Calculate the improper double integral: double integral

Calculate the improper double integral: $$ \iint_{0\leqslant x\leqslant y} e^{-(x+y)}\,dx\,dy. $$

Explanation of Method

For fixed $x\geqslant0$, $y$ ranges over $[x,\infty)$; integrating the exponential in $y$ first leaves a clean exponential in $x$, which is then integrable over $[0,\infty)$.

Set up the iterated integral with $x$ as the outer variable.
Integrate with respect to $y$ from $x$ to $\infty$.
Integrate the resulting function of $x$ from $0$ to $\infty$.

Step 1 · Set up the iterated integral

$$ I = \int_0^\infty e^{-x}\left(\int_x^\infty e^{-y}\,dy\right)dx. $$

Step 2 · Evaluate the inner integral

$$ \int_x^\infty e^{-y}\,dy = \Big[-e^{-y}\Big]_x^\infty = e^{-x}. $$

Step 3 · Evaluate the outer integral

$$ I = \int_0^\infty e^{-x}\cdot e^{-x}\,dx = \int_0^\infty e^{-2x}\,dx = \left[-\frac12e^{-2x}\right]_0^\infty = \frac12. $$

Additional Notes

By symmetry, the complementary region $0\leqslant y\leqslant x$ gives the same value $\frac12$ by the same computation with $x,y$ exchanged, and the two regions partition the full first quadrant, so together they should reproduce $\displaystyle\iint_{x,y\geqslant0}e^{-(x+y)}\,dx\,dy=\left(\int_0^\infty e^{-x}dx\right)^2=1$. Indeed $\frac12+\frac12=1$, confirming the result.

Result

The solution to the integral is:

$$ \iint_{0\leqslant x\leqslant y} e^{-(x+y)}\,dx\,dy = \frac12. $$