Trig Fresnel Integral

$\displaystyle \int_{0}^{\infty}\sin(x^{2})dx$

Let $\displaystyle u=x^{2}, \;\ x=\sqrt{u}, \;\ dx=\frac{1}{2}u^{-1/2}du$

$\displaystyle\frac{1}{2}\int_{0}^{\infty}\frac{\sin(u)}{\sqrt{u}}du$

Now, the Gamma function comes in handy:

$\displaystyle\frac{1}{2\Gamma(a)}\int_{0}^{\infty}z^{a-1}e^{-uz}\sin(u)dz$

But, since $a=1/2$, we have:

$\displaystyle\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-uz}\sin(u)}{\sqrt{z}}dudz$

$\displaystyle\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{1}{\sqrt{z}(1+z^{2})}dz$

Let $t=\sqrt{z}$:

$\displaystyle\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{1}{t^{4}+1}dt$

This is a rather famous integral and can be found here and there. I leave it's evaluation to the reader.

But, it evaluates to $\displaystyle \frac{1}{\sqrt{\pi}}\cdot \frac{\pi\sqrt{2}}{4}=\frac{\sqrt{2\pi}}{4}$, and so does the Fresnel in question.

The integral does not have an elementary antiderivative. If you want to avoid complex analysis, try this method.