Proof of $\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}$

Numerically it seems to be true that

$$ \int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}. $$

Any ideas how to prove this?

Solution 1:

Using contour integration, we get $$ \begin{align} \int_0^\infty\frac{e^{ix}}{\sqrt{x}}\,\mathrm{d}x &=\sqrt{i\,}\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,\mathrm{d}x\\ &=\frac{1+i}{\sqrt{2}}\Gamma\left(\frac12\right)\\ &=(1+i)\sqrt{\frac\pi2} \end{align} $$ Therefore, $$ \int_0^\infty\frac{\cos(x)}{\sqrt{x}}\,\mathrm{d}x=\int_0^\infty\frac{\sin(x)}{\sqrt{x}}\,\mathrm{d}x=\sqrt{\frac\pi2} $$

About the Contour Integration

If we integrate $f(z)=\dfrac{e^{iz}}{\sqrt{z}}$ around the contour $[0,R]\cup Re^{i[0,\pi/2]}\cup i[R,0]$ as $R\to\infty$, we get that $$ \int_0^R\frac{e^{ix}}{\sqrt{x}}\,\mathrm{d}x +\int_0^{\pi/2}\frac{e^{iRe^{ix}}}{\sqrt{R}e^{ix/2}}iRe^{ix}\,\mathrm{d}x -\sqrt{i\,}\int_0^R\frac{e^{-x}}{\sqrt{x}}\,\mathrm{d}x =0 $$ because there are no singularities of $f$ inside the contour. Then because $$ \begin{align} \left|\int_0^{\pi/2}\frac{e^{iRe^{ix}}}{\sqrt{R}e^{ix/2}}iRe^{ix}\,\mathrm{d}x\right| &\le\sqrt{R}\int_0^{\pi/2}e^{-R\sin(x)}\,\mathrm{d}x\\ &\le\sqrt{R}\int_0^{\pi/2}e^{-2Rx/\pi}\,\mathrm{d}x\\ &\le\frac\pi{2\sqrt{R}} \end{align} $$ vanishes as $R\to\infty$, we have $$ \int_0^\infty\frac{e^{ix}}{\sqrt{x}}\,\mathrm{d}x =\sqrt{i\,}\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,\mathrm{d}x $$

Real Method

Substituting $u^2=x$ and applying this answer, which uses only real methods, yields $$ \begin{align} \int_0^\infty\frac{\sin(x)}{\sqrt{x}}\,\mathrm{d}x &=2\int_0^\infty\sin(u^2)\,\mathrm{d}u\\ &=2\sqrt{\frac\pi8}\\ &=\sqrt{\frac\pi2} \end{align} $$

Solution 2:

That is a Fresnel integral.

Make the substitution $\sqrt{x}=u$. Then you get $dx=2udu$ from where

$$\int_0^\infty \sin(x) x^{-1/2} dx =2 \int_0^\infty \sin(u^2)du=2\sqrt{\frac{\pi}{8}}=\sqrt{\frac{\pi}{2}}$$

See this question for more details.