# Finding $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2} dx$

Calculate $$\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx$$

I have tried to put $$\displaystyle x=\frac{1}{t}$$ and $$\displaystyle dx=-\frac{1}{t^2}dt$$

$$\int^{\infty}_{0}\frac{t^2\ln^2(t)}{(t^2-1)^2}dt$$

$$\frac{1}{2}\int^{\infty}_{0}t\ln^2(t)\frac{2t}{(t^2-1)^2}dt$$

$$\frac{1}{2}\bigg[-t\ln^2(t)\frac{1}{t^2-1}+\int^{\infty}_{0}\frac{\ln^2(t)}{t^2-1}+2\int^{\infty}_{0}\frac{\ln(t)}{t^2-1}dt\bigg]$$

How can I solve it?

• Are you sure about the limits of the integral? Because the integrant has a singularity of the order $1$ at $x=1$. – Mundron Schmidt Feb 2 at 11:15
• @MundronSchmidt W|A gives $\pi^2/4$ as the result – TheSimpliFire Feb 2 at 11:19
• Note that $$\int_0^\infty\frac{\ln^2x}{(1-x^2)^2}\,dx=\int_{-\infty}^\infty\frac{x^2e^{-3x}}{(1-e^{2x})^2}\,dx$$ and \begin{align}\int_0^\infty\frac{x^2e^{-3x}}{(1-e^{2x})^2}\,dx&=\int_0^{\infty}\left(x^2e^{-3x}\sum_{n=0}^{\infty}\left(\left(1+n\right)e^{-2nx}\right)\right)\,dx\\&=\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+3)t}\,dt\\&=\sum_{n=0}^\infty\frac{2(1+n)}{(2n+3)^3}=\frac{\pi^2-7\zeta(3)}8\end{align} – TheSimpliFire Feb 2 at 11:45

You're definetly on the right track with that substitution of $$x=\frac1t$$

Basically we have: $$I=\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx=\int_0^\infty \frac{x^2\ln^2 x}{(1-x^2)^2}dx$$ Now what if we add them up? $$2I=\int_0^\infty \ln^2 x \frac{1+x^2}{(1-x^2)^2}dx$$ If you don't know how to deal easily with the integral $$\int \frac{1+x^2}{(1-x^2)^2}dx=\frac{x}{1-x^2}+C$$ I recommend you to take a look here.

Anyway we have, integrating by parts: $$2I= \underbrace{\frac{x}{1-x^2}\ln^2x \bigg|_0^\infty}_{=0} +2\underbrace{\int_0^\infty \frac{\ln x}{x^2-1}dx}_{\large =\frac{\pi^2}{4}}$$ $$\Rightarrow 2I= 2\cdot \frac{\pi^2}{4} \Rightarrow I=\frac{\pi^2}{4}$$ For the last integral see here for example.

This is a variant of TheSimpliFire's approach given in a comment.

By letting $$x=e^t$$ we get \begin{align*} \int_0^\infty\frac{\ln^2(x)}{(1-x^2)^2}\,dx &=\int_{-\infty}^\infty\frac{t^2e^{t}}{(1-e^{2t})^2}\,dt\\ &=\int_{0}^{+\infty}\frac{t^2e^{-t}}{(1-e^{-2t})^2}\,dt+\int_{0}^\infty\frac{t^2e^{-3t}}{(1-e^{-2t})^2}\,dt\\ &=\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+1)t}\,dt+\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+3)t}\,dt\\ &=\sum_{n=0}^\infty\frac{2(1+n)}{(2n+1)^3}+\sum_{n=0}^\infty\frac{2(1+n)}{(2n+3)^3}\\ &=\left(\sum_{n=0}^\infty\frac{1}{(2n+1)^2}+\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\right)+\left(\sum_{n=0}^\infty\frac{1}{(2n+1)^2}-\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\right)\\ &=2\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=2\left(\sum_{n=0}^\infty\frac{1}{n^2}-\sum_{n=0}^\infty\frac{1}{(2n)^2}\right)\\&=2\left(1-\frac{1}{4}\right)\sum_{n=0}^\infty\frac{1}{n^2}=\frac{3}{2}\cdot \frac{\pi^2}{6}=\frac{\pi^2}{4}. \end{align*}

• Could you provide a reference for those two fascinating sums (the ones with the $(2n+1)^3$ and $(2n+3)^3$ in the denominators)? Thanks – clathratus Feb 9 at 16:15
• @clathratus Actually we don't need them. See my edit! – Robert Z Feb 9 at 16:43
• Beautiful! Still though, any sources regarding those sums would be appreciated. They are intriguing indeed. – clathratus Feb 9 at 20:35
• $\sum_{n=0}^\infty\frac{1}{(2n+1)^3}=\sum_{n=1}^\infty\frac{1}{n^3}-\sum_{n=1}^\infty\frac{1}{(2n)^3}=\frac{7\zeta(3)}{8}.$ – Robert Z Feb 9 at 21:14

Here is yet another slight variation on a theme.

Let $$I = \int_0^\infty \frac{\ln^2}{(1 - x^2)^2} \, dx$$ then \begin{align} I &= \int_0^1 \frac{\ln^2 x}{(1 - x^2)^2} \, dx + \int_1^\infty \frac{\ln^2 x}{(1 - x^2)^2} \, dx = \int_0^1 \frac{(1 + x^2) \ln^2 x}{(1 - x^2)^2} \, dx \tag1, \end{align} after a substitution of $$x \mapsto 1/x$$ has been enforced in the second of the integrals.

As $$\frac{1}{1 - x^2} = \sum_{n = 0}^\infty x^{2n}, \qquad |x| < 1,$$ differentiating with respect to $$x$$ gives $$\frac{1}{(1 - x^2)^2} = \sum_{n = 1}^\infty n x^{2n - 2}.$$ On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have $$I = \sum_{n = 1}^\infty n \int_0^1 (x^{2n - 2} + x^{2n}) \ln^2 x \, dx.$$ Integrating by parts twice, we are left with $$I = \sum_{n = 1}^\infty \left (\frac{2n}{(2n - 1)^3} + \frac{2n}{(2n + 1)^3} \right ).$$ As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have \begin{align} I &= \sum_{n = 1}^\infty \left [\frac{1}{(2n - 1)^2} + \frac{1}{(2n - 1)^3} + \frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\ &= \sum_{n = 1}^\infty \left [\frac{1}{(2n - 1)^2} + \frac{1}{(2n - 1)^3} \right ] + \sum_{n = 1}^\infty \left [\frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\ &= \sum_{n = 0}^\infty \left [\frac{1}{(2n + 1)^2} + \frac{1}{(2n + 1)^3} \right ] + \sum_{n = 1}^\infty \left [\frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\ &= 2 + 2 \sum_{n = 1}^\infty \frac{1}{(2n + 1)^2}\\ &= 2 \sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}\\ &= 2 \left [\sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} \right ]\\ &= 2 \left (1 - \frac{1}{4} \right ) \sum_{n = 1}^\infty \frac{1}{n^2}\\ &= \frac{3}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\ &= \frac{3}{2} \cdot \frac{\pi^2}{6}\\ &= \frac{\pi^2}{4}, \end{align} as expected.