Arc contribution in $\int_{-\infty}^\infty \mathrm{d}z \frac{e^{-z^2}}{z-1}$

Question

Consider an improper integral with a pole on the integration contour at say $z=1$,

$$ \tag{1} I = \int_{-\infty}^\infty \mathrm{d}z\ \frac{e^{-z^2}}{z-1+i\epsilon},~~~~~\epsilon>0. $$ Let $$f(z) = \frac{e^{-z^2}}{z-1+i\epsilon}$$ then $$ \sum_{residues~inside~\Gamma} = 0 = \oint_\Gamma f(z) = I+\left(\int_{\Gamma_\epsilon}+\int_{\Gamma_\infty} \right) f(z), $$ where the total contour is $\Gamma\equiv (-R,R)+\Gamma_\epsilon+\Gamma_\infty$ with $R\rightarrow \infty$.

Thus $$ I = - \left(\int_{\Gamma_\epsilon}+\int_{\Gamma_\infty} \right) f(z). $$ The contour $\Gamma_\epsilon$ is a semicircle centered about $z = 1$ of radius $\epsilon$. Its contribution is given by $$ \int_{\Gamma_\epsilon} \mathrm{d}z ~f(z) = i (\theta_2-\theta_1)~ \mathrm{Res}(f;z=1) = \frac{-i\pi}{e}. $$

Evaluating $(1)$ in Mathematica and taking the $\epsilon\rightarrow 0 $ limit gives $$ I = e^{(\epsilon +i)^2} \left(-\pi \text{erfi}(1-i \epsilon )+\log (-1+i \epsilon )+\log \left(\frac{i}{\epsilon +i}\right)-2 i \pi \right) \\ \longrightarrow -\frac{\pi (\text{erfi}(1)+i)}{e}~~~(\text{as}~ \epsilon \rightarrow 0). $$

Thus apparently, $$ \tag{2} \int_{\Gamma_\infty} \mathrm{d}z \, f(z) = \frac{2\pi i}{e}+\frac{\mathrm{erfi}(1)}{e}. $$

Can anyone derive this contribution from the semicircle at infinity? I.e. is $(2)$ correct and how about generalizations of $(1)$ to integrals of the form

$$\tag{3} I = \int_{-\infty}^\infty \mathrm{d}z\ \frac{z^n e^{-z^2}}{(z-a+i\epsilon)(z-b-i\epsilon)},~~~~~\epsilon>0,a,b\in\mathbb{R},n\in \mathbb{N}. $$

---

Note, erfi is defined as $\mathrm{erfi}(z) \equiv \mathrm{erf}(iz)/i$ with the familiar error function.

Answer 1

The contribution from the (superfluous) deformation of the segment on $[0,2]$ to a semicircle is cancelled by other parts of the integral along the real axis. This deformation crosses no poles, so the path integral along $\Gamma_\epsilon + [0,2]$ is zero.

So what's going on? In your expression $$ I+\left(\int_{\Gamma_\epsilon}+\int_{\Gamma_\infty} \right) f(z) \text{,} $$ you have two paths from $0$ to $2$. One runs along the real axis and is contributed by $I$. The other runs along $\Gamma_\epsilon$. It should come as no surprise that this contribution is overcounted in the result you have.

Answer 2

To check your work: if you care about a fast solution, start with considering $I(a)= \int_{-\infty}^{\infty}\frac{e^{-a(x^2-1)}}{x-1}\textrm{dx}$, where by differentiating with respect to $a$ and then integrating back, we're lead to $I(1)=-\sqrt{\pi}\int_0^1 \frac{e^a}{\sqrt{a}}\textrm{d}a=-\pi\operatorname{erfi}(1).$ Hence, we have $$\int_{-\infty}^{\infty}\frac{e^{-x^2}}{x-1}\textrm{dx}=-\frac{\pi}{e}\operatorname{erfi}(1).$$

Note: the integrals above have a meaning only in the CPV sense as seen described at https://en.wikipedia.org/wiki/Cauchy_principal_value.

Literature: related to Faddeeva function (it's sometimes referred to as the plasma dispersion function) - https://en.wikipedia.org/wiki/Faddeeva_function.

Answer 3

Near the top of the arc, the integrand blows up exponentially. I would avoid using that arc.

