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 screen 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?
 A: 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
$$ \mathcal{P} \int \limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-k^2}}{a-k} \, \mathrm{d} k = \pi \mathrm{e}^{-a^2} \operatorname{erfi}(a) = 2 \sqrt{\pi} \operatorname{F}(a) \, , \, a \in \mathbb{R} \, ,$$
where the final step is just the definition of Dawson's integral $\operatorname{F}$ as per Tyma Gaidash's comment.
A: An approach via Fourier transform
The definition that will be used here is
$$
\mathcal{F}f(\xi) = \int_{-\infty}^{\infty} f(x) \, e^{-i \xi x} \, dx,
\quad
f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{F}f(\xi) \, e^{i \xi x} \, d\xi
.
$$
The integral
$$
I(x) := \mathcal{P}\int_{-\infty}^{\infty} \frac{e^{-y^2}}{x-y} \mathrm{d}y
$$
can be seen as the convolution of the tempered distribution 
$u(x) = \mathcal{P}\frac{1}{x}$
with the Schwartz function 
$\varphi(x) = e^{-x^2}$,
i.e. $I = u * \varphi.$
Taking the Fourier transform we then have
$
\mathcal{F}I 
= \mathcal{F}(u*\varphi)
= \mathcal{F}u \, \mathcal{F}\varphi
.
$
Now,
$\mathcal{F}u(\xi) = -i\pi \operatorname{sign}(\xi)$ 1
and
$\mathcal{F}\varphi(\xi) = \sqrt{\pi} e^{-\xi^2/4}$, so $\mathcal{F}I(\xi) = -i\pi^{3/2} \operatorname{sign}(\xi) \, e^{-\xi^2/4}$.
Thus,
$$\begin{align}
I(x) 
&= \frac{1}{2\pi} \int_{-\infty}^{\infty} -i\pi^{3/2} \operatorname{sign}(\xi) \, e^{-\xi^2/4} e^{i \xi x} \, d\xi \\
&= \frac{\sqrt{\pi}}{2i} \int_{-\infty}^{\infty} \operatorname{sign}(\xi) \, e^{-\xi^2/4} e^{i \xi x} \, d\xi \\
&= \sqrt{\pi} \int_{0}^{\infty} e^{-\xi^2/4} \sin\xi x \, d\xi . \\
\end{align}$$
Taking the derivative gives
$$\begin{align}
I'(x) 
&= \sqrt{\pi} \int_{0}^{\infty} e^{-\xi^2/4} \xi \cos\xi x \, d\xi \\
&= \sqrt{\pi} \left( \left[(-2 e^{-\xi^2/4}) \cos\xi x \right]_{0}^{\infty} - \int_{0}^{\infty} (-2 e^{-\xi^2/4}) \, (-x \sin\xi x) \, d\xi \right) \\
&= \sqrt{\pi} \left( 2 - 2 x \int_{0}^{\infty} e^{-\xi^2/4} \sin\xi x \, d\xi \right) \\
&= 2 \sqrt{\pi} - 2 x I(x), \\
\end{align}$$
which is easily solved using integrating factor, giving
$$
I(x) = 2 \sqrt{\pi} \, e^{-x^2} \int_0^x e^{t^2} \, dt = \pi \, e^{-x^2} \operatorname{erfi}(x).
$$
A: $$first\ we\ know\ that:\\
\\
pv\int_{-\infty }^{\infty }\frac{cos(2bx)}{1-x}dx=pv\int_{-\infty }^{\infty }\frac{cos(2b)cos(2by)+sin(2b)sin(2by)}{y}dy\\
\\
=cos(2b). \ pv\int_{-\infty }^{\infty }\frac{cos(2by)}{y}dy+sin(2b).  pv\int_{-\infty }^{\infty }\frac{sin(2by)}{y}dy\\
\\
\\
\therefore \ \ \ pv\int_{-\infty }^{\infty }\frac{cos(2bx)}{1-x}dx=\pi sin(2b)\ \ \ \ \ \ for\ b>0\\
\\
\\
\therefore pv\int_{-\infty }^{\infty }\frac{e^{-b^2}.cos(2bx)}{1-x}dx=\pi e^{-b^2}sin(2b)\\
\\
\\
