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?
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?
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.
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 $$ \int_{ - \infty}^{ \infty} e^{-x^2} \, dx = \lim_{R \to \infty} \int_{-R}^R e^{-x^2} \, dx = \lim_{R \to \infty}\int_{-R}^R e^{-(x+i)^2} \, dx = \int_{ - \infty}^{ \infty} e^{-(x+i)^2} \, dx \, . $$