Examine the convergence of the improper double integral
Examine the convergence of the improper double integral: $$ \iint_{y\geqslant x} \frac{dx\,dy}{x^2+1}. $$
Explanation of Method
The integrand $\dfrac1{x^2+1}$ does not depend on $y$ at all, yet the region $y\geqslant x$ lets $y$ range over an unbounded interval $[x,\infty)$ for every fixed $x$. Integrating a nonzero constant (in $y$) over an infinite range diverges immediately — no cancellation or decay in $y$ is available to rescue it.
- Set up the iterated integral with $y$ as the inner variable.
- Observe that the inner integral diverges for every fixed $x$.
- Conclude that the double integral diverges.
Step 1 · Set up the iterated integral
For fixed $x$, $y$ ranges over $[x,\infty)$:
$$ \iint_{y\geqslant x}\frac{dx\,dy}{x^2+1} = \int_{-\infty}^{\infty}\frac{1}{x^2+1}\left(\int_x^\infty dy\right)dx. $$
Step 2 · Evaluate the inner integral
$$ \int_x^\infty dy = \lim_{R\to\infty}(R-x) = +\infty \qquad \text{for every fixed } x. $$
Step 3 · Conclude
Since the inner integral is $+\infty$ for every $x$, and $\dfrac1{x^2+1}>0$ throughout, the outer integral is likewise $+\infty$:
$$ \iint_{y\geqslant x}\frac{dx\,dy}{x^2+1} = \int_{-\infty}^{\infty}\frac{+\infty}{x^2+1}\,dx = +\infty. $$
Additional Notes
- This divergence is of a qualitatively different (and more elementary) kind than the logarithmic divergence seen in the radially symmetric integrals in this collection: here the integrand fails to decay in the $y$-direction at all, so no amount of decay in $x$ (indeed $\frac1{x^2+1}$ decays quite fast) can compensate. Any region unbounded in a direction along which the integrand does not vanish will produce divergence in this direct way.
Result
The integral diverges: