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?

Let your integral be labelled
$$I_1=\int_{-\infty}^{\infty} e^{-x^2}\cdot \cos(x^2)\space dx,$$
and a second integral
$$I_2=\int_{-\infty}^{\infty} e^{-x^2}\cdot\sin(x^2)\space dx.$$
It follows that
$$I_1-i\cdot I_2=\int_{-\infty}^{\infty} e^{-x^2}\cdot(\cos(x^2)-i\cdot \sin(x^2))\space dx.$$
Apply Euler's formula in complex analysis:
$$I_1-i\cdot I_2=\int_{-\infty}^{\infty} e^{-x^2}\cdot e^{-ix^2} dx=\int_{-\infty}^{\infty} e^{-(1+i)x^2}dx.$$
Let $x=(1+i)^{-1/2}\ t$ such that $dx=(1+i)^{-1/2}\ dt$, where $t\in(-\infty,\infty)$:
$$I_1-i\cdot I_2=(1+i)^{-1/2}\cdot\int_{-\infty}^{\infty} e^{-t^2}dt.$$
Evaluate the Gaussian integral:
$$I_1-i\cdot I_2=(1+i)^{-1/2}\cdot\sqrt{\pi}.$$
Rewrite the expression by making use of general properties of the exponential function and logarithms:
$$I_1-i\cdot I_2=e^{\ln((1+i)^{-1/2})}\cdot\sqrt{\pi}=e^{-1/2\ln(1+i)}\cdot\sqrt{\pi}.$$
The complex number $1+i$ can be converted from cartesian notation to polar notation; $1+i=\sqrt{2}\space e^{i \pi/4}$. Taking the natural logarithm on both sides gives $\ln(1+i)=\ln(\sqrt{2}\space e^{i \pi/4})=\ln(\sqrt{2})+i \pi/4$. Therefore,
$$I_1-i\cdot I_2=e^{-1/2(\ln(\sqrt{2})+i \pi/4)}\cdot\sqrt{\pi} = e^{\ln(\frac{1}{\sqrt{\sqrt{2}}})}\cdot e^{-i \pi/8}\cdot\sqrt{\pi}=\frac{1}{\sqrt{\sqrt{2}}}\cdot e^{-i \pi/8}\cdot \sqrt{\pi}.$$
Apply Euler's formula in complex analysis:
$$I_1-i\cdot I_2=\frac{1}{\sqrt{\sqrt{2}}}\cdot (\cos(\pi/8)-i\cdot \sin(\pi/8))\cdot \sqrt{\pi}.$$
Rewrite $\cos(\pi/8)$ by making use of the half-angle formula for the cosine function, $\cos(x)=2\cos^2(x/2)-1$:
$$\cos(\pi/4)=2\cos^2(\pi/8)-1\rightarrow \cos^2(\pi/8)=\frac{1+\cos(\pi/4)}{2}=\frac{1+\sqrt{2}}{2\sqrt{2}}=\frac{2+\sqrt{2}}{4}.$$
Then, $\cos(\pi/8)=\frac{\sqrt{2+\sqrt{2}}}{2}$.
Rewrite $\sin(\pi/8)$ by making use of the half-angle formula for the sine function, $\cos(x)=1-2\sin^2(x/2)$:
$$\cos(\pi/4)=1-2\sin^2(\pi/8)\rightarrow \sin^2(\pi/8)=\frac{1-\cos(\pi/4)}{2}=\frac{1-\sqrt{2}}{2\sqrt{2}}=\frac{2-\sqrt{2}}{4}.$$
Then, $\sin(\pi/8)=\frac{\sqrt{2-\sqrt{2}}}{2}$.
Substitution into the expression gives
$$I_1-i\cdot I_2=\frac{1}{\sqrt{\sqrt{2}}}\cdot (\frac{\sqrt{2+\sqrt{2}}}{2}-i\cdot \frac{\sqrt{2-\sqrt{2}}}{2})\cdot \sqrt{\pi}= (\frac{\sqrt{\sqrt{2}+1}}{2}-i\cdot \frac{\sqrt{\sqrt{2}-1}}{2})\cdot \sqrt{\pi}.$$ 
After expanding the terms it follows that
$$I_1=\frac{\sqrt{\sqrt{2}+1}}{2}\cdot \sqrt{\pi},\space I_2=\frac{\sqrt{\sqrt{2}-1}}{2}\cdot \sqrt{\pi}.$$
The integral you wanted to evaluate is $I_1$.
