in Numerical Methods
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7 votes

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 Numerical Methods
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2 Comments

Please someone answer this question..
1
in question its not 20 but 2
1

3 Answers

10 votes

         

\\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:

\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}

= \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 \sqrt{5\pi}, the given options are incorrect. You can verify the result here: wolframalpha(Click to view)


edited by

7 Comments

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

3
How to do gamma reduction. Please provide full solution.
0
It is not needed for GATE
1
Is gamma reduction in syllabus?
0
how the limits of integration are changed while doing substitution?
0

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

0
thanks for the help!
0
3 votes
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
0 votes

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

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