<s>[INST] Error in $\int\limits_0^{\infty}dx\,\frac {\log^2 x}{a^2+x^2}$ This is my work for solving the improper integral$$I=\int\limits_0^{\infty}dx\,\frac {\log^2x}{x^2+a^2}$$I feel like I did everything write, but when I substitute values into $a$, it doesn’t match up with Wolfram Alpha.

First substitute $x=\frac {a^2}u$ so$$\begin{align*}I & =-\int\limits_{\infty}^0du\,\frac {2\log^2a-\log^2 u}{x^2+a^2}\\ & =\int\limits_0^{\infty}du\,\frac {2\log^2a}{a^2+x^2}-\int\limits_0^{\infty}du\,\frac {\log^2 u}{a^2+x^2}\end{align*}$$Hence$$\begin{align*}I & =\int\limits_0^{\infty}du\,\frac {\log^2a}{a^2+x^2}\\ & =\frac {\log^2a}{a^2}\int\limits_0^{\pi/2}dt\,\frac {a\sec^2t}{1+\tan^2t}\\ & =\frac {\pi\log^2a}{2a}\end{align*}$$However, when $a=e$ Wolfram Alpha evaluates the integral numerically as$$I\approx 2.00369$$however the input that I arrived at evaluates numerically$$\frac {\pi}{2e}\approx0.5778$$Where did I go wrong? And how would you go about solving this integral?
 [/INST] $I  =-\int_{\infty}^0du\,\frac {2\log^2a-\log^2 u}{x^2+a^2}$
You are mixing $x's$ and $u's$
$I  =\int_0^{\infty}\frac {\ln^2 x}{x^2+a^2}\ dx\\
x = \frac {a^2}{u}, dx = -\frac {a^2}{u^2} du\\
\int_{\infty}^0\frac {(\ln \frac {a^2}{u})^2}{\frac {a^4}{u^2}+a^2} (-\frac {a^2}{u^2})\ du\\
\int_0^{\infty}\frac {(2\ln a - \ln u)^2}{a^2+u^2} \ du$
Which is not 
$\int_0^{\infty}\frac {2\ln^2 a - \ln^2 u}{a^2+u^2} \ du$
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