I want to calculate the following integral : $$I = \int_{0}^{2}dx\int_{0}^{4-x^{2}}\left(\frac{xe^{2y}}{4-y}\right)dy$$ I am not sure whether this can be calculated as it is, but my reasoning was the following in order to simplify the calculation: Make this a double integral, change the order of integration while at the same time using Tonneli and Fubini to verify the existence of the integral and calculate. I managed to do that and my result is numerically correct.
However, in order to use Tonneli-Fubini, calculate the integral using the opposite order I first have to verify it exists ( so I don't get a different result form the opposite order ).
Here is my problem. From what I can see: $$f(y) = \frac{xe^{2y}}{4-y}\epsilon C_{0}[0, 4[$$ Which is tto say I have to verify integrability when $y = 4$. But: $$\frac{1}{y-4} = O\left(f(y)\right), y\rightarrow 4$$ which by the integrability criteria means the function is not integrable when $y = 4$.
I know I have to be doing something wrong, but I don't seem to be able to find what.
Let $u=x^2$. Then the integral becomes $\frac{1}{2}\int_{u=0}^4 \int_{y=0}^{4-u} \frac{e^{2y}}{4-y}dy$. Change the order of integration to get $\frac{1}{2}\int_{y=0}^4 \frac{e^{2y}}{4-y}dy\int_{u=0}^{4-y}du=\frac{1}{2}\int_{y=0}^4 e^{2y}dy= (e^8-1)/4$.