complex integrationcomplex-analysisgaussian-integralimproper-integrals
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.
Proposition. Given a symmetric complex $n\times n$ matrix $S=S^T\in {\rm Mat}_{n\times n}(\mathbb{C})$ with real part ${\rm Re}(S)>0$ positive definite, then (i) the matrix $S$ is invertible, and (ii) the Gaussian integral is $$ I~:=~\int_{\mathbb{R}^n} \! d^nx ~e^{-\frac{1}{2}x^T Sx} ~=~\sqrt{\frac{(2\pi)^n}{\det S }}. \tag{1}$$
Remark. Note that the condition$^1$ $${\rm Re}(S)~>~0\qquad\Leftrightarrow\qquad \forall x~\in~\mathbb{R}^n\backslash\{0\}:~~x^T{\rm Re}(S) x~>~0 \tag{2}$$ ensures that the integrand (1) is Lebesgue integrable.
Remark. We will below give a proof of the proposition that does not rely on analytic continuation of the $S$ matrix, as requested by OP.
Induction proof of the proposition. The case $n=1$ is e.g. proven in my Phys.SE answer here. Next consider the induction step. Write $$ S~=~\begin{pmatrix} a & b^T \cr b & C \end{pmatrix}, \qquad a~\in~\mathbb{C}, \quad b~\in~\mathbb{C}^{n-1}, \quad C~\in~{\rm Mat}_{n-1 \times n-1}(\mathbb{C}). \tag{3}$$ It follows that $${\rm Re}(a)~\stackrel{(2)+(3)}{>}~0, \tag{4}$$ which implies that $a\neq 0$. Define $$ \gamma~:=~\alpha + i \beta~:=~b/a~\in~\mathbb{C}^{n-1}, \qquad \alpha , \beta~\in~\mathbb{R}^{n-1}, \tag{5}$$ $$ S^{\prime}~:=~C-\gamma a \gamma^T ~\in~{\rm Mat}_{n-1 \times n-1}(\mathbb{C}), \tag{6}$$ and $$ \widetilde{S}^{\prime}~:=~S^{\prime}-\beta a \beta^T ~\in~{\rm Mat}_{n-1 \times n-1}(\mathbb{C}). \tag{7}$$ Then $$ S~\stackrel{(3)+(5)+(6)}{=}~\begin{pmatrix} 1 & 0 \cr \gamma & \mathbb{1} \end{pmatrix} \begin{pmatrix} a & 0 \cr 0 & S^{\prime} \end{pmatrix} \begin{pmatrix} 1 & \gamma^T \cr 0 & \mathbb{1} \end{pmatrix} ~\stackrel{(9)}{=}~\begin{pmatrix} 1 & 0 \cr \alpha & \mathbb{1} \end{pmatrix} \widetilde{S} \begin{pmatrix} 1 & \alpha^T \cr 0 & \mathbb{1} \end{pmatrix}, \tag{8}$$ where $$ \widetilde{S}~:=~\begin{pmatrix} 1 & 0 \cr i\beta & \mathbb{1} \end{pmatrix} \begin{pmatrix} a & 0 \cr 0 & S^{\prime} \end{pmatrix} \begin{pmatrix} 1 & i\beta^T \cr 0 & \mathbb{1} \end{pmatrix} ~\stackrel{(7)}{=}~\begin{pmatrix} a & i\beta^T \cr i\beta & \widetilde{S}^{\prime} \end{pmatrix} . \tag{9}$$ Eq. (8) implies that ${\rm Re}(\widetilde{S})>0$. Eq. (9) then implies that ${\rm Re}(\widetilde{S}^{\prime})>0$. Eq. (4) implies that ${\rm Re}(\beta a \beta^T)\geq 0$. Eq. (7) then implies that ${\rm Re}({S}^{\prime})>0$. [Here we have used that (semi)positive-definite matrices form a convex cone.] By induction we then know that ${S}^{\prime}$ is invertible. Eq. (8) then implies that $S$ is invertible. The integral becomes $$ \begin{align}I~\stackrel{(1)+(8)}{=}&~\int_{\mathbb{R}^n} \! d^nx ~e^{-\frac{1}{2}x^T \widetilde{S}x}\cr ~\stackrel{(9)}{=}~&\int_{\mathbb{R}^{n-1}} \! d^{n-1}x^{\prime} ~e^{-\frac{1}{2}x^{\prime T} S^{\prime}x^{\prime}} \int_{\mathbb{R}+i\beta^Tx^{\prime}} \! dz~e^{-\frac{1}{2}az^2}\cr ~=~&\int_{\mathbb{R}^{n-1}} \! d^{n-1}x^{\prime} ~e^{-\frac{1}{2}x^{\prime T} S^{\prime}x^{\prime}} \int_{\mathbb{R}} \! dz~e^{-\frac{1}{2}az^2}\cr ~=~&\sqrt{\frac{(2\pi)^{n-1}}{\det S^{\prime}}} \sqrt{\frac{2\pi}{a}}\cr ~\stackrel{(8)}{=}~&\sqrt{\frac{(2\pi)^n}{\det S}} . \end{align}\tag{10}$$ To perform the 1-dimensional $z$-integral, we have used the fact that it is possible to shift the horizontal integration contour back to the real axis $\mathbb{R}$ without changing the integral value, cf. Cauchy's integral theorem. $\Box$
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$^1$In this answer, the notation $M>0$ means a positive-definite matrix $M$; not a matrix with positive entries.
