integration - Compute an integral about error function $int_{-infty}^{infty} frac{e^{-k^2}}{1-k} mathrm{d}k$

There is an integral
$$\mathcal{P}\int_{-\infty}^{\infty} \frac{e^{-k^2}}{1-k} \mathrm{d}k$$
where $\mathcal{P}$ means Cauchy principal value.

Mathematica gives the result (as the screent shot shows)
$$\mathcal{P}\int_{-\infty}^{\infty} \frac{e^{-k^2}}{1-k} \mathrm{d}k = \frac{\pi}{e}\mathrm{erfi}(1) = \frac{\pi}{e}\cdot \frac{2}{\sqrt{\pi}} \int_0^1 e^{u^2}\mathrm{d}u$$
Mathematica screen shot

where $\mathrm{erfi}(x)$ is imaginary error function define as
$$\mathrm{erfi}(z) = -\mathrm{i}\cdot\mathrm{erf}(\mathrm{i}z)$$
$$\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^{x} e^{-t^2}\mathrm{d}t$$
How can we get the right hand side from left hand side?

Answer

For $a \in \mathbb{R}$ define
$$\begin{align} f(a) &\equiv \mathrm{e}^{a^2} \mathcal{P} \int \limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-k^2}}{a-k} \, \mathrm{d} k = \mathrm{e}^{a^2} \mathcal{P} \int \limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-(x-a)^2}}{x} \, \mathrm{d} x= \lim_{\varepsilon \to 0^+} \left[\int \limits_{-\infty}^{-\varepsilon} \frac{\mathrm{e}^{-x^2 + 2 a x}}{x} \, \mathrm{d} x + \int \limits_\varepsilon^\infty \frac{\mathrm{e}^{-x^2 + 2 a x}}{x} \, \mathrm{d} x\right] \\ &= \lim_{\varepsilon \to 0^+} \int \limits_\varepsilon^\infty \frac{\mathrm{e}^{-x^2 + 2 a x} - \mathrm{e}^{-x^2 - 2 a x}}{x} \, \mathrm{d} x = \int \limits_0^\infty \frac{\mathrm{e}^{-x^2 + 2 a x} - \mathrm{e}^{-x^2 - 2 a x}}{x} \, \mathrm{d} x \, . \end{align}$$
In the last step we have used that the integrand is in fact an analytic function (with value $4a$ at the origin). The usual arguments show that $f$ is analytic as well and we can differentiate under the integral sign to obtain

$$f'(a) = 2 \int \limits_0^\infty \left[\mathrm{e}^{-x^2 + 2 a x} + \mathrm{e}^{-x^2 - 2 a x}\right]\, \mathrm{d} x = 2 \int \limits_{-\infty}^\infty \mathrm{e}^{-x^2 + 2 a x}\, \mathrm{d} x = 2 \sqrt{\pi} \, \mathrm{e}^{a^2} \, , \, a \in \mathbb{R} \, .$$
Since $f(0) = 0$,
$$f(a) = 2 \sqrt{\pi} \int \limits_0^a \mathrm{e}^{t^2} \, \mathrm{d} t = \pi \operatorname{erfi}(a)$$
follows for $a \in \mathbb{R}$. This implies
$$\mathcal{P} \int \limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-k^2}}{a-k} \, \mathrm{d} k = \pi \mathrm{e}^{-a^2} \operatorname{erfi}(a) \, , \, a \in \mathbb{R} \, .$$