Prove that $\frac{\int_{0}^\infty x^n e^{-x^n} dx}{\int_{0}^\infty e^{-x^n} dx} = \frac{1}{n}$ What is the simplest way to prove that
$$\frac{\int_{0}^\infty x^n e^{-x^n} dx}{\int_{0}^\infty e^{-x^n} dx} = \frac{1}{n}$$
for all real $n > 0$? As an example, integration by parts can easily solve $n=2$:
$$\frac{\int_{0}^\infty x^2 e^{-x^2} dx}{\int_{0}^\infty e^{-x^2} dx} = \frac{xe^{-x^2}|_{0}^\infty - \int_{0}^\infty -\frac{1}{2}e^{-x^2}dx}{\int_{0}^\infty e^{-x^2}dx} = \frac{0+\frac{1}{2}\int_{0}^\infty e^{-x^2}dx}{\int_{0}^\infty e^{-x^2}dx} = \frac{1}{2}$$
 A: It suffices to prove that
$$\int_0^\infty (x^n - 1/n)\mathrm{e}^{- x^n}\mathrm{d} x = 0$$
or
$$\int_0^\infty \left[-\frac{1}{n} x \mathrm{e}^{-x^n}\right]' \mathrm{d} x = 0$$
or
$$\left[-\frac{1}{n} x \mathrm{e}^{-x^n}\right]\Bigg\vert_0^\infty = 0$$
which is true.
A: Consider $$ \int_0^\infty x^\alpha e^{-x^\beta}\, dx$$ with $\alpha ≥ 0,\beta > 0$. Letting $y=x^\beta$ gives \begin{align*}
\int_0^\infty x^\alpha e^{-x^\beta}\, dx= \frac 1 \beta \int_0^\infty y^{\frac{\alpha+1}\beta-1} e^{-y}\, dy = \frac 1 \beta \Gamma \bigg (\frac{\alpha+1}\beta \bigg )
\end{align*} where $\Gamma$ is the Euler-Gamma function. Hence, $$ \frac{\int_0^\infty x^n e^{-x^n}\, dx}{\int_0^\infty e^{-x^n}\, dx}=\frac{\Gamma \big (\frac{1}n+1 \big )}{\Gamma \big (\frac1n \big )}=\frac{1}n$$ for all $n>0$.
A: Using integration by parts, we get $$
\begin{aligned}
 \int_{0}^{\infty} x^{n} e^{-x^{n}} d x
&=-\frac{1}{n} \int_{0}^{\infty} x d\left(e^{-x^{n}}\right) \\
\\&=-\frac{1}{n} \left[ x e^{-x^{n}} \right]^{\infty}+\frac{1}{n} \int_{0}^{\infty} e^{-x^{n}} d x\\& =\frac{1}{n} \int_{0}^{\infty} e^{-x^{n}} d x
\end{aligned}
$$
Hence we can conclude that $$
\frac{\int_{0}^{\infty} x^{n} e^{-x^{n}} d x}{\int_{0}^{\infty} e^{-x^{n}} d x}=\frac{1}{n}
$$