Since $b$ is purely imaginary, this is a Fourier Transform.
Assuming $a>0$, $c>0$ ,and $\Re(b) = 0$; make the substitution $- \pi s = \Im (b)$ or equivalently $-i\pi s =b$ or $\pi s = ib$
$$\begin{align*}\displaystyle & \int_{-\infty}^{\infty} \frac{\exp\left[-a\left(x-b\right)^2\right]}{1+cx^2} dx\\ \\ &= \int_{-\infty}^{\infty} \frac{\exp\left[-a\left(x+i\pi s\right)^2\right]}{1+cx^2} dx\\ \\ &= \int_{-\infty}^{\infty} \frac{\exp\left[-ax^2+a(\pi s)^2-2\pi i axs\right]}{cx^2+1} dx\\ \\ &= \frac{a^2}{c}e^{a(\pi s)^2}\int_{-\infty}^{\infty} \frac{e^{-\frac{1}{a}(ax)^2}}{(ax)^2+\frac{a^2}{c}}e^{-2\pi i (ax)s} dx\\ \\ &= \frac{a}{c}e^{a(\pi s)^2}\int_{-\infty}^{\infty} \frac{e^{-\frac{1}{a}y^2}}{y^2+\frac{a^2}{c}}e^{-2\pi iys} dy\\ \\ &= \frac{a}{c}e^{a(\pi s)^2}(2\pi)^2\frac{1}{2\left(\frac{2\pi a}{\sqrt{c}}\right)}\int_{-\infty}^{\infty} \frac{2\left(\frac{2\pi a}{\sqrt{c}}\right)}{(2\pi y)^2+\left(\frac{2\pi a}{\sqrt{c}}\right)^2}\space e^{-\pi^2\left(\frac{y}{\pi\sqrt{a}}\right)^2}\space e^{-2\pi iys} dy\\ \\ &= \frac{\pi}{\sqrt{c}}e^{a(\pi s)^2}\mathscr{F}\left\{ \frac{2\left(\frac{2\pi a}{\sqrt{c}}\right)}{(2\pi y)^2+\left(\frac{2\pi a}{\sqrt{c}}\right)^2}\space e^{-\pi\left(\frac{y}{\sqrt{\pi a}}\right)^2}\right\} \\ \\ &= \frac{\pi}{\sqrt{c}}e^{a(\pi s)^2}\left[\mathscr{F}\left\{ \frac{2\left(\frac{2\pi a}{\sqrt{c}}\right)}{(2\pi y)^2+\left(\frac{2\pi a}{\sqrt{c}}\right)^2}\right\} * \mathscr{F}\left\{ e^{-\pi\left(\frac{y}{\sqrt{\pi a}}\right)^2}\right\} \right]\\ \\ &= \frac{\pi}{\sqrt{c}}e^{a(\pi s)^2}\left[e^{-\frac{2 a}{\sqrt{c}}|\pi s|} * \sqrt{\pi a} e^{-a\left(\pi s\right)^2}\right] \\ \\ &= \pi \sqrt{\frac{\pi a}{c}}e^{a(\pi s)^2}\int_{-\infty}^{\infty}e^{-\frac{2 a}{\sqrt{c}}|\pi \tau|} e^{-a\left(\pi s -\pi \tau\right)^2}\space d\tau \\ \\ &= \pi \sqrt{\frac{\pi a}{c}}e^{a(\pi s)^2}\left[\int_{-\infty}^{0}e^{\frac{2 a}{\sqrt{c}}\pi \tau} e^{-a\left(\pi s -\pi \tau\right)^2}\space d\tau +\int_{0}^{\infty}e^{-\frac{2 a}{\sqrt{c}}\pi \tau} e^{-a\left(\pi s -\pi \tau\right)^2}\space d\tau \right]\\ \\ &= \pi \sqrt{\frac{\pi a}{c}}\left(\int_{-\infty}^{0}\exp\left[-a\left([\pi\tau]^2-2\left[\pi s+\frac{1}{\sqrt{c}}\right]\pi \tau\right)\right]\space d\tau +\int_{0}^{\infty}\exp\left[-a\left([\pi\tau]^2-2\left[\pi