Limit $\lim _{t\to 0 }\frac{ \int_0^{\infty} \frac{e^{-xt}}{\pi^2+(\log x)^2}dx }{ \int_0^{1/t}\frac{dx}{(\log x)^2}}$ $$
\mbox{Prove}\qquad
\lim _{t \to 0}{\displaystyle\int_{0}^{\infty}{\mathrm{e}^{-xt} \over
\pi^{2} + \log^{2}\left(x\right)}\,\mathrm{d}x
\over
\displaystyle
\int_{0}^{1/t}{\mathrm{d}x \over \ln^{2}\left(x\right)}}
\qquad\mbox{tends to a finite limit.}
$$

*

*( It might be the case that the limit is in fact $1$, but I am not sure. )

*I was not able to find any theorem regarding a situation in which both the range and function depends on $t$.

*The Monotone Convergence theorems I have seen are the closest related, but they are either only for "function independent of $t$ and range dependent on $t$" or "function dependent on $t$ and range independent of $t$".

Any help appreciated !.
 A: First, noting that $\frac1{\log^2(x)}\sim \frac1{(x-1)^2}$ as $x\to 1$, the integral $\int_0^{1/t}\frac{1}{\log^2(x)}\,dx$ diverges for all $t<1$ due to the sharp singularity at $x=1$.
So, let's look instead at the integral
$$f(t)=\int_C^{1/t}\frac{1}{\log^2(x)}\,dx\tag1$$
for $C>1$, $t<1$.  Using integration by parts on the integral in $(1)$, we see that
$$\begin{align}
f(t)&=\int_C^{1/t}\frac1{\log^2(x)}\,dx\\\\
&\overbrace{=}^{\text{IBP}}\frac{1}{t\log^2(t)}-\frac{C}{\log^2(C)}-\int_{C}^{1/t}\frac1{\log^3(x)}\,dx\\\\
&=\frac{1}{t\log^2(t)}+O\left(\frac1{t\log^3(t)}\right)\\\\
&=\frac{1+o(1)}{t\log^2(t)}\tag2
\end{align}$$


Now, let's investigate the integral $g(t)=\int_0^\infty \frac{e^{-xt}}{\pi^2+\log^2(x)}\,dx$.  Enforcing the substitution $xt\mapsto x$ reveals
$$\begin{align}
g(t)&=\int_0^\infty \frac{e^{-xt}}{\pi^2+\log^2(x)}\,dx\\\\
&=\frac1{t \log^2(t)}\int_0^\infty \frac{\log^2(t)}{\pi^2+\log^2(x/t)}e^{-x}\,dx\\\\
&=\frac1{t \log^2(t)}\int_0^{\sqrt{t}} \frac{\log^2(t)}{\pi^2+\log^2(x/t)}e^{-x}\,dx\\\\
&+\frac1{t \log^2(t)}\int_{\sqrt{t}}^\infty \frac{\log^2(t)}{\pi^2+\log^2(x/t)}e^{-x}\,dx\tag3
\end{align}$$
For the first integral on the right-hand side of $(3)$, we make use of the estimate $\frac{1}{\pi^2+\log^2(x/t)}\le \frac{1}{\pi^2}$. Then, we have for the first integral
$$\begin{align}
0&\le \frac1{t \log^2(t)}\int_0^{\sqrt{t}} \frac{\log^2(t)}{\pi^2+\log^2(x/t)}e^{-x}\,dx\\\\
&\le \frac1{t\log^2(t)} \frac{\left(1-e^{-\sqrt{t}}\right)\log^2(t)}{\pi^2 }\\\\
&=\frac{o(1)}{t\log^2(t)}
\end{align}$$
For the second integral we make use of the estimate $\left|\xi_{x\ge \sqrt t}(x)\frac{\log^2(t)}{\pi^2+\log^2(x/t)}e^{-x}\right|\le \frac{\log^2(t)}{\pi^2+\frac14\log^2(t)}e^{-x}\le 4e^{-x}$. Then, we apply the Dominated Convergence Theorem to reveal
$$\begin{align}
\lim_{t\to 0}\int_{\sqrt{t}}^\infty \frac{\log^2(t)}{\pi^2+\log^2(x/t)}e^{-x}\,dx&=\lim_{t\to 0}\int_{0}^\infty \xi_{x\ge \sqrt t}(x)\frac{\log^2(t)}{\pi^2+\log^2(x/t)}e^{-x}\,dx\\\\
&\int_{0}^\infty \lim_{t\to 0}\left(\xi_{x\ge \sqrt{t}}(x)\frac{\log^2(t)}{\pi^2+\log^2(x/t)}\right)e^{-x}\,dx\\\\
&=\int_0^\infty e^{-x}\,dx\\\\
&=1
\end{align}$$
from which we conclude
$$\frac1{t\log^2(t)}\int_{\sqrt{t}}^\infty \frac{\log^2(t)}{\pi^2+\log^2(x/t)}e^{-x}\,dx=\frac{1+o(1)}{t\log^2(t)}$$
Now just put everything together and finish it off.
A: Here I present a couple of observations.


*

*The numerator $g(t):=\int^\infty_0\frac{e^{-xt}}{\pi^2+\log^2(x)}\,dx$ is finite for any $0<t<\infty$, decreasing and $g(t)\xrightarrow{t\rightarrow0}\int^\infty_0\frac{dx}{\pi^2+\log^2(x)}=\infty$ by monotone convergence.

*The denominator $h(t):=\int^{1/t}_0\frac{dx}{\log^2(x)}$ is non decreasing in $t$, finite for $t>1$ and $\infty$ for $0<t\leq1$ since $\frac{1}{\log^2(x)}\sim\frac{1}{(1-x)^2}$ as $x\rightarrow1$.


All this means that  $\frac{g(t)}{h(t)}=0$ for all $0<t<1$ and so, $\lim_{t\rightarrow0}\frac{g(t)}{h(t)}=0$.

As Marc Viola pointed out, a more interesting problem is to analyze the existence (or lack of) of the limit when $h(t)$ is changed for $h_c(t)=\int^{1/t}_c\frac{dx}{\log^2(x)}$ for $c>1$ and $0<t\leq c$. The function $h_c$ is finite an decreasing in $t$, and $h_c(t)\xrightarrow{t\rightarrow0}\infty$ since $x>\log^2(x)$ for all $x$ large enough.  
Here Mark Viola's details asymptotic will be useful.
