Characteristic Function of Gaussian/ Change of Variables in Complex Plane

Question

In the process of finding the characteristic function of a Guassian random variable, one must evaluate the following integral: $$ \int_{-\infty}^{\infty} e^{-(x- it)^2} dx. $$ One way to evaluate this is to use a contour integral over a rectangle and then use the residue theorem. This is presumably rigorous. Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line. Note that $dy/dx = 1$. My question is whether this method is rigorous or can be rigorously justified?

Answer 1

Here is one approach that does not require a change of variables and contour deformation. First we write

$$\begin{align} \int_{-\infty}^\infty e^{-(x-it)^2}\,dx&=e^{t^2}\int_{-\infty}^\infty e^{-x^2+i2xt}\,dx\\\\ &=e^{t^2}\int_{-\infty}^\infty e^{-x^2}\cos(2tx)\,dx\tag1 \end{align}$$

Let $f(t)$ be given by the integral

$$f(t)=\int_{-\infty}^\infty e^{-x^2}\cos(2tx)\,dx\tag2$$

Differentiation reveals of $(2)$

$$f'(t)=\int_{-\infty}^\infty (-2xe^{-x^2})\sin(2tx)\,dx\tag3$$

Integrating by parts the integral in $(3)$ with $u=\sin(2tx)$ and $v=e^{-x^2}$ we obtain the ODE

$$f'(t)=-2tf(t) \tag4$$

Using $f(0)=\sqrt \pi$, we find from $(4)$ that

$$f(t)=\sqrt\pi e^{-t^2}\tag5$$

Substituting $(5)$ into $(1)$ yields the coveted result

$$\int_{-\infty}^\infty e^{-(x-it)^2}\,dx=\sqrt \pi$$

Answer 2

Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line.

Unless $t = 0$, the substitution transforms the integral into an integral over the horizontal line $\operatorname{Im}(z) = -t$, not over the real line. This method is really part of the contour integral method. If $t > 0$, we consider the contour integral $\displaystyle\oint_{\Gamma(R)} e^{-z^2}\, dz$, where $\Gamma(R)$ is a clockwise-oriented rectangle in the complex plane with vertices at $-R, R, R - it$, and $-R - it$. The integrals of $e^{-z^2}$ along the vertical edges can be shown to be $O(e^{-R^2})$ as $R \to \infty$. Since $e^{-z^2}$ is entire, Cauchy's theorem gives $\displaystyle\oint_{\Gamma(R)} e^{-z^2}\, dz = 0$. Hence $\displaystyle\lim_{R\to \infty} \int_{-R - it}^{R - it} e^{-z^2}\, dz = \lim_{R\to \infty} \int_{-R}^R e^{-x^2}\, dx$, or $$\int_{-\infty}^\infty e^{-(x-it)^2}\, dx = \int_{-\infty}^\infty e^{-x^2}\, dx$$A similar argument holds when $t < 0$.