s-\frac{1}{\sqrt{c}}\right]\pi \tau\right)\right]\space d\tau \right)\\ \\ &= \pi \sqrt{\frac{\pi a}{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\int_{-\infty}^{0}\exp\left[-a\left(\pi\tau-\left[\pi s+\frac{1}{\sqrt{c}}\right]\right)^2\right]\space d\tau +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\int_{0}^{\infty}\exp\left[-a\left(\pi\tau-\left[\pi s-\frac{1}{\sqrt{c}}\right]\right)^2\right]\space d\tau \right)\\ \\ &= \pi \sqrt{\frac{\pi a}{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\frac{1}{\pi\sqrt{a}}\int_{-\infty}^{-\sqrt{a}\left(\pi s+\frac{1}{\sqrt{c}}\right)}e^{-u^2}\space du +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\frac{1}{\pi\sqrt{a}}\int_{-\sqrt{a}\left(\pi s-\frac{1}{\sqrt{c}}\right)}^{\infty}e^{-u^2}\space du\right)\\ \\ &= \sqrt{\frac{\pi}{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\int_{-\infty}^{-\sqrt{a}\left(\pi s+\frac{1}{\sqrt{c}}\right)}e^{-u^2}\space du +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\int_{-\sqrt{a}\left(\pi s-\frac{1}{\sqrt{c}}\right)}^{\infty}e^{-u^2}\space du\right)\\ \\ &= \frac{\pi}{2}\frac{1}{\sqrt{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\mathrm{erf}(u)\biggr{|}_{-\infty}^{-\sqrt{a}\left(\pi s+\frac{1}{\sqrt{c}}\right)} +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\mathrm{erf}(u)\biggr{|}_{-\sqrt{a}\left(\pi s-\frac{1}{\sqrt{c}}\right)}^{\infty}\right)\\ \\ &= \frac{\pi}{2}\frac{1}{\sqrt{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\left[\mathrm{erf}\left(-\sqrt{a}\left[\pi s+\frac{1}{\sqrt{c}}\right]\right)+1\right] +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\left[1 -\mathrm{erf}\left(-\sqrt{a}\left[\pi s-\frac{1}{\sqrt{c}}\right]\right)\right]\right)\\ \\ &= \frac{\pi}{2}\frac{1}{\sqrt{c}}\left(e^{\left[-\sqrt{a}\left(ib+\frac{1}{\sqrt{c}}\right)\right]^2}\left[\mathrm{erf}\left(-\sqrt{a}\left[ib+\frac{1}{\sqrt{c}}\right]\right)+1\right] +e^{\left[-\sqrt{a}\left(ib-\frac{1}{\sqrt{c}}\right)\right]^2}\left[1 -\mathrm{erf}\left(-\sqrt{a}\left[ib-\frac{1}{\sqrt{c}}\right]\right)\right]\right)\\ \end{align*}$$
Best Answer
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, $$\int^\infty_{-\infty}\exp(-(x+i)^2)dx=\lim_{T\to+\infty}\int^{T+i}_{-T+i}\exp(-u^2)du=\lim_{T\to+\infty}\int^T_{-T}\exp(-u^2)du=\sqrt\pi$$