\therefore pv\int_{-\infty }^{\infty }\frac{1}{1-x}\int_{0}^{\infty }e^{-b^2}cos(2bx)db dx=\pi \int_{0}^{\infty }e^{-b^2}sin(2b)db$$
$$\therefore \frac{\sqrt{\pi }}{2}pv\int_{-\infty }^{\infty }\frac{e^{-x^2}}{1-x}dx=\pi \int_{0}^{\infty }e^{-x^2}.sin(2b)db\\
\\
\\
\therefore pv.\int_{-\infty }^{\infty }\frac{e^{-x^2}}{x-1}dx=-2\sqrt{\pi }\int_{0}^{\infty }e^{-x^2}sin(2x)dx\\
\\
\\
let\ x=-y\ \ \ ,   \therefore pv\int_{-\infty }^{\infty }\frac{e^{-x^2}}{x-1}dx=pv.\int_{-\infty }^{\infty }\frac{-e^{-x^2}}{x+1}dx\\
\\
\\
\therefore 2pv\int_{-\infty }^{\infty }\frac{e^{-x^2}}{x-1}dx=2pv\int_{-\infty }^{\infty }\frac{e^{-x^2}}{x^2-1}dx\\
\\
\\
\therefore pv\int_{-\infty }^{\infty }\frac{e^{-x^2}}{x-1}dx=2pv\int_{0}^{\infty }\frac{e^{-x^2}}{x^2-1}dx$$
                               now let 

$$F(b)=2pv\int_{0}^{\infty }\frac{e^{-(x^2-1)b^2}}{x^2-1}dx\ \ \Rightarrow F(0)=0\\
\\
\\
\therefore F'(b)=-4b.pv\int_{0}^{\infty }e^{-(x^2-1)b^2}dx=-4b.e^{b^2}\int_{0}^{\infty }e^{-x^2b^2}dx\\
\\
\\
\therefore F'(b)=-2\sqrt{\pi }e^{b^2}\ \ \ \ \ , since\ \ F(0)=0\ \Rightarrow F(1)=\int_{0}^{1}F'(b)db\\
\\
\\
\therefore F(1)=-2\sqrt{\pi }\int_{0}^{1}e^{x^2}dx\\
\\
\\
\therefore 2pv\int_{0}^{\infty }\frac{e^{-(x^2-1)}}{x^2-1}dx=2e.pv.\int_{0}^{\infty }\frac{e^{-x^2}}{x^2-1}dx\\
\\
\\
\therefore pv.\int_{-\infty }^{\infty }\frac{e^{-x^2}}{x-1}dx=\frac{-2\sqrt{\pi }}{e}\int_{0}^{1}e^{x^2}dx\\
\\
\\
\therefore pv.\int_{-\infty }^{\infty }\frac{e^{-x^2}}{1-x}dx=\frac{\pi }{e}erfi(1)$$
A: Another approach:
$$I=-pv.\int_{-\infty }^{\infty }\frac{e^{-x^2}}{x-1}dx=-\sum_{n=1}^{\infty }\Gamma (\frac{1-2n}{2})\\
\\
\\
=-\sum_{n=1}^{\infty }(-1)^n\frac{2^n.\sqrt{\pi }}{\prod_{n=1}^{\infty }(2m-1)}=-\sum_{n=1}^{\infty }(-1)^n\frac{2^n\sqrt{\pi }}{\frac{2^n.\Gamma (n+\frac{1}{2})}{\sqrt{\pi }}}\\
\\
\\
\therefore I=-\sum_{n=1}^{\infty }(-1)^n\frac{\pi }{\Gamma (n+\frac{1}{2})}=-\sqrt{\pi }\sum_{n=1}^{\infty }(-1)^n\frac{\Gamma (n).\Gamma (\frac{1}{2})}{\Gamma (n)\Gamma (n+\frac{1}{2})}\\
\\
\\
=-\sqrt{\pi }\sum_{n=1}^{\infty }(-1)^n\int_{0}^{1}\frac{(1-x)^{n-1}}{\sqrt{x}(n-1)!}dx\\
\\
\\
\therefore I=-\frac{\pi }{e}.ierf(i)=\frac{\pi }{e}erfi(1$$
A: You can solve this using shwinger trick.
Use the second result from wiki
This is easier to use many times.
Shwinger parameterization
A: $f(t)=\displaystyle\int_{-\infty }^{+\infty }\frac{e^{-tk^{2}}}{1-k}dk:t\gt 0,I=f(1)\\
f'(t)=\displaystyle\int_{-\infty }^{+\infty }\frac{-k^{2}e^{-tk^{2}}}{1-k}dk=\int_{-\infty }^{+\infty }\frac{(1-k^{2})e^{-tk^{2}}}{1-k}dk-f(t)\\
=\displaystyle\int_{-\infty }^{+\infty }(1+k)e^{-tk^{2}}dk-f(t)=\frac{1}{\sqrt{t}}\int_{-\infty }^{+\infty }e^{-(\sqrt{t}k)^{2}}d(\sqrt{t}k)-\frac{1}{2t}\int_{-\infty }^{+\infty }(-2tk)e^{-tk^{2}}-f(t)\\
\Rightarrow f'(t)=\displaystyle\frac{\sqrt{\pi}}{\sqrt{t}}-f(t)\Rightarrow f'(t)+f(t)=\sqrt{\frac{\pi}{t}}\\
$
