I ran into a problem today that I don't really understand and was hoping someone could help me out. I'm sure it's basic, but I must be missing something from elementary calculus.
I was trying to integrate a probability density function to determine the normalizing constant $c$:
$$f_{XY}(x,y)={{cx^2}\over{y}}$$ for $0<x<y<1$ and $(c>0)$
I can easily find that $c=9$ when I compute the inner most integral with respect to $x$ first and the outer most integral with respect to $y$. However, To check my work, I reversed the order of integration and stumbled into a problem: toward the end of my solution I find that I'm needing to evaluate $\int_{0}^{1}x^{2}\log(x)dx=\frac{1}{3}x^{3}\ln(x)\left|_{0}^{1}\right.-{1\over{9}}$ after performing $u$,$v$ substitution. The upper bound of this can clearly be calculated and it evaluates to $0$, but I'm finding that the lower bound should be undefined since $log(0)$ is undefined. What am I missing? It this because the bound of support is strictly less than $1$ so for the lower bound I should be really evaluating $\lim_{x\to 0}\frac{1}{3}x^{3} \ln(x)=0$?
Thanks.
Note that $$\int_0^1 x^2 \log x \, \mathrm{d}x = \frac{x^3}{3}\ln x\bigg]_0^1 -\int_0^1 \frac{x^3}{3x} \, \mathrm{d}x = -\lim_{\epsilon \to 0} \frac{\epsilon^3}{3}\ln \epsilon - \bigg[\frac{x^3}{9}\bigg]_0^1 = -\frac{1}{9}$$
using integration by parts with $u=\ln x$ and $\mathrm{d}v =x^2$. The fact that it's negative doesn't bode well if this has to do with probability but it expected given the integrand is always negative in $[0,1]$.
The reason we take the limit is because $x^2 \log x$ isn't defined at $0$, so this is known as an improper integral. The notation $\int_0^1 x^2 \log x \, \mathrm{d}x$ really means $\lim_{\epsilon \to 0} \int_{\epsilon}^1 x^2 \log x \, \mathrm{d}x$ in this case, assuming the integral is convergent.