Let $T > 0$ and $T \ll 1$. Let $f(t, s) = \frac{1}{(t-s)(\ln(T - s))^2}> 0$ and $$E(t) = \int_{-T}^{t - \lambda^2(t)}f(t, s)ds,$$ for some strictly positive and decreasing function $\lambda$ that goes to zero for $t \to T$ and that can be chosen arbitrarily small by choosing $T$ small enough. Now consider $\varepsilon > 0$ such that $\varepsilon < T$. Now I would like to know if it is possible to find the following bound $$E(t) \le C(\varepsilon) \int_{t - \lambda(t)^2 - \varepsilon}^{t - \lambda^2}f(t,s)ds$$ for $C(\varepsilon)$ a constant depending only on $\varepsilon$ (not on $t)$. I feel like it should be possible to find such a constant but I am struggling to compute it explicitly.
EDIT. I actually have found something, could you let me know if you agree?
Note that for every admissible we have $f(t, s) > 0$ for every admissible $(t, s)$, i.e. $$S = \{(t, s)~|~ -T \le s \le t - \lambda(t)^2, -T\le t \le T \},$$ the function $f(t, s) > 0$ and moreover, $\exists \delta > 0$ such that $f(t, s) > \delta$ as the only possible point at which we could have $f(t,s) \to 0$ is for $s \to T$ but we actually have $f(s, T)\to \infty$ in that case. Therefore, $$0< \delta \varepsilon \le \int_{t - \lambda^2 - \varepsilon}^{t - \lambda^2} f(t, s)ds, $$ such that $$E(t) \le \frac{K}{\delta \varepsilon}\delta \varepsilon \le \frac{K}{\delta \varepsilon} \int_{t - \lambda^2 - \varepsilon}^{t - \lambda^2} f(t, s)ds,$$ for $K = \sup_{t\le T}E(t).$