Evaluating $\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$

Question

I want to evaluate $$\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$$ My attempt: Letting $$I(\alpha,\beta)=\int_0^1\frac{\ln(1+\alpha^2x^2)\ln(1+\beta^2x^2)}{x^4}dx$$ $$ \begin{aligned} I_{12}''(\alpha,\beta)&=\int_0^1\frac{4\alpha\beta}{(1+\alpha^2x^2)(1+\beta^2x^2)}dx\\ &=\frac{4\alpha\beta}{\alpha^2-\beta^2}\int_0^1\frac{\alpha^2}{1+\alpha^2x^2}-\frac{\beta^2}{1+\beta^2x^2}dx\\ &=\frac{4\alpha\beta}{\alpha^2-\beta^2}(\alpha\arctan\alpha-\beta\arctan\beta) \end{aligned} $$ $$ I=\int_0^1\int_0^1I_{12}''(\alpha,\beta)d\beta d\alpha $$ But I can't go further.

Answer 1

Taking integration by parts,

$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx = -\frac{1}{3}\log^2 2 + \frac{4}{3}\int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx. $$

Now

$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx = \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx - \int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx, $$

and the first integral is easily computed by integration by parts:

$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx = -\log 2 + \frac{\pi}{2}. $$

The second integral is trickier, and plugging $x=\tan\theta$ and utilizing the expansion

\begin{align*} \log \sec\theta = -\log \left| \frac{e^{i\theta} + e^{-i\theta}}{2} \right| &= \log 2 - \operatorname{Re}\log(1+e^{2i\theta}) \\ &= \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\cos(2n\theta), \end{align*}

we have

\begin{align*} \int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx &= 2 \int_{0}^{\frac{\pi}{4}} \log \sec\theta \, d\theta \\ &= \frac{\pi}{2}\log 2 - \underbrace{ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2} }_{=G,} \end{align*}

where $G$ is Catalan's constant. Combining altogether, we obtain

$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx = \frac{1}{3}\left( 4G - \log^2 2 - 4\log2 - 2\pi\log2 + 2\pi \right). $$

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Remark. Of course, some CAS can deal with this integral. For instance, Mathematica 11 yields

Answer 2

The function does not have a elementary anti derivative.

Approximately (by numerical integration) the answer is $0.779611255707666$.