Find $\lim \limits_{y\rightarrow\infty}\left (\ln^2y\,-2\int_{0}^y\frac{\ln x}{\sqrt{x^2+1}}dx\right)$ I have difficulty with this limit. Where to start?
$$\lim_{y\rightarrow\infty}\left (\ln^2y\,-2\int_{0}^y\frac{\ln x}{\sqrt{x^2+1}}dx\right)$$
 A: An alternative approach: Write $ \ln^2y $ as the integral of its derivative:
$$ \ln^2 y = 2\int_1^y \frac{\ln x}{x}dx,$$
and then combine with the integral that explicitly appears in $F(y)$, where $F$ is the function whose limit you want to calculate. Combining yields
$$ F(y) = 2\int_1^y\frac{\ln x}{x\sqrt{x^2+1}(x+\sqrt{x^2+1})}dx - 2\int_0^1\frac{\ln x}{\sqrt{x^2+1}}dx,$$
and so the limit is 
$$L = 2\int_1^\infty\frac{\ln x}{x\sqrt{x^2+1}(x+\sqrt{x^2+1})}dx - 2\int_0^1\frac{\ln x}{\sqrt{x^2+1}}dx,$$
where the improper infinite integral clearly converges due to $\ln x/x^3$ fall-off as $x\to\infty$. This at least shows that the limit exists. As for evaluating those integrals...Mathematica yields a bunch of log, polylog and arcsinh terms eventually yielding the same result that @sos440 found in a much nicer analytic way. 
A: By simple integration by parts, we have
$$ \int_{0}^{y} \frac{\log x}{\sqrt{1+x^2}} \; dx = \log y \, \sinh^{-1}y - \int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx. $$
Now by the substitution
$$x = \frac{u^2-1}{2u} \quad \Longleftrightarrow \quad u = x + \sqrt{x^2+1},$$
and the easy equality $ \sinh^{-1} y = \log \left( y + \sqrt{y^2+1} \right)$, we have
$$ \int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx = -\int_{1}^{y + \sqrt{y^2+1}} \left( \frac{2u}{1-u^2} + \frac{1}{u} \right) \log u \; du = -F\bigg(y+\sqrt{y^2+1}\bigg),$$
where 
$$F(s) := \int_{1}^{s} \left( \frac{2u}{1-u^2} + \frac{1}{u} \right) \log u \; du.$$
Now simple observation shows that for $s > 0$ we have
$$F\left( \frac{1}{s} \right) = -F(s).$$
Since $y + \sqrt{y^2+1} \gg 1$ whenever $y \gg 1$, in view of the identity above, we may calculate $F(s)$ for $s = \left( y + \sqrt{y^2+1} \right)^{-1} \in (0, 1)$ instead since
$$ F\left(y+\sqrt{y^2+1}\right) = -F\left(\frac{1}{y+\sqrt{y^2+1}}\right) = -F(s) .$$
Now we introduce the dilogarithm function, defined by
$$ \mathrm{Li}_{2} (x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2} = -\int_{0}^{x} \frac{\log(1-t)}{t} \; dt.$$
Then
$$ \begin{align*}
F(s)
&= \frac{1}{2} \int_{1}^{s} \frac{\log(u^2)}{1-u^2} \; (2udu) + \int_{1}^{s} \frac{\log u}{u} \; du \\
&= \frac{1}{2} \int_{1}^{s^2} \frac{\log v}{1-v} \; dv + \frac{1}{2} \log^2 s \qquad (v = u^2) \\
&= - \frac{1}{2} \int_{0}^{1-s^2} \frac{\log (1-w)}{w} \; dw + \frac{1}{2} \log^2 s \qquad (w = 1-v) \\
&= \frac{1}{2} \mathrm{Li}_{2}(1-s^2) + \frac{1}{2} \log^2 s. \qquad (w = 1-v)
\end{align*}$$
Thus plugging back, we have
$$ \int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx = \frac{1}{2} \left[ \mathrm{Li}_{2} \left( \frac{2y}{y+\sqrt{y^2+1}}\right) + \log^2 \left(y+\sqrt{y^2+1}\right) \right].$$
This shows that
$$ \begin{align*}
\log^2 y - 2\int_{0}^{y} \frac{\log x}{\sqrt{x^2+1}}\;dx
&= \log^2 y - 2 \log y \, \log \left( y + \sqrt{y^2+1} \right) \\
& \qquad + \mathrm{Li}_{2} \left( \frac{2y}{y+\sqrt{y^2+1}}\right) + \log^2 \left(y+\sqrt{y^2+1}\right) \\
&= \mathrm{Li}_{2} \left( \frac{2y}{y+\sqrt{y^2+1}}\right) + \log^2 \left(1+\sqrt{1+y^{-2}}\right),
\end{align*}$$
which clearly converges to
$$ \mathrm{Li}_{2}(1) + \log^2 2 = \zeta(2) + \log^2 2 = \frac{\pi^2}{6} + \log^2 2.$$
Numerical experiment shows that this converges to its limit relatively fast.

