Help with integral $\int_{-\infty}^\infty e^{iαx-α^2 t} dα$ Solve by integration
$$\int_{-\infty}^\infty e^{-iαx-α^2 t}   dα$$
solve by integration , this integral is - infinity to infinity and exponent value

This question has a short answer and a long one.
completing the square you have:
$$
e^{−iαx−α^2t}=e^{-t(\alpha+\frac{ix}{2t})} \,e^{\frac{-x^2}{4t}} \\
\Rightarrow \int_{-\infty}^{\infty} e^{−iαx−α^2t}\, d \alpha=e^{\frac{-x^2}{4t}} \, \int_{-\infty}^{\infty}e^{-t(\alpha+\frac{ix}{2t})^2} \,d\alpha \\=e^{\frac{-x^2}{4t}} \, \int_{-\infty}^{\infty}e^{-t\alpha^2} \,d\alpha \\=e^{\frac{-x^2}{4t}} \sqrt{\frac{\pi}{t}}
$$
But in the complete answer, we have to prove that
$$
\int_{-\infty}^{\infty}e^{-t(\alpha+b)^2} \,d\alpha=\int_{-\infty}^{\infty}e^{-t\alpha^2} \,d\alpha
$$
For that we can just use a change of variable, but $b$ is imaginary and that means that the limits of integration would change to a line in the complex plane.
So to prove that we need another approach and what is commonly done is to use the fact $e^{z^2}$ is an entire function, so any closed line integral in the complex plane will be $0$. So you choose a rectangle whose segments parallel to the real line go from $-\infty$ to $\infty$ , and the other segments go from $0$ to some $ic$. Then you solve the 4 integrals for each segment and using that the sum of the 4 of them is $0$, you should get the desired result, that is, it is fine to complete the squares even if you have to deal with an imaginary term.
