I am studying the following integral with parameter $$E(t) = \int_{-T}^{t - \lambda(t)^2} \frac{1}{|\ln(T - s)|^2(t - s)}ds, \quad \text{for } t \in [-T/4, T].$$ $T$ is a constant that can be chosen arbitrarily small and $\lambda$ is defined as $$\lambda(t) = \frac{T - t}{|\ln(T - t)|^2}.$$ We see that, at $t = T$, $$E(T) = \frac{1}{|\ln(2T)|},$$ where we used the substitution $u = |\log(T - s)|$ in the integral to compute it explicitly. Now the thing is that for $t< T$ this function $E(t)$ is harder to manipulate as we cannot compute the integral anymore. I would like to prove that $$E(t) - E(T) \ge 0$$ for every $t$ in $[-T/4, T]$. I computed $E(t) - E(T)$ using Python for $T = 0.04$ and it looks like on the picture below. The inequality seems obvious in this case.
I tried to see what was the shape of the derivative but $E'(t)$ is extremely ugly so I couldn't conclude anything from there. Moreover, I tried to rewrite $E(t) - E(T)$ as $$E(t) - E(T) = \underbrace{\int_{-T}^{t - \lambda(t)^2} \frac{1}{|\ln(T - s)|^2} \left(\frac{1}{t - s} - \frac{1}{T - s}\right)ds}_{\equiv A} - \underbrace{\int_{t - \lambda(t)^2}^T \frac{1}{|\log(T - s)|^2(T - s)}ds}_{\equiv B}$$ and then show that $A \ge B$ but again, it isn't obvious. Anyone has an idea about this ?
Let's try to figure it out from the other end. Namely, let's see for what positive functions $\varphi(\sigma)$ ($\sigma=T-t$) we have that $$ F(\sigma)=\int_{\sigma+\varphi(\sigma)}^{2T}\frac{ds}{(s-\sigma)|\log s|^2} $$ is increasing on $\sigma\in(0,\frac{5T}4)$. I'll assume that $T<\frac 1{2e}$, so all logarithms are negative and greater than $1$ in absolute value.
Taking the derivative in $\sigma$, we see that we need the inequality $$ \frac{1+\varphi'}{\varphi |\log(\sigma+\varphi)|^2} \le \int_{\sigma+\varphi}^{2T}\frac{ds}{(s-\sigma)^2|\log s|^2} $$ We write the last integral as the sum of $$ \int_{\sigma+\varphi}^{2T}\frac{ds}{(s-\sigma)^2|\log(\sigma+\varphi)|^2} =\frac{1}{|\log(\sigma+\varphi)|^2}\left(\frac 1\varphi-\frac{1}{2T-\sigma}\right) $$ and $$ \int_{\sigma+\varphi}^{2T}\frac{ds}{(s-\sigma)^2}\left[ \frac 1{|\log s|^2}-\frac 1{|\log(\sigma+\varphi)|^2} \right] \\ \ge 2\frac 1{|\log(\sigma+\varphi)|^3}\int_{\sigma+\varphi}^{2T}\frac{ds}{(s-\sigma)^2} \frac {(s-\sigma)-\varphi}{s} $$ where we used the inequality $$ |\log a|-|\log b|\ge \frac{b-a}b $$ for $0<a<b<1$ and its consequence $$ \frac 1{|\log b|^2}\ge \frac 1{|\log a|^2}\frac 1{(1-\frac{(b-a)/b}{|\log a|})^2} \ge \frac 1{|\log a|^2}\left(1+2\frac{(b-a)/b}{|\log a|}\right)\,. $$ The last integral can be computed explicitly, but I'm not a big fan of partial fractions, so I'll just make the change of variable $s-\sigma=u$ and estimate it as $$ \int_{\varphi}^{\infty}\frac{u-\varphi}{u^2(\sigma+u)}\,du-\int_{2T-\sigma}^\infty\frac{u-\varphi}{u^2(\sigma+u)}\,du= \\ \int_{\varphi}^{\infty}\frac{u}{u^2(\sigma+u)}\,du -\int_{\varphi}^{\infty}\frac{\varphi}{u^2(\sigma+u)}\,du -\int_{2T-\sigma}^{\infty}\frac{u-\varphi}{u^2(\sigma+u)}\,du =I_1-I_2-I_3\,. $$ Now $I_1=\frac 1\sigma \log\frac{\sigma+\varphi}\varphi$, $ I_2\le \frac 1\sigma $ (ignore $u$ in $\sigma+u$) and $I_3\le \int_{2T-\sigma}^\infty \frac{du}{u^2}=\frac{1}{2T-\sigma}$.
Thus, bringing everything together and multiplying by $|\log(\sigma+\varphi)|^2$, we see that we will be happy if $$ \frac{\varphi'}{\varphi}\le -\frac{1}{2T-\sigma}\left(1+\frac{2}{|\log(\sigma+\varphi)|}\right) +\frac{2}\sigma\frac 1{|\log(\sigma+\varphi)|}\left[\log\frac{\sigma+\varphi}\varphi -1\right]\,. $$ Assuming that $\varphi\le e^{-1}(\sigma+\varphi)^2$ and recalling that all $|\log|$'s are $\ge 1$ and $\sigma<\frac 54T$, we see that under these restrictions, we just need $$ \frac{\varphi'}\varphi\le -\frac{4}T+\frac 2\sigma\, $$ which calls for $\varphi(\sigma)=\sigma^2 e^{-1-\frac{4\sigma}T}$, which dominates $c(T-t)^2$ for $c=e^{-6}$, say, on the required interval. And you threw the whole $|\log\sigma|^4$ into the denominator of your $\lambda^2$, which was way more than really needed and which was the reason for such a sharp rise at the endpoint $t=T$ on your graph.
Technically you should also show that with my $\varphi$, one has $F(\sigma)\to F(0)$ as $\sigma\to 0$, but I leave that to you (hint: use the uniform equi-integrability criterion, not the DCT!)
The End.