How to evaluate the following integral which cannot be simply evaluated by residue theorem.

Question

Evaluate $$\int_{-\infty}^{+\infty} \frac{e^{-x^2}}{x-i\omega} dx$$ where $\omega$ is real and $\omega\ne 0$.

Original image

Answer 1

We can use the Fourier transform: $$\mathcal Ff(\xi)=\int_{\mathbb R}f(x)e^{-2i\pi\xi x}dx$$ Then let $f(x)=e^{-x^2}$ and $g(x)=\frac 1 {x+i\omega}$. Then, by Plancherel's theorem, $$\int_{\mathbb R}f(x)\overline{g(x)}dx=\int_{\mathbb R}\mathcal F(f)(\xi)\overline{\mathcal F(g)(\xi)}d\xi$$ Let's assume first that $\omega > 0$ (to compute the value for a negative $\omega$, replace $\omega$ by $-\omega$ and conjugate the result). Now, it's well known that $$\mathcal F(f)(\xi)=\sqrt{\pi}e^{-\pi^2\xi^2} \quad\text{and}\quad \mathcal F(g)(\xi)=-2\pi ie^{-2\pi\omega\xi}H(\xi)$$ where $H$ is the Heaviside step function. Therefore, $$\begin{split} \int_{\mathbb R}\frac{e^{-x^2}}{x-i\omega}dx &= 2i\pi^{\frac 3 2}\int_0^{+\infty}e^{-\pi^2\xi^2-2\pi\omega\xi}d\xi\\ &=2\pi^{\frac 3 2}e^{\omega^2}\int_0^{+\infty}e^{-(\pi\xi+\omega)^2}d\xi\\ &=2i\sqrt{\pi}e^{\omega^2}\int_{\omega}^{+\infty}e^{-u^2}du\\ &=2i\sqrt{\pi}e^{\omega^2}\left(\frac{\sqrt{\pi}}2\text{erfc}(\omega)\right) \end{split}$$ where erfc is the complementary error function. Remembering that for negative $\omega$, we can replace by $-\omega$ and conjugate, we conclude that

$$\boxed{ \int_{\mathbb R}\frac{e^{-x^2}}{x-i\omega}dx =\text{sgn}(\omega)i\pi e^{\omega^2}\text{erfc}(|\omega|)}$$

Answer 2

First, note that $\frac{1}{x-i\omega}=\frac{x+i\omega}{x^2+\omega^2}$. Hence, exploiting symmetry we find that for $\omega\ne0$

$$\int_{-\infty}^\infty \frac{e^{-x^2}}{x-i\omega}\,dx=i2\omega \int_0^\infty \frac{e^{-x^2}}{x^2+\omega^2}\,dx\tag1$$

Next, we let $F(a)$ be given by

$$F(a)=\int_0^\infty \frac{e^{-ax^2}}{x^2+\omega^2}\,dx\tag2$$

Differentiating $(2)$ with respect to $a$ reveals

$$\begin{align} F'(a)&=-\int_0^\infty \frac{x^2e^{-ax^2}}{x^2+\omega^2}\,dx\\\\ &=-\int_0^\infty \frac{(x^2+\omega^2-\omega^2)e^{-ax^2}}{x^2+\omega^2}\,dx\\\\ &=-\int_0^\infty e^{-ax^2}\,dx+\omega^2 F(a) \end{align}$$

Therefore, $F(a)$ satisfies the ODE

$$F'(a)-\omega^2 F(a)=-\frac{\sqrt\pi}{2\sqrt a}\tag3$$

with $F(0)=\frac{\pi}{2|\omega|}$.

It is straightforward to find the solution of $(3)$ with the given initial condition is can be written

$$F(a)=-\frac{\sqrt \pi}{2}e^{\omega^2 x}\int_0^a \frac{e^{-\omega^2 x}}{\sqrt{x}}\,dx+\frac{\pi}{2|\omega|}e^{\omega^2 x}\tag4$$

Enforcing the substitution $x\to x^2/\omega^2$ in the integral on the right-hand side of $(4)$ yields

$$\begin{align} F(a)&=-\frac{\sqrt \pi}{\omega}e^{\omega^2 a}\text{sgn}(\omega)\int_0^{|\omega| \sqrt a} e^{-x^2}\,dx+\frac{\pi}{2|\omega|}e^{\omega^2 a}\\\\ &=-\frac{\pi}{2\omega}e^{\omega^2 a}\text{sgn}(\omega)\text{erf}(|\omega |\sqrt a)+\frac{\pi}{2|\omega|}e^{\omega^2 a}\tag5 \end{align}$$

Setting $a=1$ in $(5)$ and substituting the result into $(1)$, we arrive at the identity for $\omega\ne0$

$$\int_{-\infty}^\infty \frac{e^{-x^2}}{x-i\omega}\,dx=i\pi e^{\omega^2}\text{sgn}(\omega)\left(1-\text{erf}(|\omega|)\right)=i\pi e^{\omega^2}\text{sgn}(\omega)\text{erfc}(|\omega|)$$

Answer 3

$$ \begin{align} \int_{-\infty}^\infty\frac{e^{-x^2}}{x-i\omega}\,\mathrm{d}x &=\int_{-\infty}^\infty\frac{(x+i\omega)\,e^{-x^2}}{x^2+\omega^2}\,\mathrm{d}x\tag1\\ &=i\omega\int_{-\infty}^\infty\frac{\,e^{-x^2}}{x^2+\omega^2}\,\mathrm{d}x\tag2\\ &=i\int_{-\infty}^\infty\frac{\,e^{-\omega^2x^2}}{x^2+1}\,\mathrm{d}x\tag3 \end{align} $$ Explanation:
$(1)$: make denominator real
$(2)$: use symmetry; at this point, the function is odd
$(3)$: substitute $x\mapsto\omega x$; this makes the function even

For $\omega\gt0$, define $$ I(\omega)=\int_{-\infty}^\infty\frac{\,e^{-\omega^2x^2}}{x^2+1}\,\mathrm{d}x\tag4 $$ Then, we have $$ I'(\omega)-2\omega I(\omega)=-2\sqrt\pi\tag5 $$ with an integrating factor of $e^{-\omega^2}$, we get $$ \left[e^{-\omega^2}I(\omega)\right]'=-2\sqrt\pi e^{-\omega^2}\tag6 $$ Integrating, we have $$\newcommand{\erfc}{\operatorname{erfc}}\newcommand{\sgn}{\operatorname{sgn}} I(\omega)=\pi e^{\omega^2}\erfc(\omega)\tag7 $$ Making the function odd yields, for all $\omega\ne0$, $$ \bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty\frac{e^{-x^2}}{x-i\omega}\,\mathrm{d}x=i\sgn(\omega)\,\pi e^{\omega^2}\erfc(|\omega|)}\tag8 $$ Note that $\sgn(0)=0$ also gives the proper Cauchy Principal Value for $\omega=0$.