Integration

Question

Value of an Integral :

I = $\frac{1}{\sqrt{2\Pi }} \int_{0}^{\infty } e^{\frac{-x^{2}}{8}}dx$

Answer given is 1.

Comments

@joshi_nitish Thankyou so much!

Is this gamma function and all in our syllabus??

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no, it is not in syllabus

i also took long time to recall that this qsn can be transformed into gamma integral.

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what is gamma integral (or function)?
any material?

Answer 1

I = $\frac{1}{\sqrt{2\Pi }}\int_{0}^{\infty }e^{\frac{-x^{2}}{8}}dx$ ,

rewriting integral

I = $\int_{0}^{\infty }\frac{1}{\sqrt{2\Pi }}e^{-\frac{1}{2}(\frac{x}{2})^{2}}dx$ ,

let $\frac{x}{2} = t \Rightarrow dx = 2 dt$,

I = $2\int_{0}^{\infty }\frac{1}{\sqrt{2\Pi }}e^{-\frac{1}{2}(t)^{2}}dt$

$\int_{0}^{\infty }\frac{1}{\sqrt{2\Pi }}e^{-\frac{1}{2}(t)^{2}}dt$  is Standard  Normal distribution

So,

I = $2\int_{0}^{\infty }\frac{1}{\sqrt{2\Pi }}e^{-\frac{1}{2}(t)^{2}}dt$

        = $2*\frac{1}{2}$ = 1

Standard Normal Distribution is normal distribution with $\mu = 0$ and $\sigma = 1$

We know that standard normal graph is symmetrical and area under graph is $1$

reference

Comments

This is crazy!