How to find this improper integral? $\int_{0}^{\pi}{\frac{\sin{x}}{\sqrt{x}}dx}$ How to calculate improper integral? $$\int_{0}^{\pi}{\frac{\sin{x}}{\sqrt{x}}dx}.$$
 A: This is related to Fresnel integrals $$\int{\frac{\sin{x}}{\sqrt{x}}dx}=\sqrt{2 \pi } S\left(\sqrt{\frac{2 x}{\pi }} \right)$$ $$\int_{0}^{\pi}{\frac{\sin{x}}{\sqrt{x}}dx}=\sqrt{2 \pi } S\left(\sqrt{2}\right) \simeq 1.789662939$$
Without special function, you can compute the integral starting with the Taylor development of $\sin(x)$ which then leads to  $$\int_{0}^{\pi}{\frac{\sin{x}}{\sqrt{x}}dx}=\int_{0}^{\pi} \sum_{n=0}^\infty \frac{(-1)^n x^{2 n+\frac{1}{2}}}{(2 n+1)!}~dx=2\pi^{\frac{3}{2}}\sum_{n=0}^\infty \frac{ (-1)^n \pi ^{2 n}}{(4 n+3) (2 n+1)!}$$ Using $5$ terms, the result is $1.789604144$ while using $10$ terms leads to $1.789662939$.
A: $\newcommand{\+}{^{\dagger}}
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With $\ds{x \equiv {\pi t^{2} \over 2}}$:

\begin{align}
\color{#66f}{\large\int_{0}^{\pi}{\sin\pars{x} \over \root{x}}\,\dd x}&=
\root{2\pi}\int_{0}^{\root{2}}\sin\pars{\pi t^{2} \over 2}\,\dd t
=\color{#66f}{\large\root{2\pi}{\rm S}\pars{\root{2}}}
\approx 1.7897
\end{align}

wehe $\ds{{\rm S}\pars{x}}$ is a
Fresnel Integral.
