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

Solution 1:

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.

Solution 2:

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 $(12)$.