Prove that <msubsup> ∫ <mrow class="MJX-TeXAtom-ORD"

Question

Prove that

\frac{\int_{0}^{\infty} x^{n} e^{- x^{n}} d x}{\int_{0}^{\infty} e^{- x^{n}} d x} = \frac{1}{n}

Answer 1

It suffices to prove that

\int_{0}^{\infty} (x^{n} - 1 / n) e^{- x^{n}} d x = 0

or

\int_{0}^{\infty} {[- \frac{1}{n} x e^{- x^{n}}]}^{'} d x = 0

or

\int_{0}^{\infty} {[- \frac{1}{n} x e^{- x^{n}}]}^{'} d x = 0

or

[- \frac{1}{n} x e^{- x^{n}}] |_{0}^{\infty} = 0

which is true.

Answer 2

Consider

\int_{0}^{\infty} x^{α} e^{- x^{β}} d x

with

α \geq 0, β > 0 . Letting y = x^{β} gives

\begin{aligned} \int_{0}^{\infty} x^{α} e^{- x^{β}} d x = \frac{1}{β} \int_{0}^{\infty} y^{\frac{α + 1}{β} - 1} e^{- y} d y = \frac{1}{β} Γ (\frac{α + 1}{β}) \end{aligned}

Where

Γ

is the Euler-Gamma function. Hence,

\frac{\int_{0}^{\infty} x^{n} e^{- x^{n}} d x}{\int_{0}^{\infty} e^{- x^{n}} d x} = \frac{Γ (\frac{1}{n} + 1)}{Γ (\frac{1}{n})} = \frac{1}{n}

for all

n > 0

Answers (2)