TITLE: The expected value of $\ln(X)$ if $X$ is chi square distributed
QUESTION [5 upvotes]: $X$ is a chi-squared random variable which is the square of two standard normal variables. How can I find the expected value of $\ln(X)$ in this case.
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REPLY [2 votes]: Consider $Z = X^2 + Y^2$ where X and Y are independent standard normals.
Then the joint density $f_{(X,Y)}$ of X and Y is the product of the densities and so
$$\begin{align*}E[\ln(Z)] &= E[\ln(X^2 + Y^2)] \\ &= \int_{-\infty}^\infty \int_{-\infty}^\infty \ln(x^2 + y^2) f_{(X,Y)}(x,y)\; dx\; dy \\ &= \frac{1}{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty \ln(x^2 + y^2)\exp\left(-\frac{x^2 + y^2}{2}\right) \; dx\; dy \\ &= -\gamma + \ln(2) \end{align*}$$
where $\gamma$ is the Euler-Mascheroni constant and the last equation was calculated by Mathematica…