Question

3) Evaluate integral 0 to infinity x^4e^-(x^4)dx​

Answer 1

$\large\underline{\sf{Solution-}}$

We know that,

By definition of gamma function, we have

$\boxed{\displaystyle\int_0^ \infty \tt \: {e}^{ - x} {x}^{n - 1} = \Gamma n}$

and

$\boxed{ \bf \Gamma \: \dfrac{1}{2} \: = \sqrt{\pi} }$

Consider,

$\rm :\longmapsto\:Let \: I \: = \displaystyle\int _{0}^\infty \sf \: {x}^{4} \: {e}^{ - {x}^{4}}dx$

To solve this integral, we use method of Substitution

$\red{\rm :\longmapsto\:Let \: {x}^{4} = y} \\ \\ \red{\rm :\longmapsto\: {4x}^{3} = \dfrac{dy}{dx}} \\ \\ \red{\rm :\longmapsto\:dx = \dfrac{dy}{4 {x}^{3} }} \\ \\ \red{\rm :\longmapsto\:dx = \dfrac{dy}{4 {y}^{ \frac{3}{4} } }}$

So, above integral can be rewritten as

$\rm :\longmapsto \: I \: = \displaystyle\int _{0}^\infty \sf \: y \: {e}^{ -y} \: \dfrac{1}{4 {y}^{ \frac{3}{4} } } \: dy$

$\rm :\longmapsto \: I \: =\dfrac{1}{4} \displaystyle\int _{0}^\infty \sf \: {y}^{ \frac{1}{4} } \: {e}^{ -y} \:\: dy$

can be rewritten as

$\rm :\longmapsto \: I \: =\dfrac{1}{4} \displaystyle\int _{0}^\infty \sf \: {y}^{ \frac{5}{4} - 1 } \: {e}^{ -y} \:\: dy$

$\rm :\longmapsto \: I \: =\dfrac{1}{4} \Gamma \:\bigg(\dfrac{5}{4}\bigg)$

$\rm :\longmapsto \: I \: =\dfrac{1}{4} \Gamma \:\bigg(\dfrac{1}{4} + 1\bigg)$

$\rm :\longmapsto \: I \: =\dfrac{1}{4} \times \dfrac{1}{4} \: \: \Gamma \:\bigg(\dfrac{1}{4}\bigg)$

$\rm :\longmapsto \: I \: =\dfrac{1}{16} \Gamma \:\bigg(\dfrac{1}{4}\bigg)$

Additional Information :-

$\green{\boxed{ \tt \:\Gamma \:(n + 1) = n \: \Gamma \:(n) }}$

$\green{\boxed{ \tt \: \Gamma \:(n + 1) = n! \: \: where \: n \in \: natural \: number}}$

$\green{\boxed{ \tt \: \Gamma \:(1) = 1}}$