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The following definite integral evaluates to

$$\int_{-\infty}^{0} e^ {-\left(\frac{x^2}{20} \right )}dx$$

1. $\frac{1}{2}$
2. $\pi \sqrt{10}$
3. $\sqrt{10}$
4. $\pi$

in question its not 20 but 2

$\dpi{120} \\Let: x = \sqrt{20t} \qquad\text{Then, } \mathrm{d}x = \frac{\sqrt 5}{\sqrt t}\,\mathrm{d}t$

Now, the given integral can be written as:

$\dpi{120} \int\limits_{0}^{\infty}{ e^{-t} \cdot \frac{\sqrt 5}{\sqrt t}\, \mathrm{d}t}\qquad =\qquad \sqrt{5} \int\limits_{0}^{\infty}{\frac1{e^t\sqrt{t}}\,\mathrm{d}t}$

$\dpi{120} = \sqrt{5} \cdot \sqrt \pi \qquad \qquad \Bigg\{\int\limits_{0}^{\infty}{\frac1{e^t\sqrt{t}}\,\mathrm{d}t} = \sqrt{\pi} \text{ is a famous } \gamma \text{-reduction}$

Since none of the options are equal to $\dpi{120} \sqrt{5\pi}$, the given options are incorrect. You can verify the result here: wolframalpha(Click to view)

The original answer, in case needed, can be obtained from what follows:

let put x = (20t)1/2
Equation become     dx = ((51/2)/t1/2)dt
51/2e-tt-1/2dt================== very famous gamma reduction
51/2pie1/2

How to do gamma reduction. Please provide full solution.
It is not needed for GATE
Is gamma reduction in syllabus?
how the limits of integration are changed while doing substitution?

We can solve it without Gamma Function (which is termed as $\gamma$-reduction here) to solve this question.

This might help.

@siddhartha983, this answer is solved by considering limits from $0$ to $\infty$ and not for $-\infty$ to $0$ (which is in the original question) and I think that’s why you didn’t get it.

$I = \int_{x = -\infty}^{x = 0} e^{\frac{-x^2}{20}}dx$

$I = -\int_{x=0}^{x =-\infty} e^{\frac{-x^2}{20}}dx$

Let, $x = -u \implies dx = -du$

When $x=0, u = 0$ and when $x = -\infty , u = \infty$

So, $I = -\int_{u=0}^{u =\infty} e^{\frac{-(-u)^2}{20}}(-du)$

$I = \int_{u=0}^{u =\infty} e^{\frac{-(u)^2}{20}}du$

$I = \int_{x=0}^{x =\infty} e^{\frac{-(x)^2}{20}}dx$

Now, do the substitution as mentioned in the answer.

$\sqrt{5}\int_{0}^{\infty} t^{\frac{-1}{2}}e^{-t} dt$

Now, according to the definition of the gamma function i.e. $\Gamma (z) = \int_{0}^{\infty} e^{-t} t^{z-1} dt$

So, $\sqrt{5}\int_{0}^{\infty} t^{\frac{-1}{2}}e^{-t} dt = \sqrt{5}\int_{0}^{\infty} t^{\frac{1}{2} -1}e^{-t} dt = \sqrt{5}\Gamma(\frac{1}{2}) = \sqrt{5} \sqrt{\pi}$

thanks for the help!
You can simply view $\int_{-\infty}^{0} e^ {-\left(\frac{x^2}{20} \right )}dx$ as normal distribution  with mean = 0 and $\sigma =\sqrt{10}$. Multiply and divide by $\sqrt{2\pi}\sigma$.
by

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, $I = \frac{\sqrt{\pi}}{2}$