$D(x,y)=\frac{xy}{x^3+y^3}$ verifies $\int_0^1 D(x,y) dx \leq c$

Question

Let $D(x,y)=\frac{xy}{x^3+y^3}$. In order to prove that the operator $$T_D : L^2(0,1) \to L^2(0,1), \quad f \mapsto T_Df(x)=\int_0^1 D(x,y)f(y)dy $$ is bounded, I need to show that there exists some $c\geq 0$ such that for almost every $y \in (0,1)$ we have $$ \int_0^1 D(x,y) dx \leq c. $$ Solving this integral and finding directly a bound is turning out to be complicated. I would really appreciate any ideas to find a solution.

Answer 1

Notice that $\frac{xy}{x^3+y^3}\,dx=\frac{(x/y)}{(x/y)^3+1}\cdot\frac{dx}{y}$.

Now since $y\neq 0$ we can let $u=x/y$ so that $du=dx/y:$

$$\int_0^1 D(x,y)\,dx =\int_0^{1/y} \frac{u}{u^3+1}\,du\le \int_0^\infty \frac{u}{u^3+1}\,du.$$

Clearly this integral converges since $$\int_1^\infty \frac{u}{u^3+1}\,du\le \int_1^\infty \frac{u}{u^3}\,du=1,$$ and similarly, $$\int_0^1 \frac{u}{u^3+1}du \le \int_0^1 \frac{u}{1} \,du=.5.$$

Therefore $c=1.5$ works, and it's relatively close to the sharp bound of approximately $1.2$ from the comments.