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Solve the following $\int_{0}^{\infty}e^{-x^2}x^4dx$
in Calculus
recategorized | 202 views

put x^2=t
and use improper integrals.
it will give $\frac{3\sqrt{\pi }}{8}/$
by Active (4.8k points)

Let $I = \displaystyle{}\int_{0}^{\infty}e^{-x^2}x^4dx\rightarrow (1)$

let $x^{2} = t\implies x= \sqrt{t}$

$2xdx = dt\implies dx = \dfrac{dt}{2x} \implies dx = \dfrac{dt}{2\sqrt{t}} = \dfrac{t^{-1/2}\:dt}{2}$

Put the values in equation $(1),$ we get

$I = \displaystyle{}\int_{0}^{\infty}e^{-t}. t^{2}\bigg( \frac{t^{-1/2}}{2}\bigg)dt$

$I = \displaystyle{}\frac{1}{2}\int_{0}^{\infty}e^{-t} .t^{2}t^{-1/2}dt$

$I = \displaystyle{}\frac{1}{2}\int_{0}^{\infty}e^{-t}.t^{3/2}dt$

We know that $\Gamma(n) = \displaystyle{}\frac{1}{2}\int_{0}^{\infty}e^{-x}.x^{n-1}dx\:\:\:;n>0$

$I = \displaystyle{}\frac{1}{2}\bigg[\int_{0}^{\infty}e^{-t}.t^{(5/2-1)}dt\bigg]$

$\therefore I = \dfrac{1}{2}\:\Gamma \left(\frac{5}{2}\right)$

Standard Results:

• $\Gamma(1) = 1$
• $\Gamma(\frac{1}{2}) = \sqrt{\pi}$
• $\Gamma(n+1) = n\Gamma(n)$
• $\Gamma(n+1) = n!$

Now$,I = \dfrac{1}{2}\:\Gamma \left(\frac{5}{2}\right) = \dfrac{1}{2}\cdot\Gamma(\frac{3}{2} + 1) = \dfrac{1}{2}\cdot\frac{3}{2}\Gamma(\frac{3}{2}) = \dfrac{1}{2}\cdot\frac{3}{2}\Gamma(\frac{1}{2} + 1) = \dfrac{1}{2}\cdot \frac{3}{2}\cdot \frac{1}{2}\Gamma (\frac{1}{2}) = \frac{3}{8} \cdot \sqrt{\pi}$

So, the correct answer is  $I = \dfrac{3\:\sqrt{\pi}}{8}$

by Veteran (59.5k points)
edited