Calculate the improper iterated integral: integral from 0
Calculate the improper iterated integral: $$ \int_0^\infty dx\int_x^\infty e^{-y^2}\,dy. $$
Explanation of Method
The inner integral $\displaystyle\int_x^\infty e^{-y^2}\,dy$ has no elementary antiderivative in closed form, so integrating in the given order is a dead end. However, reversing the order of integration — a standard technique when one direction is intractable — converts the problem into one where $y$ is integrated last and $x$ first, and $\displaystyle\int_0^y e^{-y^2}\,dx = y\,e^{-y^2}$ is elementary and directly integrable.
- Identify the region of integration: $0\leqslant x\leqslant y<\infty$.
- Reverse the order, integrating with respect to $x$ first.
- Integrate the resulting $y\,e^{-y^2}$ with respect to $y$.
Step 1 · Identify the region
The given iterated integral is $\displaystyle\iint_D e^{-y^2}\,dx\,dy$ over the region $D=\{(x,y):x\geqslant0,\ y\geqslant x\}$, i.e. $0\leqslant x\leqslant y<\infty$.
Step 2 · Reverse the order of integration
Describing $D$ with $y$ as the outer variable: for each fixed $y\geqslant0$, $x$ ranges over $[0,y]$:
$$ \int_0^\infty dx\int_x^\infty e^{-y^2}\,dy = \int_0^\infty e^{-y^2}\left(\int_0^y dx\right)dy = \int_0^\infty y\,e^{-y^2}\,dy. $$
Step 3 · Evaluate the resulting integral
With $w=y^2$, $dw=2y\,dy$:
$$ \int_0^\infty y\,e^{-y^2}\,dy = \frac12\int_0^\infty e^{-w}\,dw = \frac12\Big[-e^{-w}\Big]_0^\infty = \frac12. $$
Additional Notes
- Verified numerically: direct numerical evaluation of the original iterated integral (integrating $y$ from $x$ to a large cutoff, then $x$ from $0$ to a large cutoff) converges to $\approx0.5$, confirming the closed-form answer obtained by reversing the order.
- This is a standard illustration of why reversing the order of integration is often the most efficient — sometimes the only practical — route to an answer whenever one of the two possible inner integrals lacks an elementary antiderivative (here, $\int e^{-y^2}\,dy$, related to the error function) while the other does not.
Result
The solution to the integral is: