another nice integral problem

[galactus]

Show that:

$$\int_{0}^{\infty}\frac{(x+1)\ln^{2}(x+1)}{(4x^{2}+8x+5)^{3/2}}dx=\frac{\pi^{2}}{40}$$

[Tolaso J Kos]

Show that:

$$\int_{0}^{\infty}\frac{(x+1)\ln^{2}(x+1)}{(4x^{2}+8x+5)^{3/2}}dx=\frac{\pi^{2}}{40}$$

First we will quote some lemmata that we are needing here:

Lemma 1: It holds that:

\begin{equation} \frac{x\cosh x}{\sinh x}= \frac{x\left ( e^x+e^{-x} \right )}{e^x-e^{-x}} = x + \frac{2xe^{-x}}{e^x-e^{-x}}= x + 2\sum_{n=1}^{\infty}xe^{-2nx} \end{equation}

Lemma 2: The polylog functions are defined as (Liebniz - Euler):

\begin{equation} \operatorname{Li}_m(z)= \sum_{n=1}^{\infty}\frac{z^n}{n^m}, \; \left | z \right |< 1 \end{equation}

For $m=1$ we get that: $\displaystyle \operatorname{Li}_1 (x) = \sum_{n=1}^{\infty}\frac{z^n}{n}= - \ln (1-z)$ , while for $m=2$ we get that $\displaystyle \operatorname{Li}_2(x)= \sum_{n=1}^{\infty}\frac{z^n}{n^2}, \; \left | z\right |\leq 1$. This function is what we call the dilogarithm function.

This is what we call the principal branch of the dilog. function which is a single-valued analytic function in the cut plane. If, however, we permit the variable $z$ to wander around the complex plane without restriction, we then create the general branch of the function. Then $\ln (1-z)$ at $z=0$ may no longer have its principal value but instead $2\kappa \pi i$ with a non zero value of the integer $\kappa$. Thus, if we begin with the principal branch of $\operatorname{Li}_2(z)$ at any point $z$ in the plane and integrate along a closed contour that goes in a continuous manner once around the branch point at $z=1$ in the positive rotational sense then the value of the function on returning to $z$ is $\operatorname{Li}_2(z)= 2\pi i \ln z$. For this branch and, more generally, for all of the branches of the dilogarithm other than the principal branch, the point $z=0$ is a "hidden" branch of a logarithmic type. Hence, if the principal branch is denoted as $\operatorname{Li}_2^*(z)$ then the relation

\begin{equation} \operatorname{Li}_2(z)= \operatorname{Li}_2^*(z) + 2m \pi i \ln z + 4 \kappa \pi^2 \end{equation}

holds, where $m , \kappa \in \mathbb{Z}$. Note, also, that $m , \kappa$ depend critically on the path of integration. The above relation actually allows us to evaluate the special values:

1. $\operatorname{Li}_2(-\phi)= -\frac{\pi^2}{10} - \ln^2 \phi$
2. $\operatorname{Li}_2\left ( - \frac{1}{\phi} \right )= -\frac{\pi^2}{15}+ \frac{\ln^2 \phi}{2}$
3. $\operatorname{Li}_2 \left ( \frac{1}{\phi^2} \right )= \frac{\pi^2}{15}- \ln^2 \phi$
4. $\operatorname{Li}_2 \left ( \frac{1}{\phi} \right )= \frac{\pi^2}{10} - \ln^2 \phi$
5. Similarly, we can evaluate the value $\operatorname{Li}_1 \left ( \frac{1}{\phi^2} \right )$.

where $\phi$ is the golden ratio. Of course these results are quoted in Levin's book (Polylogs and Associated functions). The above formulae are extracted via $(3)$ in combination with the known functional identities.

Now back to the integral:

\begin{align*}
\int_{0}^{\infty}\frac{(x+1)\ln^2 (x+1)}{\left ( 4x^2+8x+5 \right )^{3/2}}\, {\rm d}x &\overset{\frac{1}{u}=x+1}{=\! =\! =\!} \int_{0}^{1}\frac{\ln^2 u}{\left ( 4+u^2 \right )^{3/2}}\, {\rm d}u\\
&= \cancelto{0}{\left [ \frac{u \ln^2 u}{4\sqrt{u^2+4}} \right ]_0^1} - \frac{1}{2}\int_{0}^{1}\frac{\ln u}{\sqrt{u^2+4}}\, {\rm d}u\\
&=\cancelto{0}{\left [ -\frac{1}{2}\ln u \operatorname{arcsinh} \left ( \frac{u}{2} \right ) \right ]_0^1} +\frac{1}{2} \int_{0}^{1}\frac{\operatorname{arcsinh } \frac{u}{2}}{u} \, {\rm d}u \\
&= \frac{1}{2}\int_{0}^{\ln \phi} \frac{u \cosh u}{\sinh u}\, {\rm d}u\\
&\overset{(1)}{=}\frac{1}{2}\int_{0}^{\ln \phi}\left ( u + 2 \sum_{n=1}^{\infty}ue^{-2n u} \right )\, {\rm d}u \\
&= \frac{\ln^2 \phi}{4}+ \sum_{n=1}^{\infty}\int_{0}^{\ln \phi} ue^{-2n u} \, {\rm d}u \\
&\! \! \overset{(2), (3)}{=\! =\! =\! } \frac{\ln^2 \phi}{4} + \frac{\pi^2}{24} - \frac{1}{4}\operatorname{Li}_2 \left ( \frac{1}{\phi^2} \right ) - \frac{\ln \phi}{2} \operatorname{Li}_1 \left ( \frac{1}{\phi^2} \right ) \\
&=\cdots \\
&=\frac{\pi^2}{40}
\end{align*}

Actually before this approach I had subbed in $\displaystyle \int_{0}^{1}\frac{\ln u}{\sqrt{u^2+4}}\, {\rm d}u$ the substitution $u\mapsto 2\tan \theta$ and went on to try Fourier series since

$$\ln \tan \theta = 2 \sum_{n=0}^{\infty}\frac{\cos (4n+2)\theta}{2n+1}$$

but I did not make any headway. Maybe else can?

Edit: Actually, taking a better look there are more elementary proofs for the quoted results above that those I found in the book, by just using the functional equations $(3), (4), (5)$ listed here. The difference is that the author actuall defines the polylogs in the interval $0<x<1$ so he actually needs an argument of analytic expansion to define it in $(-1, 0)$.

[galactus]

Wow, cool beans T.

It's fun discovering other things while researching another.

That is where I have found some cool stuff I submit.