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I've been trying to solve this question in many ways, at first I tried a graphical approach, which, didn't bring me many results, not any correct ones at least… However, searching around, I found out about the "Heaviside function method", so this is my second attempt: $$ \int_{0}^{1}\int_{e^{y-1}}^{e^y} f(x,y)dxdy=\int_{0}^{1}\int_{1}^{e^y}f(x,y)dxdy+\int_{0}^{1}\int_{e^y-1}^{1}f(x,y)dxdy\\ =\int_{0}^{1}\int_{1}^{e}\Theta (e^y-x)f(x,y)dxdy+\int_{0}^{1}\int_{1/e}^{1}\Theta (x-e^{y-1})f(x,y)dxdy\\ =\int_{1}^{e}\int_{0}^{1}\Theta (y-\ln(x))f(x,y)dydx+\int_{1/e}^{1}\int_{0}^{1}\Theta (\ln(x)+1-y)f(x,y)dydx\\ =\int_{1}^{e}\int_{\ln(x)}^{1}f(x,y)dydx+\int_{1/e}^{1}\int_{0}^{\ln(x)+1}f(x,y)dydx $$ The first part(first integral) seems to be right, the second, however, does not, can anyone please help me find my mistake? Is it in the boundaries?

I assume having a hard time figuring out how to calculate and adjust the boundaries to the Heaviside function ""conversion"", due to the lack of information I found about this part. After reading this[link] I thought I had it figured out, but, after this result I'm not so sure anymore…

Mohammad Riazi-Kermani
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Iorpim
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    All is good, as far as I can see. – Julián Aguirre Apr 13 '18 at 12:37
  • Hm… I've been using integral calculators to check my results, by doing $\int _0^1:\int _{e^{y-1}}^1:xdxdy-\int _{\frac{1}{e}}^1:\int _0^{ln\left(x\right)+1}:xdydx$ for example, in this case, could this just be a flawed method for answer checking? I mean, it seems to have been working so far, but, all the other functions were much simpler. – Iorpim Apr 13 '18 at 12:42
  • Now that I have looked more closely, are you sure of the order of integration? Since the inner limits are functions of $y$, shouldn't it be $dxdy$? – Julián Aguirre Apr 13 '18 at 13:11
  • Oh, definitely, sorry, my mistake, I inverted it by accident when transcribing to LaTeX, the orders are correct on paper, and now, also on the question, thank you. – Iorpim Apr 13 '18 at 13:42

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