Let phi be continuous and positive on [-c,c]. Prove that
Let $\varphi$ be continuous and positive on $[-c,c]$. Prove that: $$ \iint_{x^2+y^2<c^2} \frac{a\varphi(x)+b\varphi(y)}{\varphi(x)+\varphi(y)}\,dx\,dy = \frac{(a+b)c^2\pi}{2}. $$
Explanation of Method
The disk $D:x^2+y^2<c^2$ is symmetric under the swap $(x,y)\to(y,x)$. Splitting the integrand into two pieces, $\dfrac{\varphi(x)}{\varphi(x)+\varphi(y)}$ and $\dfrac{\varphi(y)}{\varphi(x)+\varphi(y)}$, the swap exchanges these two pieces with each other — so their integrals over $D$ are equal. Since the two pieces add up to the constant function $1$, each integral must equal exactly half the area of $D$.
- Define $J=\displaystyle\iint_D\frac{\varphi(x)}{\varphi(x)+\varphi(y)}\,dx\,dy$ and $K=\displaystyle\iint_D\frac{\varphi(y)}{\varphi(x)+\varphi(y)}\,dx\,dy$.
- Show $J=K$ using the symmetry $(x,y)\to(y,x)$ of $D$.
- Use $J+K=\operatorname{area}(D)$ to solve for $J$ and $K$, then assemble the answer.
Step 1 · Split the integrand
Write the integrand as a sum of two fractions and split the integral accordingly:
$$ I = \iint_D \frac{a\varphi(x)+b\varphi(y)}{\varphi(x)+\varphi(y)}\,dx\,dy = a\underbrace{\iint_D\frac{\varphi(x)}{\varphi(x)+\varphi(y)}\,dx\,dy}_{J} + b\underbrace{\iint_D\frac{\varphi(y)}{\varphi(x)+\varphi(y)}\,dx\,dy}_{K}. $$
Step 2 · Show $J=K$ by symmetry
The disk $D:x^2+y^2<c^2$ is invariant under relabeling the integration variables $x\leftrightarrow y$ (the region is unchanged, since $x^2+y^2=y^2+x^2$). Relabeling in $J$:
$$ J = \iint_D\frac{\varphi(x)}{\varphi(x)+\varphi(y)}\,dx\,dy \;\xrightarrow{\ x\leftrightarrow y\ }\; \iint_D\frac{\varphi(y)}{\varphi(y)+\varphi(x)}\,dy\,dx = K. $$
Since relabeling a dummy integration variable does not change the value of a definite integral, $J=K$.
Step 3 · Use $J+K=\operatorname{area}(D)$
Since $\varphi(x)+\varphi(y)>0$ on $D$ (as $\varphi>0$), the two fractions in $J$ and $K$ add to $1$ pointwise:
$$ J+K = \iint_D\frac{\varphi(x)+\varphi(y)}{\varphi(x)+\varphi(y)}\,dx\,dy = \iint_D 1\,dx\,dy = \pi c^2. $$
Combined with $J=K$ from Step 2:
$$ J = K = \frac{\pi c^2}{2}. $$
Step 4 · Assemble the result
$$ I = aJ+bK = a\cdot\frac{\pi c^2}{2}+b\cdot\frac{\pi c^2}{2} = \frac{(a+b)\pi c^2}{2}. $$
Additional Notes
- The result is independent of the specific function $\varphi$ (as long as it is continuous and positive) — only the values $a$, $b$, and the radius $c$ matter. This can be checked directly by taking $\varphi\equiv1$, which makes the integrand identically $\frac{a+b}{2}$, immediately giving $\frac{a+b}{2}\cdot\pi c^2$.
- The same symmetry trick (splitting into two pieces exchanged by a symmetry of the domain, then using that they sum to something simple) is a standard device for evaluating integrals of the form $\displaystyle\int\frac{f(x)}{f(x)+f(a-x)}\,dx$ in one variable.
Result
This establishes the claim: