How to find this improper integral? $\int_{0}^{\pi}{\frac{\sin{x}}{\sqrt{x}}dx}$

Question

How to calculate improper integral? $$\int_{0}^{\pi}{\frac{\sin{x}}{\sqrt{x}}dx}.$$

Answer 1

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 $1.789662939$.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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.