I have the following case:
\begin{equation} \int_\limits{0}^1 \int_\limits{0}^1 y\ln(x)dxdy \end{equation}
My professor says that integral should not exist because of the improper integral, my question is why if she is right, we are talking about volume, however, I am just talking about the definite integral? My answer on wolfram Alpha says the other. The answer on wolfram Can Anyone explain? I also know how to do the work to get the answer.
Having a double integral here just adds confusion. The question is: Does the improper single integral $$\int_0^1 \ln x\,dx$$ exist? The answer is yes, since $$\lim_{a\to 0^+} (x\ln x-x)\Big|_a^1 = \lim_{a\to 0^+} (-1 -a\ln a) = -1$$ certainly exists.
If the question is not about improper integrals but about whether the function $$f(x)=\begin{cases}\ln x, & x>0 \\ 0, & x=0\end{cases}$$ is (Riemann) integrable on $[0,1]$, the answer is NO. (Why?)