What is general formulation of an integration i.e., $\int_0^\infty e^{-x^n} dx$.
Well I did some calculation but the answer is wrong, but am not getting where did I got wrong.
But if that's correct, it doesn't work for anything except $n=1$.
What is the value of integration over zero to infinity of $exp^{-x^n}$?
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$\begingroup$
definite-integrals
exponential-function
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1So you make the substitution $y = \sqrt{x}$, so $dx = 2ydy$. Then the integral becomes $\int_0^{\infty} 2ye^{-y^{2n}}dy$, which is not what you have written. – 2017-01-20
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0The substitution $y=x^n$ yields $dy=nx^{n-1}dx$ hence $dx=y^{1/n-1}dy/n$ and your integral is $\int_0^\infty y^{1/n-1}e^{-y}dy/n=\Gamma(1/n)/n=\Gamma(1+1/n)$. – 2017-01-20
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0Sorry didnt got your point, where did I got wrong again? – 2017-01-20
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0I thought functions of this form didn't have closed-form integral. – 2017-01-21
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0Yeah I got it already, and found solution as gamma {(1+n)/n} – 2017-01-21
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0Yeah I got it already, and found solution as gamma {(1+n)/n} – 2017-01-21
1 Answers
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Your error is that $e^{-x^n} =e^{-y^{n/2}} $, not $e^{-y}$.
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0Yeah I got it already, and found solution as gamma {(1+n)/n} – 2017-01-21
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0I like to write that as (1/n)! – 2017-01-21