Discuss the convergence of the double integral of sin(x)sin(y)/(x+y)^p over the region x >= 0, y >= 0, x + y >= 1.

Step-by-step solution

1. 1

   Change variables so the weight depends on one variable. The integrand only sees $x+y$ through the denominator, so set $u=x+y$, $v=x$. Then $y=u-v$ and the Jacobian is $1$. The constraints $x\ge0$, $y\ge0$ become $0\le v\le u$, and $x+y\ge1$ becomes $u\ge1$, so $I(p)=\int_1^\infty \frac{1}{u^p}\left(\int_0^u \sin v\,\sin(u-v)\,dv\right)du.$

2. 2

   Collapse the inner integral with a product-to-sum identity. Using $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$ with $A=v$ and $B=u-v$ gives $\sin v\sin(u-v)=\tfrac12[\cos(2v-u)-\cos u]$. Integrating over $v\in[0,u]$, the first piece contributes $\sin u$ and the second contributes $u\cos u$, so $\int_0^u \sin v\,\sin(u-v)\,dv=\frac{\sin u-u\cos u}{2}.$

3. 3

   Split the result into two oscillatory tails. Substituting back, $I(p)=\frac12\int_1^\infty\frac{\sin u}{u^p}\,du-\frac12\int_1^\infty\frac{\cos u}{u^{p-1}}\,du.$ The two-dimensional convergence question is now two one-dimensional improper integrals of the same standard shape.

4. 4

   Apply the Dirichlet test to each tail. The partial integrals $\int_1^X\sin u\,du$ and $\int_1^X\cos u\,du$ stay bounded, so $\int_1^\infty \sin u/u^p\,du$ converges whenever $u^{-p}$ decreases to $0$, that is $p>0$, and $\int_1^\infty \cos u/u^{p-1}\,du$ converges whenever $p-1>0$. The second condition is the binding one. At $p=1$ the second tail is $\int_1^\infty\cos u\,du$, which oscillates without settling, and for $p<1$ its amplitude $u^{1-p}$ even grows, so convergence holds exactly when $p>1$.

5. 5

   Test absolute convergence separately. Taking absolute values destroys the cancellation: $A(u)=\int_0^u|\sin v\,\sin(u-v)|\,dv$ grows linearly, $cu\le A(u)\le Cu$ for large $u$, because $|\sin v\sin(u-v)|$ has a positive average over each period. Hence $\iint_D|\cdot|\asymp\int_1^\infty u^{1-p}\,du$, which converges exactly when $1-p<-1$, that is $p>2$.

6. 6

   Assemble the trichotomy. Absolute convergence needs $p>2$ and plain convergence needs $p>1$, so for $1<p\le 2$ the integral converges but not absolutely, and for $p\le 1$ it diverges.