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I want to compute the integral

$$(*)\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) \mathrm d y \mathrm dx$$ for \begin{align} f(x,y) = \begin{cases} e^{y-x}, & \mbox{if } 0 \leq y \leq x \\ -e^{x-y}, & \mbox {if } 0 \leq x < y \\ 0, & \mbox{else} \end{cases}. \end{align}

However, I'm struggling to find out how this integral looks like. I tried to rewrite the inner integral as follows $$\int_{-\infty}^{\infty} f(x,y) \mathrm d y = \int_{0}^x e^{y-x} \mathrm dy + \int_{x}^{\infty} -e^{x-y} \mathrm dy$$ given the boundaries of $f$. But then, $(*)$ is not finite (but it should be).

Can someone help me finding the correct integral limits?

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    I assume $0 \leq y \leq y$ should be $0 \leq x \leq y$? By symmetry, the integral *should* be 0, if the limit exists. – 2017-01-11
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    Yes, even $0 \leq x < y$. I edited. – 2017-01-11
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    I did not compute rigorously, but I guess that $f \notin L^1(\mathbb{R}^2)$. Anyway, the integral probably exists as an improper Riemann integral. – 2017-01-11
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    Thank you. The goal of this exercise is to show that the integrals $\int \int f dy dx$ and $\int \int f dx dy$ exist, but Fubini doesn't apply. Therefore, I'm a bit confused if the Lebesgue-integrals wouldn't exist. – 2017-01-11

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