# Moments of $\text{exp}(-|x|^{1/2})$

I'm supposed to show that all of the moments of the density $$\text{exp}(-|x|^{1/2})$$ are finite.

I'm not convinced this is true though. The $$p$$th moment is \begin{align*} \mathbb{E}[X^p] &= \int_{-\infty}^\infty x^p \text{exp}(-|x|^{1/2}) dx \\ &= - \int_0^\infty x^p e^u x^{1/2}dx \\ &=\int_0^\infty u^{2p+1} e^u du \end{align*}

where the second equality comes from u-substitution with $$u = -x^{1/2}$$, $$du = -1/2 x^{-1/2}dx$$, and the fact that the integrand is even (so that we consider the range $$0$$ to $$\infty$$).

This last integral is infinite since I believe that $$\int_0^\infty u^p e^u du = \infty$$ for any $$p> 1$$. What am I doing wrong?

• I get by substitution (for even p) $\int_0^\infty u^{2p+1} e^{-u} du$, which is a Gamma function Commented Nov 11, 2021 at 23:38
• @kjetilbhalvorsen oh.... so you let $u = x^{1/2}$? This would make sense Commented Nov 11, 2021 at 23:38

First note that the factor necessary to make this a density function on the real line is $$1/4$$. So \begin{align} \int_{-\infty}^\infty x^p e^{- |x|^{1/2}}/4 \; dx = \\ \frac14 \int_0^\infty \left[ (-1)^p+1\right] u^p e^{-u^{1/2}} \; du = \\ \begin{cases} 0, \text{odd p} \\ \int_0^\infty u^{2p+1} e^{-u}\; du, \text{even p} \end{cases} =\\ \begin{cases} 0, \text{odd p} \\ (2p+1)!, \text{even p} \end{cases} \end{align} so is clearly finite for all positive integers $$p$$. The integrations above are done by successive substitutions. The last equality is by the definition of the Gamma function, and the well-known equality $$\Gamma(n)=(n-1)!$$, which can be proved by induction.
• Perfect thank you! Could you explain why $\int_0^\infty u^{2p+1} e^{-u} du = (2p+1)!$? Commented Nov 12, 2021 at 11:46