---

Real Method

By substituting $z\mapsto-z$, we get $$ \operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2}}{z-1}\mathrm{d}z =-\operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2}}{z+1}\mathrm{d}z\tag1 $$ Therefore, $$ \begin{align} \operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2-1}e^{-2z}}{z}\mathrm{d}z &=-\operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2-1}e^{2z}}{z}\mathrm{d}z\tag2\\ &=-\frac1e\int_{-\infty}^\infty e^{-z^2}\frac{\sinh(2z)}z\,\mathrm{d}z\tag3 \end{align} $$ Explanation:
$(2)$: substitute $z\mapsto z+1$ on the left and $z\mapsto z-1$ on the right of $(1)$
$(3)$: average the right and left of $(2)$

Setting $$ f(a)=\int_{-\infty}^\infty e^{-z^2}\frac{\sinh(az)}z\,\mathrm{d}z\tag4 $$ we have $f(0)=0$ and $$ \begin{align} f'(a) &=\int_{-\infty}^\infty e^{-z^2}\cosh(az)\,\mathrm{d}z\tag5\\ &=\int_{-\infty}^\infty e^{-z^2}e^{az}\,\mathrm{d}z\tag6\\ &=e^{a^2/4}\int_{-\infty}^\infty e^{-z^2}\,\mathrm{d}z\tag7\\[3pt] &=\sqrt\pi\,e^{a^2/4}\tag8 \end{align} $$ Explanation:
$(5)$: take the derivative under the integral
$(6)$: $\cosh(ax)$ is the even part of $e^{ax}$
$(7)$: substitute $z\mapsto z+a/2$
$(8)$: evaluate the integral

Thus, $$ \begin{align} \operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2}}{z-1}\mathrm{d}z &=-\frac1e\int_{-\infty}^\infty e^{-z^2}\frac{\sinh(2z)}z\,\mathrm{d}z\tag9\\ &=-\frac{\sqrt\pi}e\int_0^2e^{a^2/4}\,\mathrm{d}a\tag{10}\\ &=-\frac{2\sqrt\pi}e\int_0^1e^{a^2}\,\mathrm{d}a\tag{11}\\[3pt] &=-\frac\pi{e}\,\operatorname{erfi}(1)\tag{12} \end{align} $$ Explanation:
$\phantom{0}(9)$: apply $(3)$
$(10)$: apply $(8)$
$(11)$: substitute $a\mapsto2a$
$(12)$: evaluate the integral

---

A Cleaner, But Still Real, Approach $$ \begin{align} \mathrm{PV}\int_{-\infty}^\infty\frac{e^{-x^2}}{x-1}\,\mathrm{d}x &=-\mathrm{PV}\int_{-\infty}^\infty\frac{e^{-x^2}}{x+1}\,\mathrm{d}x\tag{13}\\ &=\mathrm{PV}\int_{-\infty}^\infty\frac{e^{-x^2}}{x^2-1}\,\mathrm{d}x\tag{14}\\ &=\left.\frac1e\,\mathrm{PV}\int_{-\infty}^\infty\frac{e^{-a(x^2-1)}}{x^2-1}\,\mathrm{d}x\,\right]_{a=1}\tag{15}\\ &=\frac1e\,\mathrm{PV}\int_{-\infty}^\infty\frac1{x^2-1}\,\mathrm{d}x-\frac1e\int_0^1\int_{-\infty}^\infty e^{-a(x^2-1)}\,\mathrm{d}x\,\mathrm{d}a\tag{16}\\ &=0-\frac1e\int_0^1\sqrt{\frac\pi a}\,e^a\,\mathrm{d}a\tag{17}\\ &=-\frac{2\sqrt\pi}e\int_0^1e^{a^2}\,\mathrm{d}a\tag{18}\\ &=-\frac{2\sqrt\pi}e\frac{\sqrt\pi}2\,\operatorname{erfi}(1)\tag{19}\\[6pt] &=-\frac\pi e\,\operatorname{erfi}(1)\tag{20} \end{align} $$ Explanation:
$(13)$: substitute $x\mapsto-x$
$(14)$: average the left and right sides of $(13)$
$(15)$: set up for differentiation under the integral
$(16)$: write as the integral of the derivative under the integral
$(17)$: the PV can be evaluated using a simple contour integration
$\phantom{(17)\text{:}}$ evaluate the inner integral on the right
$(18)$: substitute $a\mapsto a^2$
$(19)$: evaluate the integral
$(20)$: simplify

