What are the detailed steps of the integral $\int_{-\infty}^{+\infty}{{\rm e}^{(-\frac{(x-(a+bc^2))^2} {2c^2})}}\,{\rm d}x$?

Question

What are the detailed steps of calculating the integral $\int_{-\infty}^{+\infty}{{\rm e}^{(-\frac{(x-(a+bc^2))^2} {2c^2})}}\,{\rm d}x$, I got an answer of $\sqrt{\pi b}$ from WolframAlpha but couldn't get a detailed answer of how it is derived. Also, the indefinite integral is $-\frac{1}{2}\sqrt{\pi b}{\rm erf}(\frac{a-x}{\sqrt{b}})+C$, I wonder how it is derived too. Thanks.

Answer 1

put $x-(a+bc^2)=t$

then,

$I=\displaystyle\int_{-\infty}^{\infty}e^{-t^2/2c^2} dt$

we know,

$\mathcal{F}[e^{-{at^2}}]=\displaystyle\int_{t=-\infty}^{t=\infty}e^{-at^2}e^{-j\omega t}dt=\sqrt{\dfrac{\pi}{a}}\exp\left(\dfrac{-\omega^2}{4a}\right)$

for your case put $\omega =0$ and $a=\dfrac{1}{2c^2}$

you'll get

$I=c\sqrt{2\pi}$

note:

in general

$\displaystyle\int_{-\infty}^{\infty} \exp(-mx^n)dx=\dfrac{2\Gamma(\frac{1}{n})}{n(m)^\frac{1}{n}}$