Integrating around a triangular contour for Fresnel integral

$$\int^\infty_0\frac{\cos(x)}{\sqrt{x}}\,dx =\int^\infty_0\frac{\sin(x)}{\sqrt{x}}\,dx =\sqrt{\frac{\pi}{2}}$$

$$\textit{solution}$$

Consider the following function

$$f(z)=z^{-1/2}\,e^{iz}$$

Where we choose the principle root for \( z^{-1/2}=e^{-1/2\log(z)}\). By integrating around the following contour

$$\int_{C_r}f(z)\,dz+\int_{r}^R f(x)\,dx+\int_{\gamma}f(z)\,dz+\int^{iR}_{ir}f(x)\,dx = 0$$

Taking the integral around the small quarter circle with $r\to 0$
$$\left| \int_{C_r}f(z)\,dz\right|\leq \left|\sqrt{r}\int^{\pi/2}_{0}e^{it/2} e^{rie^{it}}\,dt\right| \leq \sqrt{r}\int^{\pi/2}_{0}\left|e^{-r\sin(t)}\right|\,dt\sim 0$$

On \(\gamma(t)=(1-t)R+iRt\) where \(0\leq t \leq 1\)

$$\left|\int_{\gamma}f(z)\,dz\right| = \left| R(i-1)\int^1_0e^{-1/2\log(R(1-t)+iRt)}e^{i(1-t)R-Rt}\,dt\right| \\ \leq \frac{\sqrt{2}}{\sqrt{R}} \int^1_0 \frac{e^{-Rt}}{\sqrt[4]{(1-t)^2+t^2}}\,dt$$

Hence we have

$$\left|\int_{\gamma}f(z)\,dz\right| \leq \frac{\sqrt{2}}{\sqrt{R}} \int^1_0 e^{-Rt}\,dt=\frac{\sqrt{2}}{R\sqrt{R}}\left(1-e^{-R}\right)\sim_{\infty}0$$

Finally what is remaining when \(r\to 0\) and \(R \to \infty\)

$$\int^\infty_0 \frac{e^{ix}}{\sqrt{x}}\,dx =i \int^{\infty}_{0}(ix)^{-1/2}e^{-x}\,dx$$

Note that \(i^{-1/2}=e^{-i\pi/4}\)

$$\int^\infty_0\frac{e^{ix}}{\sqrt{x}}\,dx = ie^{-i\pi/4}\int^{\infty}_{0}x^{-1/2}e^{-x}\,dx =ie^{-i\pi/4} \sqrt{\pi} $$

Using that we have

$$\boxed{\int^\infty_0\frac{\cos(x)}{\sqrt{x}}\,dx =\int^\infty_0\frac{\sin(x)}{\sqrt{x}}\,dx =\sqrt{\frac{\pi}{2}}}$$

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