Find: lim as alpha to 0+ I(a), where I(a)=double integral
Find: $$ \lim_{\alpha\to0^+} I(a), \qquad \text{where } I(a)=\iint_D (x+y)^{-a}\,dx\,dy, $$ and $D$ is the triangular region $x\geqslant0$, $y\geqslant0$, $\alpha<x+y<1$, with $\alpha\to0^+$.
Explanation of Method
Note on notation: the region's lower bound is written here as $\alpha$, kept distinct from the exponent $a$ in the integrand $(x+y)^{-a}$ — the two symbols must not be confused, since it is $\alpha$ that tends to $0^+$ while $a$ is a fixed parameter of the integrand. Switching to the substitution $u=x+y$, $v=y$ collapses the double integral to a single integral in $u$, whose convergence as $\alpha\to0^+$ depends on the size of $a$ relative to $2$.
- Change variables to $u=x+y$, $v=y$, turning the triangular strip into a rectangle in $(u,v)$.
- Integrate out $v$, reducing to a single integral $\int_\alpha^1 u^{1-a}\,du$.
- Take the limit $\alpha\to0^+$, distinguishing the cases $a<2$, $a=2$, and $a>2$.
Step 1 · Change of variables
Let $u=x+y$, $v=y$, so $x=u-v$, $y=v$, with Jacobian $|\partial(x,y)/\partial(u,v)|=1$. The region $x\geqslant0,\,y\geqslant0,\,\alpha<x+y<1$ becomes $\alpha<u<1$, $0\leqslant v\leqslant u$:
$$ I(a) = \int_\alpha^1\int_0^u u^{-a}\,dv\,du = \int_\alpha^1 u^{1-a}\,du. $$
Step 2 · Evaluate the remaining integral
For $a\neq2$:
$$ I(a) = \left[\frac{u^{2-a}}{2-a}\right]_\alpha^1 = \frac{1-\alpha^{2-a}}{2-a}. $$
For $a=2$:
$$ I(2) = \int_\alpha^1 \frac{du}{u} = -\ln\alpha. $$
Step 3 · Take the limit $\alpha\to0^+$
If $a<2$, then $2-a>0$, so $\alpha^{2-a}\to0$ as $\alpha\to0^+$:
$$ \lim_{\alpha\to0^+} I(a) = \frac{1}{2-a}. $$
If $a=2$, then $-\ln\alpha\to+\infty$ as $\alpha\to0^+$, so the limit diverges.
If $a>2$, then $2-a<0$, so $\alpha^{2-a}\to+\infty$ as $\alpha\to0^+$, and the limit again diverges (to $+\infty$).
Summary Table
| Case | $\displaystyle\lim_{\alpha\to0^+} I(a)$ |
|---|---|
| $a<2$ | $\dfrac{1}{2-a}$ |
| $a\geqslant2$ | diverges ($+\infty$) |
Additional Notes
- The divergence for $a\geqslant2$ reflects the non-integrable singularity of $(x+y)^{-a}$ at the origin once $a$ reaches or exceeds the dimension-related threshold $2$ for a two-dimensional region.
Result
The solution to the integral is: