Substituting a real variable for a complex variable (integration by substitution) If I have an integral such as:
$$\int_{ - \infty}^{ \infty} e^{-(x+i)^2}dx$$
Now of course there is the weird substitution $u = x + i$.
Does this substitution always work?
If not when does it not work?
 A: This is valid, but you have to be careful. We need some complex analysis to do this. Consider the function $f : \mathbb C \rightarrow \mathbb C$ such that $f(z) = e^{-z^2}$, and let the contour $\gamma_R$ be defined by $\gamma_R : [-R,R] \rightarrow \mathbb C, \gamma_R(t)=t+i$. Then we can say $$\int_{-\infty}^{\infty}f(t+i)dt=\lim_{R\to\infty}\int_{\gamma_R}f(z)dz,$$ where the former is the integral you want. So, if interpreted as an integral over a contour, then yes, the substitution is valid, but it has to be treated with the proper formalism of complex analysis. You cannot non-chalantly replace $u=x+i$ as you would in a real-analytic calculus course, since then you would have the problem of giving meaning to the bounds $-\infty+i$ and $\infty+i$, which you cannot really do in a way that is consistent.
A: This happens to work in this case because the integral $\int_\gamma e^{-z^2} \, dz$ over any closed curve is zero, according to Cauchy's integral theorem.
If you choose $\gamma$ as the boundary of the rectangle with corners $-R, +R, +R+i, -R+i$ then
$$
 0 = \int_{-R}^R e^{-x^2} \, dx + \int_0^1 e^{(R+iy)^2} i\, dy - \int_{-R}^R e^{-(x+i)^2} \, dx - \int_0^1 e^{(-R+iy)^2} i\, dy \, .
$$
The integrals of the “vertical” parts of the curve converge to zero for $R \to \infty$, and that is why
$$
 \int_{ - \infty}^{ \infty} e^{-x^2} \, dx = \lim_{R \to \infty} \int_{-R}^R e^{-x^2} \, dx = \lim_{R \to \infty}\int_{-R}^R e^{-(x+i)^2} \, dx
= \int_{ - \infty}^{ \infty} e^{-(x+i)^2} \, dx \, .
$$
