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$\displaystyle\int_0^{\infty}\dfrac{\ln^2x}{1+x^2}dx$

$x=\arctan\alpha$

$dx=\dfrac{1}{1+\alpha^2}d\alpha$

and try a lot items, but didnt arrived anywhere.

Micheal Brain Hurts
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3 Answers3

6

We may prove through Euler's beta function and the reflection formula for the $\Gamma$ function that for any $a\in(-1,1)$ we have $$ I(a)=\int_{0}^{+\infty}\frac{x^a}{x^2+1}\,dx = \frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)} \tag{1}$$ hence: $$ I''(0) = \int_{0}^{+\infty}\frac{\log^2(x)}{x^2+1}\,dx = \lim_{a\to 0}\frac{\pi^3(3-\cos(a\pi))}{16\cos^3\left(\frac{\pi a}{2}\right)}=\color{red}{\frac{\pi^3}{8}}.\tag{2}$$

I would like to point out a related fact, too.

Community
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Jack D'Aurizio
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2

Hint. A possible route. One may write, with the change of variable $x \to \frac1x$, $$ \int_0^{\infty}\dfrac{\ln^2x}{1+x^2}dx=\int_0^1\dfrac{\ln^2x}{1+x^2}dx+\int_1^{\infty}\dfrac{\ln^2x}{1+x^2}dx=2\int_0^1\dfrac{\ln^2x}{1+x^2}dx $$ then one may expand the integrand and integrate termwise, getting $$ 2\int_0^1\dfrac{\ln^2x}{1+x^2}dx=2\sum_{n=0}^\infty(-1)^n\int_0^1x^{2n}\ln^2xdx=4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}=\frac{\pi ^3}{8}. $$

Olivier Oloa
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  • This approach is fine, but requires one to know the value of the series. An alternative is to evaluate the classical keyhole contour integral of $\frac{\log^3(z)}{z^2+1}$. Or even use Feynman's Trick and evaluate the second derivative of $\int_0^\infty \frac{x^a}{x^2+1},dx$ at $a=0$. ;-)) -Mark – Mark Viola Aug 10 '16 at 20:33
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Let us evaluate it via the Mellin transform \begin{equation} \mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x \label{eq:1608132} \tag{2} \end{equation}

where \begin{equation} f(x) = \frac{1}{1+x^{2}} \label{eq:1608133} \tag{3} \end{equation}

Applying the Mellin transform, yields \begin{equation} \mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} \frac{x^{s-1}}{1+x^{2}} \mathrm{d} x = \frac{\pi}{2}\csc\left(\frac{\pi}{2}s\right) \label{eq:1608134} \tag{4} \end{equation}

And thus \begin{align} \int\limits_{0}^{\infty} \frac{\mathrm{ln}^{2}(x)}{1+x^{2}} \mathrm{d} x & = \frac{\mathrm{d}^{2}}{\mathrm{d}s^{2}} \int\limits_{0}^{\infty} \frac{x^{s-1}}{1+x^{2}} \mathrm{d} x |_{s=1} \\ & = \frac{\mathrm{d}^{2}}{\mathrm{d}s^{2}} \frac{\pi}{2}\csc\left(\frac{\pi}{2}s\right) |_{s=1} \\ & = \frac{\pi^{3}}{16}[\cos(\pi s) + 3]\csc^{3}\left(\frac{\pi}{2}s\right) |_{s=1} \\ & = \frac{\pi^{3}}{8} \label{eq:1608135} \tag{5} \end{align}

poweierstrass
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