Since question is based on a famous Gaussian Integral, we can use: $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$ and $\int_{-\infty}^{0} e^{-x^2} dx = \int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$
(Proof is at the end of the answer)
Here,
$J = \int_{-\infty}^{0} e^{-\left(\frac{x}{\sqrt{20}}\right)^2} dx $
Let, $\frac{x}{\sqrt{20}} = t \implies dx = \sqrt{20} dt$
When $x = -\infty, t = -\infty$ and
when $x = 0, t = 0$
So, $J = \sqrt{20} \int_{-\infty}^{0} e^{-t^2} dt = \sqrt{20} \int_{-\infty}^{0} e^{-x^2} dx$
$J = \sqrt{20} \times \frac{\sqrt{\pi}}{2}$
Hence,
$J = \sqrt{20} \times \frac{\sqrt{\pi}}{\sqrt{4}} = \sqrt{5\pi}$
Proof of Gaussian Integral with some limits:
Proof 1 (By Polar Coordinates) : Watch it from 29:35
Proof 2:
$I = \int_{-\infty}^{0} e^{-x^2} dx = \int_{-\infty}^{0} e^{-y^2} dy$
$I^2 = I \times I$
$I^2 = \left(\int_{-\infty}^{0} e^{-x^2} dx \right) \left(\int_{-\infty}^{0} e^{-y^2} dy \right)$
$I^2 = \int_{y= -\infty}^{y=0} \left(\int_{x = -\infty}^{x = 0} e^{-x^2 – y^2} dx \right) dy $
Now, Let $u = \frac{x}{y} \implies x = uy \implies dx = y du$ (To eliminate ‘$x$’ from the inner integral, consider $y$ as a constant)
When $x = -\infty$, $u = \infty$ because $-\infty = yu$ and $y$ is between $0$ and $-\infty$ i.e. negative and
When $x = 0$, $u = 0$ because $0 = yu$ and $y$ is strictly between $0$ and $-\infty$
Hence,
$I^2 = \int_{y= -\infty}^{y=0} \left(\int_{u = \infty}^{u = 0} e^{-u^2y^2 – y^2} ydu \right) dy $
$I^2 = \int_{u= \infty}^{u=0} \left(\int_{y = -\infty}^{y = 0} e^{-u^2y^2 – y^2} ydy \right) du $
$I^2 = \int_{u= \infty}^{u=0} \left(\int_{y = -\infty}^{y = 0} e^{-y^2 (u^2 + 1)} ydy \right) du $
$I^2 = \int_{u= \infty}^{u=0} \left(\frac{e^{-y^2(u^2 + 1)}}{-2(u^2 + 1)} \right)_{y=-\infty}^{y=0} du $
$I^2 = \int_{u= \infty}^{u=0} \frac{-1}{2(u^2 + 1)} du$
$I^2 = \frac{-1}{2} (\tan^{-1}(u))_{u=\infty}^{u=0}$
$I^2 = \frac{-1}{2}(0 – \frac{\pi}{2})$
$I^2 = \frac{\pi}{4}$
Since, $I$ is a positive function for real interval $(-\infty,0]$
Therefore, $I = \frac{\sqrt{\pi}}{2}$