$u_{tt}=u_{xx}$ given $u_0(x)=e^{-x^2}$
I know that I just need to follow the formula $$u(x,t)=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}e^{\frac{-(x-y)^2}{4t}}u_0(y)dy $$
but I get to a point in the integration where I'm not sure how to proceed. $$\frac{e^{-x^2}}{\sqrt{4\pi t}}\int_{\infty}^{\infty}e^{2xy-2y^2}dy$$
Thanks for the help
you have to complete the squares in the exponent
$$ 2xy-2y^2=-2(y^2-xy)=-2((y-\frac{x}{2})^2-\frac{x^2}{4})$$
Then use properties of exponensiation separete the exponent $\frac{x^2}{4}$ in order to take it outside the integral and proced to integrate only
$$ \int e^{-2(y-\frac{x}{2})^2 $$
which easier.