# Non-integrability of the pdf of a squared normal random variable

If $X$ is a normal random variable with mean $0$ and variance $1$, then the pdf of $Y=X^2$ is $f_Y(y)=(2\pi y)^{-\frac{1}{2}}e^{-\frac{y}{2}}$ (for $y\ge0$). But $f_Y(y)$ is not integrable at $0$, since it has a divergence. So we can't use it to compute probabilities in any interval of the form $[0,a]$ (for $a>0$). Am I wrong? If so, where am I wrong? And if not, then what is the "reason" that the pdf is behaving this way?

• No, it is integrable at $0.$ – zhw. Nov 7 '15 at 20:19

Yes, I was wrong. The pdf of $Y$ is indeed integrable. A function with a singularity might indeed be integrable, as in the example: $$\int_0^a{y^{-\frac{1}{2}}dy}=2a^{\frac{1}{2}}$$