求极限:

\[ \lim_{y\rightarrow+\infty}{\left(\ln^{2}{y}-2\int_{0}^{y}{\frac{\ln{y}}{\sqrt{x^2+1}}dx}\right)} \]

(Proposed by tian275461, Solution by sos440)

\[ \begin{align*} I&=\ln^{2}y-2\int_{0}^{y}{\frac{\ln{x}}{x}\cdot\frac{x}{\sqrt{x^2+1}}dx}\\
&=\ln^{2}{y}-\frac{y}{\sqrt{y^2+1}}\cdot\ln^{2}{y}+\int_{0}^{y}{\frac{\ln^{2}{x}}{(x^2+1)^{\frac{3}{2}}}dx}                                          \end{align*}\]
notice that
\[ \lim_{y\rightarrow+\infty}{\left(\ln^{2}{y}-\frac{y}{\sqrt{y^2+1}}\cdot\ln^{2}{y}\right)}=0 \]
So,just find the value of the following integral
\[ \int_{0}^{\infty}{\frac{\ln^{2}{x}}{(x^2+1)^{\frac{3}{2}}}dx} \]
First note that
\[\begin{align*}
I(p, q) := \int_{0}^{\infty} \frac{x^{p}}{(1+x^2)^{q}} \, dx
&= \int_{0}^{\frac{\pi}{2}} \frac{\tan^{p} \theta}{\sec^{2q} \theta} \, \sec^2 \theta \, d\theta \qquad (x = \tan\theta)\\
&= \int_{0}^{\frac{\pi}{2}} \sin^{p} \theta \cos^{2q-p-2} \theta \, d\theta \\
&= \frac{1}{2}\beta\left( \frac{p+1}{2}, \frac{2q-p-1}{2} \right) \\
&= \frac{1}{2}\frac{\Gamma \left( \frac{p+1}{2} \right) \Gamma \left( \frac{2q-p-1}{2} \right)}{\Gamma(q)}.
\end{align*}\]
Thus
\[ \int_{0}^{\infty} \frac{\log^2 x}{(1+x^2)^{3/2}} \, dx = \frac{\partial^2 I}{\partial p^2}\left( 0, \tfrac{3}{2} \right) \]
and the rest is a mere calculation. Note first that
\[ I\left(p, \tfrac{3}{2} \right) = \frac{\Gamma \left(\frac{1+p}{2}\right) \Gamma \left(1 - \frac{p}{2}\right) }{\Gamma \left(\frac{1}{2}\right) }. \]
Differentiating, we have
\[ \frac{\partial I}{\partial p}\left(p, \tfrac{3}{2} \right) = \frac{\Gamma \left(\frac{1+p}{2}\right) \Gamma \left(1 - \frac{p}{2}\right) }{2\Gamma \left(\frac{1}{2}\right) } \left[ \psi_{0}\left( \frac{1+p}{2} \right) - \psi_{0}\left( 1 - \frac{p}{2} \right) \right]. \]
Differentiating once again, we have
\[ \frac{\partial^2 I}{\partial p^2} \left(p, \tfrac{3}{2} \right) = \frac{\Gamma \left(\frac{1+p}{2}\right) \Gamma \left(1 - \frac{p}{2}\right) }{4\Gamma \left(\frac{1}{2}\right) } \left[ \left\{\psi_{0}\left( \frac{1+p}{2} \right) - \psi_{0}\left( 1 - \frac{p}{2} \right) \right\}^2 + \left\{\psi_{1}\left( \frac{1+p}{2} \right) + \psi_{1}\left( 1 - \frac{p}{2} \right) \right\} \right] . \]
Plugging $p = 0$, we have
\[ \frac{\partial^2 I}{\partial p^2} \left(0, \tfrac{3}{2} \right) = \frac{1}{4} \left[ \left\{\psi_{0}\left( \tfrac{1}{2} \right) - \psi_{0}(1) \right\}^2 + \left\{\psi_{1}\left( \tfrac{1}{2} \right) + \psi_{1}(1) \right\} \right] . \]
In view of the expansion formula
\[ \psi_{0}(s) = -\gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+s} \right), \]
we obtain
\[\begin{align*}
\psi_{0}\left( \tfrac{1}{2} \right) - \psi_{0}(1)
&= \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+\frac{1}{2}} \right) \\
&= -2 \sum_{n=0}^{\infty} \left( \frac{1}{2n+1} - \frac{1}{2n+2} \right) \\
&= -2 \ln{2}
\end{align*}\]
and likewise
\[\begin{align*}
\psi_{1}\left( \tfrac{1}{2} \right) + \psi_{1}(1)
&= \sum_{n=0}^{\infty} \left( \frac{1}{\left( n+\frac{1}{2} \right)^2} + \frac{1}{(n+1)^2} \right) \\
&= 4 \sum_{n=0}^{\infty} \left( \frac{1}{(2n+1)^2} + \frac{1}{(2n+2)^2} \right) \\
&= 4 \zeta(2) .
\end{align*}\]
Therefore we have
\[ \frac{\partial^2 I}{\partial p^2} \left(0, \tfrac{3}{2} \right) = \zeta(2) + \ln^{2}{2}.\]

$\square$