Write this integral in terms of gamma function

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SUMMARY

The integral \(\int _{0}^{\infty }e^{-x^4}dx\) can be expressed in terms of the gamma function as \(\Gamma \left( \dfrac {5} {4}\right)\). The substitution \(u = x^4\) is crucial, leading to the differential \(du = 4x^3dx\). This transformation allows for the integral to be rewritten in a form that utilizes the properties of the gamma function, specifically the relation \(t\Gamma(t) = \Gamma(t+1)\).

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Hey!

So I'm self studying mary boas's mathematical methods book and I've come across this integral:

<br /> \int _{0}^{\infty }e^{-x^4}dx<br />

and I'm suppose to write this using the gamma function. The hint given states to let x^4 = u. And the answer is:

<br /> \Gamma \left( \dfrac {5} {4}\right) <br />

I tried substituting u = x^4 and du = 4x^3dx, but that doesn't give the correct answer.

I'm a bit confused as to how the book got that answer. Any ideas would be great.

Thanks!
 
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After substitution you should have something you can express as ## t \Gamma(t) ## which you can rewrite using the relation ## t\Gamma(t) = \Gamma(t+1)##.
 

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