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I can't determine a sign of an integral written below and it has hit a dead end.

My setting is rather special. Let $a\in(0,1)$ be a given constant and $(x_{\varepsilon},y_{\varepsilon})\in[0,a)\times[0,a)$ be a given sequence such that $(x_{\varepsilon},y_{\varepsilon})\to(a,a)$ as $\varepsilon\to0$. Assume that $f,g:[0,1]\times[0,1]\to\mathbb{R}$ are continuous functions satisying the following conditions: (1) $f(a,z)>g(a,z)$ for all $z\in[0,a)$, (2) $f(a,a)=g(a,a)$

Then, what I want to know is a sign of $\lim_{\varepsilon\to0}I(\varepsilon)$, where $$ I(\varepsilon):=\int_{0}^{x_{\varepsilon}-\varepsilon}\frac{f(x_{\varepsilon},z)}{(x_{\varepsilon}-z)^{3/2}}dz-\int_{0}^{y_{\varepsilon}-\varepsilon}\frac{g(y_{\varepsilon},z)}{(y_{\varepsilon}-z)^{3/2}}dz. $$

Formally, we know $$ I(0)=\int_{0}^{a}\frac{f(a,z)-g(a,z)}{(a-z)^{3/2}}dz>0, $$ but we cannot prove at least by some convergence theorems, e.g., dominated convergence theorem, Fatou's Lemma and so on.

I'm glad if you give some ideas, solutions or examples such that a sign of $I$ is not determined if exists.

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It is easy to construct examples where the limit is positive.

Below is an example with negative limit. I will assume $a=1$ everywhere. Let $x_\epsilon = a-\sqrt{\epsilon}$, $y_\epsilon = a-\epsilon$ and $$f(x,y)=2x^2-y,\quad g(x,y) = x.$$

Note that $f(1,z) - g(1,z) = 1-z >0$ whenever $z\in [0,1)$.

With these expressions, $I(\epsilon)$ becomes (after asking Mathematica to compute it): $$ I(\epsilon) = 2 \left(2 \sqrt{e} \left(\sqrt{1-\sqrt{e}}+1\right)+\sqrt{1-e}-3\right) $$

We can compute that $\lim_{\epsilon \to 0} I(\epsilon) = -4$.

This was found using Mathematica:

a = 1;
x[e_] := a - Sqrt[e];
y[e_] := a - e;
f[x_, y_] := 2 x^2 - y;
g[x_, y_] := x;
if = Integrate[ f[ x[e], z] / (x[e] - z)^(3/2), {z, 0, x[e] - e}, 
   Assumptions :> {1/10 > e > 0}];
ig = Integrate[ g[ y[e], z] / (y[e] - z)^(3/2), {z, 0, y[e] - e}, 
   Assumptions :> {1/10 > e > 0}];
Limit[if - ig, e -> 0]
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  • $\begingroup$ Thank you for giving a nice example. Where did the idea of this example with negative limit come from? $\endgroup$
    – user
    Commented Dec 18, 2015 at 5:13
  • $\begingroup$ @user: That's a good question. I played around with the parameters until it worked, basically, but intuition sort of suggests that you want the $x_\epsilon$ to approach 0 slower than the $y_\epsilon$, since we want the integral on the right to be larger. $\endgroup$
    – Per Alexandersson
    Commented Dec 18, 2015 at 17:39

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