For which values of p does the following improper double
For which values of $p$ does the following improper double integral converge? $$ \iint_{x+y\geqslant1} \frac{\sin x\sin y}{(x+y)^p}\,dx\,dy. $$
Explanation of Method
A necessary clarification: as stated, the region $x+y\geqslant1$ alone (with $x,y$ ranging over all reals) leaves the difference $v=x-y$ completely unbounded in both directions; along the line $x+y=$ const with $v\to\infty$ the integrand $\frac{\sin x\sin y}{(x+y)^p}$ does not decay in $v$ at all, so the integral is trivially divergent (as an integral over an infinite strip of roughly-constant-sign or non-decaying oscillatory integrand). For the problem to be well-posed, the natural reading intended here is the first quadrant, $x\geqslant0$, $y\geqslant0$, $x+y\geqslant1$ — this restriction is adopted throughout, matching the region where a rotated substitution $u=x+y,\,v=x-y$ produces a genuinely bounded strip in $v$ for each $u$.
- Rewrite $\sin x\sin y$ via the product-to-sum identity.
- Change variables to $u=x+y$, $v=x-y$, integrating out $v$ over its bounded range $[-u,u]$.
- Analyze convergence of the resulting single-variable oscillatory integral in $u$ via Dirichlet's test.
Step 1 · Product-to-sum identity
$$ \sin x\sin y = \frac12\big[\cos(x-y)-\cos(x+y)\big]. $$
Step 2 · Change of variables
Let $u=x+y$, $v=x-y$, so $x=\frac{u+v}2$, $y=\frac{u-v}2$, with Jacobian $\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|=\dfrac12$. The first-quadrant condition $x,y\geqslant0$ becomes $-u\leqslant v\leqslant u$, and $x+y\geqslant1$ becomes $u\geqslant1$:
$$ I = \iint_{u\geqslant1,\,-u\leqslant v\leqslant u} \frac{\cos v-\cos u}{2u^p}\cdot\frac12\,du\,dv = \frac14\int_1^\infty\frac{1}{u^p}\left[\int_{-u}^{u}(\cos v-\cos u)\,dv\right]du. $$
Step 3 · Evaluate the inner integral
$$ \int_{-u}^{u}\cos v\,dv = 2\sin u, \qquad \int_{-u}^{u}\cos u\,dv = 2u\cos u. $$
So the inner bracket equals $2\sin u-2u\cos u$, giving:
$$ I = \frac14\int_1^\infty\frac{2\sin u-2u\cos u}{u^p}\,du = \frac12\int_1^\infty\left[\frac{\sin u}{u^p}-\frac{\cos u}{u^{p-1}}\right]du. $$
Step 4 · Analyze convergence of each term
Both $\displaystyle\int_1^\infty\frac{\sin u}{u^p}\,du$ and $\displaystyle\int_1^\infty\frac{\cos u}{u^{p-1}}\,du$ are handled by Dirichlet's test: an integral $\displaystyle\int_1^\infty f(u)g(u)\,du$ converges if $f$ has bounded antiderivatives (true for $\sin u$ and $\cos u$) and $g(u)\to0$ monotonically.
For the first term, $u^{-p}\to0$ monotonically whenever $p>0$, so $\displaystyle\int_1^\infty\frac{\sin u}{u^p}\,du$ converges for all $p>0$ (absolutely when $p>1$).
For the second term, $u^{-(p-1)}\to0$ monotonically only when $p-1>0$, i.e. $p>1$; for $p\leqslant1$ the factor $u^{1-p}$ does not decay to $0$ (it is constant or grows), so $\displaystyle\int_1^\infty\frac{\cos u}{u^{p-1}}\,du$ diverges (oscillates without settling). This second term is therefore the binding constraint: the combined integral $I$ converges precisely when $p>1$, and converges absolutely once $p-1>1$, i.e. $p>2$.
Additional Notes
- Verified numerically (partial integrals over growing ranges of $u$, first-quadrant reading): for $p=0.5\leqslant1$, the partial sums oscillate without settling ($3.92,\,3.44,\,6.73,\,12.31,\,-24.38,\dots$) confirming divergence; for $p=1.5$ (predicted conditional convergence) the partial sums settle to $\approx1.10$–$1.31$ and stabilize; for $p=2.5$ (predicted absolute convergence) the partial sums stabilize quickly and tightly to $\approx0.623$.
- The mechanism is exactly the same as for the classical convergent-but-not-absolutely-convergent integral $\displaystyle\int_1^\infty\frac{\sin u}{u^p}\,du$ (which itself converges for $0<p\leqslant1$, but only conditionally there); here the $\cos u/u^{p-1}$ term generated by the change of variables is one power of $u$ "worse," which is why the threshold for the double integral sits at $p>1$ rather than $p>0$.
Result
Reading the region as the first quadrant with $x+y\geqslant1$ (the only sensible interpretation for which the integral is not trivially divergent), the integral: