integral and sines and stuff

$\begin{aligned} I & = \int_0^\infty \frac {\sin x^\frac 23}{x^\frac 23} dx & \small \color{#3D99F6} \text{Let }t = x^\frac 13 \implies dt = \frac {dx}{3x^\frac 23} \\ & = 3 \color{#3D99F6} \int_0^\infty \sin (t^2)\ dt & \small \color{#3D99F6} \text{Fresnel integral }S(x) = \int_0^x \sin (t^2)\ dt \\ & = 3 \color{#3D99F6} \lim_{x \to \infty} S(x) & \small \color{#3D99F6} \text{and that }\lim_{x \to \infty} S(x) = \sqrt {\frac \pi 8} \\ & = 3 \color{#3D99F6} \sqrt {\frac \pi 8} \\ & = \frac 32 \sqrt {\frac \pi 2} \end{aligned}$

Therefore, $a+b = 3+2 = \boxed{5}$ .

This is more of an elaboration of Chew-Seong Cheong's solution rather than a different one.

As he did we first substitute $t=x^{1/3}$ . So we have :-

$I=3\int_{0}^{\infty}sin(t^{2})dt$

$I= -3\Im[\int_{0}^{\infty}e^{-it^{2}}dt]$ where $\Im[.]$ denotes the imaginary part . and $i=\sqrt{-1}$

So let us again substitute $\sqrt{i}t = z$

So we have $dt = \frac{dz}{\sqrt{i}}$

So we have our integral as :-

$I= -\Im[\frac{3}{\sqrt{i}}(\int_{0}^{\infty}e^{-z^{2}}dz])$

Now we evaluate $\int_{0}^{\infty}e^{-z^{2}}dz$

It is the Gaussian integral

Just substitute $z^2 = y$ . to see that the Gaussian Integral is nothing but $\frac{1}{2}\Gamma(\frac{1}{2})$ .

In case you are wondering how we evaluate the integral......there is a great video by Dr.Peyam

Here's the link to it

So we have $I = -\Im\frac{3}{\sqrt{i}}\frac{\sqrt{\pi}}{2}$

Now $\sqrt{i} = e^{-i\frac{\pi}{4}} = cos(\frac{\pi}{4}) -isin(\frac{\pi}{4})$

we have $I = \Im[\frac{3\sqrt{\pi}}{2}( -cos(\frac{\pi}{4}) + isin(\frac{\pi}{4})$

So we have our answer as $I = \frac{3\sqrt{\pi}}{2}sin(\frac{\pi}{4})$