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I want to compute $$\int_{0}^{+\infty} \frac{\sin x}{\sqrt{x}}\ dx$$ using Gamma function.

I know that by change of variable, $y=\sqrt{x}$, one gets $$\int_{0}^{+\infty} \frac{\sin x}{\sqrt{x}}\ dx=2\int_{0}^{+\infty}\sin y^2\ dy=\frac{\sqrt{2\pi}}{2}$$ by Fresnel's integral.

I try it by considering this: $$\int_{0}^{+\infty}x^{-\frac{1}{2}}e^{ix}\ dx$$ It converges for both real and imaginary part using Dirichlet test, and $0$ is not a problem here. Let the square root take the pricipal branch where $\sqrt{1}=1$. Let $y=-ix$, then $$\int_{0}^{+\infty}x^{-\frac{1}{2}}e^{ix}\ dx=\sqrt{i}\int_{0}^{-i\infty}y^{-\frac{1}{2}}e^{-y}\ dy=(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i)\Gamma(\frac{1}{2})=\frac{\sqrt{2\pi}}{2} + \frac{\sqrt{2\pi}}{2}i$$ And it coincides with the final answer!

My problem is, suppose $L$ is a ray starting from $0$ and has an angle $\phi$ with the $x$-axis, and let $\phi\in(0,2\pi)$. I want to argue that (maybe it is incorrect though) $$\Gamma(z)=\int_{L}t^{z-1}e^{-t}\ dt$$

I firstly know that it converges when $\Re(z)>0$. Choose a contour like sector, let $L_1=\{z=x+iy:y=0, r<x<R\}$, $L_2=\{z=Re^{i\theta}:0<\theta<\phi\}$, $L_3=\{z=xe^{i\phi}:r<x<R\}$ and $L_4=\{z=re^{i\theta}:0<\theta<\phi\}$, where $r<R$ and the contour is counterclockwise. By Cauchy theorem we have the contour integral should be $0$. Easy to see that (let $z=x+iy$) $$ \lim_{r\to0+, R\to+\infty}\int_{L_1}t^{z-1}e^{-t}\ dt=\Gamma(z) $$ $$ |\int_{L_4}t^{z-1}e^{-t}\ dt|=|\int_{\phi}^{0} e^{-re^{i\theta}} (re^{i\theta})^{z-1}ire^{i\theta}\ d\theta| \leq \int_{0}^{\phi} e^{-r\cos\theta} |r^{z}e^{i\theta(z-1)}|\ d\theta=\int_{0}^{\phi} e^{-r\cos\theta} r^{x}e^{-\theta y}\ d\theta \to 0, r \to 0+ $$ But when considering $L_2$: $$ |\int_{L_2}t^{z-1}e^{-t}\ dt|\leq\int_{0}^{\phi} e^{-R\cos\theta} R^{x}e^{-\theta y}\ d\theta $$ when for example, $\frac{3\pi}{2}>\phi>\frac{\pi}{2}$, we have $\cos\theta<0$ and I failed to prove the above integral goes to zero when $R\to +\infty$.

Is my usage of Gamma function to compute the original integral a coincidence to get the correct result, or there is a way to prove my argument?

Thank you so much!

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    $\begingroup$ $\int_0^\infty \sin y^2dy=\int_0^\infty\sin(y^2)dy\ne\int_0^\infty\sin^2ydy$. $\endgroup$ Jan 1, 2019 at 8:07

2 Answers 2

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There's a bit of a problem because square roots are ambiguous. While $\int_0^\infty x^{-1/2}\exp -zxdx=\sqrt{\pi}z^{-1/2}$ is easily proven on $\Bbb R^+$, if you try an analytic continuation (whether your proof approach is as detailed as you've attempted or otherwise) the intended theorem isn't even clear. Which kind of $z^{-1/2}$ should be used for the desired case $z=-i$?

Funnily enough, there's a completely different approach using the gamma function: $$\begin{align}\int_0^\infty\frac{\sin x dx}{\sqrt{x}}&=\frac{1}{\sqrt{\pi}}\Im\int_0^\infty dx\int_0^\infty y^{-1/2}\exp -x(y-i)dy\\&=\frac{1}{\sqrt{\pi}}\Im\int_0^\infty dy\frac{y^{-1/2}}{y-i}\\&=\frac{1}{\sqrt{\pi}}\int_0^\infty\frac{y^{-1/2}dy}{y^2+1}\\&=\frac{1}{\sqrt{\pi}}\int_0^{\pi/2}\tan^{-1/2}\theta d\theta\\&=\frac{1}{2\sqrt{\pi}}\operatorname{B}\left(\frac{1}{4},\,\frac{3}{4}\right)\\&=\frac{1}{2\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)\\&=\frac{\sqrt{\pi}}{2}\csc\frac{\pi}{4}=\sqrt{\frac{\pi}{2}}.\end{align}$$

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$$ \begin{aligned} \int_0^{\infty} \frac{\sin x}{\sqrt{x}} d x & =-\operatorname{Im} \int_0^{\infty} \frac{e^{-i x}}{\sqrt{x}} d x \\ & =-\operatorname{Im} \int_0^{\infty}\left(\frac{x}{i}\right)^{-\frac{1}{2}} e^{-x} \frac{d x}{i}\quad \textrm{ via } ix\mapsto x\\ & =-\operatorname{Im}\left[\frac{1}{\sqrt{i}} \int_0^{\infty} x^{-\frac{1}{2}} e^{-x} d x\right] \\ & =-\Gamma\left(\frac{1}{2}\right) \operatorname{Im}\left(e^{-\frac{\pi}{4} i}\right)\\&=\sqrt{\frac{\pi}{2}} \end{aligned} $$ In general, for any $0<a<1$, we can similarly prove that $$ I(a)=\int_0^{\infty} \frac{\sin x}{x^a} d x =\frac{\pi}{2 \Gamma(a) \sin \left(\frac{a}{2}\right)} $$ Proof: $$ \begin{aligned} I(a) & =-\operatorname{Im} \int_0^{\infty} x^{-a} e^{-x i} d x \\ & =-\operatorname{Im} \int_0^{\infty}\left(\frac{x}{i}\right)^{-a} e^{-x} \frac{d x}{i}\\&= - \Gamma(1-a) \operatorname{Im}\left(i^{-a-1}\right)\\&= \frac{\pi}{2 \Gamma(a) \sin \left(\frac{a \pi}{2}\right)} \end{aligned}\\ $$ where the last answer comes from the Reflection Property of Gamma function

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