which agrees with $(12)$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{P.V.}\int_{-\infty}^{\infty}{\expo{-z^{2}} \over z - 1}\,\dd z & = \mrm{P.V.}\int_{-\infty}^{\infty}{\expo{-\pars{z + 1}^{2}} \over z}\,\dd z = \int_{0}^{\infty} {\expo{-\pars{z + 1}^{2}} - \expo{-\pars{z - 1}^{2}} \over z}\,\dd z \\[5mm] & = \sum_{\sigma = \pm 1}\sigma\int_{0}^{\infty} \expo{-\pars{z + \sigma}^{2}}\pars{\int_{0}^{\infty}\expo{-zx}\dd x}\,\dd z \\[5mm] & = \sum_{\sigma = \pm 1}\sigma\int_{0}^{\infty}\int_{0}^{\infty} \exp\pars{-\bracks{z^{2} + 2\sigma z + 1 + xz}}\,\dd z\,\dd x \\[5mm] & = \sum_{\sigma = \pm 1}\sigma\int_{0}^{\infty}\int_{0}^{\infty} \exp\pars{-\bracks{z + \sigma + {x \over 2}}^{2} - 1 + \sigma^{2} + \sigma x + {x^{2} \over 4}}\,\dd z\,\dd x \\[5mm] & = \sum_{\sigma = \pm 1}\sigma\int_{0}^{\infty} \exp\pars{\sigma x + {x^{2} \over 4}}\int_{\sigma + x/2}^{\infty} \exp\pars{-z^{2}}\,\dd z\,\dd x \\[5mm] & = \int_{0}^{\infty}\exp\pars{-z^{2}} \sum_{\sigma = \pm 1}\sigma\int_{0}^{2z - 2\sigma} \exp\pars{{1 \over 4}\braces{\bracks{x + 2\sigma}^{\, 2} - 4\sigma^{2}}} \,\dd x\,\dd z \\[5mm] & = \expo{-1}\int_{0}^{\infty}\exp\pars{-z^{2}} \sum_{\sigma = \pm 1}\sigma\int_{2\sigma}^{2z}\exp\pars{x^{2} \over 4} \,\dd x\,\dd z \\[5mm] & = 2\expo{-1}\int_{0}^{\infty}\exp\pars{-z^{2}} \sum_{\sigma = \pm 1}\sigma\int_{\sigma}^{z}\exp\pars{x^{2}} \,\dd x\,\dd z \\[5mm] & = -2\expo{-1}\int_{0}^{\infty}\exp\pars{-z^{2}} \bracks{\int_{-1}^{z}\exp\pars{x^{2}}\,\dd x - \int_{1}^{z}\exp\pars{x^{2}}\,\dd x}\,\dd z \\[5mm] & = -2\expo{-1}\ \overbrace{\bracks{\int_{0}^{\infty}\exp\pars{-z^{2}}\dd z}} ^{\ds{\root{\pi} \over 2}}\ \overbrace{\int_{-1}^{1}\exp\pars{x^{2}}\,\dd x} ^{\ds{\mbox{Set}\ x = -\ic t}} \\[5mm] & = -2\expo{-1}\root{\pi}\int_{0}^{\ic}\exp\pars{-t^{2}}\pars{-\ic}\,\dd t \\[5mm] & = \ic\pi\expo{-1} \bracks{{2 \over \root{\pi}}\int_{0}^{\ic}\exp\pars{-t^{2}}\dd t} = \bbx{\ic\pi\expo{-1}\mrm{erf}\pars{\ic}} \end{align}

Note that $\ds{\,\mrm{erf}\pars{\ic} = \ic\,\mrm{erfi}\pars{1}}$.

Answer 5

$$I=-pv.\int_{-\infty }^{\infty }\frac{e^{-x^2}}{x-1}dx=-\sum_{n=1}^{\infty }\Gamma (\frac{1-2n}{2})\\ \\ \\ =-\sum_{n=1}^{\infty }(-1)^n\frac{2^n.\sqrt{\pi }}{\prod_{n=1}^{\infty }(2m-1)}=-\sum_{n=1}^{\infty }(-1)^n\frac{2^n\sqrt{\pi }}{\frac{2^n.\Gamma (n+\frac{1}{2})}{\sqrt{\pi }}}\\ \\ \\ \therefore I=-\sum_{n=1}^{\infty }(-1)^n\frac{\pi }{\Gamma (n+\frac{1}{2})}=-\sqrt{\pi }\sum_{n=1}^{\infty }(-1)^n\frac{\Gamma (n).\Gamma (\frac{1}{2})}{\Gamma (n)\Gamma (n+\frac{1}{2})}\\ \\ \\ =-\sqrt{\pi }\sum_{n=1}^{\infty }(-1)^n\int_{0}^{1}\frac{(1-x)^{n-1}}{\sqrt{x}(n-1)!}dx\\ \\ \\ \therefore I=-\frac{\pi }{e}.ierf(i)=\frac{\pi }{e}erfi(1$$