Integral Substitution with complex numbers

Question

I want to show that $\forall k,x\in \mathbb{R}$ $$ \int_{\mathbb{R}}e^{-(x+ik)^2/2}\mathrm{d}x=\int_{\mathbb{R}}e^{-(k-ix)^2/2}\mathrm{d}k $$ Can I just $u$ substitute these two integrals? Is it allowed to let $$ u=x+ik\Rightarrow \frac{du}{dx}=1 $$ $$ v=k-ix\Rightarrow\frac{dv}{dk}=1 $$ $$ \Rightarrow \int_{\mathbb{R}}e^{-(x+ik)^2/2}\mathrm{d}x=\int_{\mathbb{R}}e^{-u^2/2}\mathrm{d}u=\int_{\mathbb{R}}e^{-v^2/2}\mathrm{d}v=\int_{\mathbb{R}}e^{-(k-ix)^2/2}\mathrm{d}k? $$

Answer 1

One needs to apply a bit of caution upon the change of variables. Note that we can write

$$\begin{align} \int_{\mathscr{R}}e^{-(x+ik)^2/2}\,dx&=\lim_{(-L,U)\to (-\infty,\infty)}\int_{-L}^U e^{-(x+ik)^2/2}\,dx\\\\ &=\lim_{(-L,U)\to (-\infty,\infty)}\int_{-L+ik}^{U+ik} e^{-x^2/2}\,dx\tag1 \end{align}$$

Tp proceed, the key is applying Cauchy's Integral Theorem to $(1)$ to deform the contour from $-L+ik$ to $U+ik$ to the straight-line contour from $-L$ to $U$. Proceeding accordingly yields

$$\int_{\mathscr{R}}e^{-(x+ik)^2/2}\,dx=\lim_{(-L,U)\to (-\infty,\infty)}\int_{-L}^{U} e^{-x^2/2}\,dx=\sqrt{2\pi}$$

and we are done!

Answer 2

Note that $I=\int e^{-{1 \over 2} (x+ik)^2} dx = \int e^{-{1 \over 2} (-t+ik)^2} dt= \int e^{-{1 \over 2} (-x+ik)^2} dx$ using the substitution $t=-x$. Since $I = \int e^{-{1 \over 2} (x-ik)^2} dx$ (since $y^2=(-y)^2$), you have the desired result.