Prove that :
$$PV\displaystyle\int_{\mathbb{R}}\frac{e^{-x^2}}{x+2}dx=\frac{π\text{erfi}{2}}{e^4}$$
I know that for $a \le p \le b$
$$PV\int_{[a,b]}f(x)dx =\lim_{t\to 0^+} \left(\int_{[a,c-p]}f(x)dx+\int_{[c+p,b]}f(x)dx \right)$$
I think I should use the series expansion of
$$\frac{1}{x+2}=\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}}$$
But I don't know how I find
$$\int_{\mathbb{R}}x^{n}e^{-x^2}dx$$?
Let $$f(y) = PV\int_{-\infty}^{\infty} \frac{e^{-x^2 + (x+2)y}}{x+2} \, \mathrm{d}x.$$ Then $$f'(y) = \int_{-\infty}^{\infty} e^{-x^2 + (x+2)y} \, \mathrm{d}x = \sqrt{\pi} e^{y^2/4 + 2y}$$ so $$f(y) = \frac{\pi}{e^4} \mathrm{erfi}(2 + y/2) + C$$ for some constant $C$. Since $PV \int_{-\infty}^{\infty} \frac{e^{-(x+2)^2}}{x+2} \, \mathrm{d}x = 0$ it follows that $f(-4) = 0$ so $C = 0$.
Now set $y=0$.