In fact, reflection formula for dilagorithm gives the following estimate.
$$ \log^2 y - 2\int_{0}^{y} \frac{\log x}{\sqrt{x^2+1}}\;dx = \frac{\pi^2}{6} + \log^2 2 - \frac{\log y}{2y^2} + O\left(\frac{1}{y^2}\right).$$
A: Take $y=e^{1/t}$,then this limit becomes $$\lim_{t\to0^+}\frac{1-2t^2\int_0^{e^{1/t}}\frac{\ln x}{\sqrt{x^2+1}}dx}{t^2}$$.Now you can solve it easily using L' Hopital's Rule.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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The Question:
  $\ds{\lim_{y \to \infty}\bracks{\ln^{2}\pars{y} -
      2\int_{0}^{y}{\ln\pars{x}\over \root{x^{2} + 1}}\,\dd x} =\, ?}$

\begin{align}
&\color{#f00}{\lim_{y \to \infty}\bracks{\ln^{2}\pars{y} - 2\int_{0}^{y}{\ln\pars{x}\over \root{x^{2} + 1}}\,\dd x}} =
\lim_{y \to \infty}\bracks{\ln^{2}\pars{y} - \int_{x = 0}^{x = y}
{x \over \root{x^{2} + 1}}\,\dd\bracks{\ln^{2}\pars{x}}}
\\[3mm] = &
\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{x^{2} + 1}^{3/2}}\,\dd x =
\lim_{\mu \to 0}\partiald[2]{}{\mu}
\int_{0}^{\infty}{x^{\mu} \over \pars{x^{2} + 1}^{3/2}}\,\dd x\
\stackrel{x^{2}\ \to x}{=}\
\half\,\lim_{\mu \to 0}\partiald[2]{}{\mu}
\int_{0}^{\infty}{x^{\mu/2 - 1/2} \over \pars{x + 1}^{3/2}}\,\dd x
\end{align}

With the identity
$\ds{\int_{0}^{\infty}{t^{a - 1} \over \pars{1 + t}^{a + b}}\,\dd t =
     {\Gamma\pars{a}\Gamma\pars{b} \over \Gamma\pars{a + b}}}$:
\begin{align}
&\color{#f00}{\lim_{y \to \infty}\bracks{\ln^{2}\pars{y} - 2\int_{0}^{y}{\ln\pars{x}\over \root{x^{2} + 1}}\,\dd x}} =
\half\,\lim_{\mu \to 0}\partiald[2]{}{\mu}\bracks{%
{\Gamma\pars{\mu/2 + 1/2}\Gamma\pars{1 - \mu/2} \over \Gamma\pars{3/2}}}
\\[3mm] = &\
\color{#f00}{{\pi^{2} \over 6} + \ln^{2}\pars{2}} \approx 2.1254
\end{align}
