How to evaluate the complex integral $\int\limits_0^\infty {e^{is} \over \sqrt{s}} ds$?

Question

What is the general procedure for evaluating this integral $$\int\limits_0^\infty {e^{is} \over\sqrt{s}} ds$$?

In the book, I saw that it is equal to $\sqrt{\pi}e^{ i\pi/4}$, but I could not see why. How do we do the calculation?

Answer 1

HINT : what is the value of $\int_{0}^{\infty} \frac{\cos(x)}{\sqrt{x}} $? Also for the same with $\sin(x)$.

Answer 2

Through Euler’s formula, we can rewrite your integral as$$I=\int\limits_0^{\infty}dx\,\frac {\cos x}{\sqrt x}+i\int\limits_0^{\infty}dx\,\frac {\sin x}{\sqrt x}$$To evaluate both integrals, consider the identity$$\frac 1{\sqrt\pi}\int\limits_0^{\infty}ds\,\frac {e^{-xs}}{\sqrt s}=\frac 1{\sqrt x}$$Therefore$$\begin{align*}\int\limits_0^{\infty}dx\,\frac {\cos x}{\sqrt x} & =\frac 1{\sqrt{\pi}}\int\limits_0^{\infty}ds\,\frac 1{\sqrt s}\int\limits_0^{\infty}dx\,e^{-sx}\cos x\\ & =\frac 1{\sqrt\pi}\int\limits_0^{\infty}ds\,\frac {\sqrt s}{1+s^2}\\ & =\frac 1{\sqrt\pi}\frac {\pi}{\sqrt2}=\sqrt{\frac {\pi}2}\end{align*}$$Repeating the same process for the imaginary part, we have that$$\int\limits_0^{\infty}dx\,\frac {\cos x}{\sqrt x}=\int\limits_0^{\infty}dx\,\frac {\sin x}{\sqrt{x}}=\color{blue}{\sqrt{\frac {\pi}2}}$$Therefore$$I=\sqrt{\pi}\left(\frac 1{\sqrt2}+\frac i{\sqrt2}\right)=\color{red}{\sqrt{\pi}e^{\pi i/4}}$$