n=1}^\infty \frac{\Gamma (2 n+1)}{4^n \,\Gamma \left(2 n+\frac{3}{2}\right)}\,a_n$$
Using the $a_n$'s given in the table, the definite integral is then
$$\frac{4 \pi ^{3/2} \left(46190338425-595324620 \pi ^2+1781520 \pi ^4-704 \pi
   ^6\right)}{503889568875}$$ which is $1.789662938921$ while the exact value is                               $1.789662938968$
</s>