Calculate the following formula by using complex integration. $$\int _{-\infty} ^{+\infty} \exp(-(x+i)^2) dx$$
It looks similar to the Gaussian-Integral $\int _{-\infty} ^{+\infty} \exp(-x^2) dx = \sqrt{\pi}$.
I tired divide it into real part and imaginary part by using Euler Formula, but ended up with $$\int _{-\infty} ^{+\infty} \exp(-x^2)\cos(2x)dx - i\int _{-\infty} ^{+\infty} \exp(-x^2)\sin(2x) dx$$ the $\sin$ and $\cos$ are too tricky for me to solve.
On
$\int_{-\infty} ^{\infty} e^{-(x+i)^{2}}dx=e\int_{-\infty} ^{\infty}e^{-x^{2}} e^{-2ix}dx$. Put $y =\sqrt 2 x$ and use the following well known formula for the characteristic function of Gaussian distribution
$\frac 1 {\sqrt {2\pi}} \int_{-\infty} ^{\infty} e^{-x^{2}/2} e^{itx}dx=e^{-t^{2}/2}$.
As you requested ‘using complex integration’, I think using Cauchy’s integral theorem is what you are expecting.
By the substitution $u=x+i$, $$\int^\infty_{-\infty}\exp(-(x+i)^2)dx=\lim_{T\to+\infty}\int^{T+i}_{-T+i}\exp(-u^2)du$$
Consider a rectangular contour $C$, with vertexes at $(T,0),(-T,0),(T,1),(-T,1)$.
By Cauchy’s integral theorem, $$\oint_C \exp(-z^2)dz=0$$
As an exercise, prove by youself that the intergrals along the vertical lines vanishes as $T\to \infty$.
Then, $$\int^\infty_{-\infty}\exp(-(x+i)^2)dx=\lim_{T\to+\infty}\int^{T+i}_{-T+i}\exp(-u^2)du=\lim_{T\to+\infty}\int^T_{-T}\exp(-u^2)du=\sqrt\pi$$