<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] It appears that you wrote
$$
\log\left(\frac{a^2}u\right)^2=2\log(a)^2-\log(u)^2
$$
when it should be
$$
4\log(a)^2-4\log(a)\log(u)+\log(u)^2
$$

Using the Dirichlet beta function as evaluated in this answer
$$
\begin{align}
&\int_0^\infty\frac{\log(x)^2}{a^2+x^2}\,\mathrm{d}x\tag1\\
&=\frac1a\int_0^\infty\frac{\log(x)^2+2\log(a)\log(x)+\log(a)^2}{1+x^2}\,\mathrm{d}x\tag2\\
&=\frac1a\int_{-\infty}^\infty\frac{x^2+2\log(a)x+\log(a)^2}{2\cosh(x)}\,\mathrm{d}x\tag3\\
&=\frac1a\int_0^\infty\frac{x^2+\log(a)^2}{\cosh(x)}\,\mathrm{d}x\tag4\\
&=\frac2a\int_0^\infty\left(x^2+\log(a)^2\right)\sum_{k=0}^\infty (-1)^ke^{-(2k+1)x}\,\mathrm{d}x\tag5\\
&=\frac2a\sum_{k=0}^\infty\int_0^\infty\frac{(-1)^k}{(2k+1)^3} x^2e^{-x}\,\mathrm{d}x+\frac{2\log(a)^2}a\sum_{k=0}^\infty\int_0^\infty \frac{(-1)^k}{2k+1}e^{-x}\,\mathrm{d}x\tag6\\
&=\frac{2\Gamma(3)}a\beta(3)+\frac{2\Gamma(1)\log(a)^2}a\beta(1)\tag7\\[6pt]
&=\frac{\pi^3}{8a}+\frac{\pi\log(a)^2}{2a}\tag8
\end{align}
$$
Explanation:
$(2)$: substitute $x\mapsto ax$
$(3)$: substitute $x\mapsto e^x$
$(4)$: remove the integral of the odd function and apply symmetry
$(5)$: write $\frac1{\cosh(x)}$ for $x\gt0$ as a series
$(6)$: split the sum and substitute $x\mapsto\frac{x}{2k+1}$
$(7)$: apply the beta sum and Gamma integral
$(8)$: evaluate the beta and Gamma functions
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