About Integration $\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx$ What i want to prove is following integral 
\begin{align}
\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx=\frac{1}{2} \sqrt{\left(1+\sqrt{2}\right) \pi }
\end{align}
can you give some explicit method to obtain this result?

So let us follow through on Eclipse Sun's suggestion. In doing so we will find that while the double integral can indeed be separated into two parts (two double integrals), neither of them turns out to be equal to zero.
Let
$$I = \int^\infty_{-\infty} e^{-x^2} \cos (x^2) \, dx = 2 \int^\infty_0 e^{-x^2} \cos (x^2) \, dx.$$
on squaring the integral we have
\begin{align*}
I^2 &= 4 \left(\int^\infty_0 e^{-x^2} \cos(x^2) \, dx\right)\times\left(\int_{0}^{\infty} e^{-y^2} \cos(y^2) \, dy\right)\\
&= 4 \int^\infty_0 \int^\infty_0 e^{-(x^2 + y^2)} \cos (x^2) \cos (y^2) \, dx dy.
\end{align*}
After making use of $2\cos x\cos y= \cos(x-y)+\cos(x+y)$ we have
$$I^2 = 2 \int^\infty_0 \int^\infty_0 e^{-(x^2 + y^2)} \cos (x^2 + y^2) \, dx dy + 2 \int^\infty_0 \int^\infty_0 e^{-(x^2 + y^2)} \cos (x^2 - y^2) \, dx dy.$$
Converting to polar coordinates $(x,y) \mapsto (r \cos \theta, r \sin \theta)$ we have
$$I^2 = 2 \int^{\frac{\pi}{2}}_0 \int^\infty_0 r e^{-r^2} \cos (r^2) \, dr d\theta + 2 \int^{\frac{\pi}{2}}_0 \int^\infty_0 r e^{-r^2} \cos (r^2 \cos 2\theta ) \, dr d\theta = 2 I_1 + 2 I_2.$$
For the first of these integrals, the $r$-integral can be found by preforming integration by parts twice before performing the $\theta$-integration. The result is $I_1 = \pi/8$.
For the second of these integrals, finding the $r$-integral first by performing integration by parts twice, leads to
$$I_2 = \frac{1}{2} \int^{\frac{\pi}{2}}_0 \frac{d\theta}{1 + \cos^2 (2\theta)} = \frac{\pi}{4\sqrt{2}}.$$
Thus
$$I^2 = 2 \left [\frac{\pi}{4} + \frac{\pi}{2\sqrt{2}} \right ] = \frac{\pi}{4} (1 + \sqrt{2}),$$
or
$$I = \frac{\sqrt{\pi}}{2} \sqrt{1 + \sqrt{2}},$$
as required. 
