Prove that:
$$\int_{0}^{\infty}\frac{(x+1)\ln^{2}(x+1)}{(4x^{2}+8x+5)^{3/2}}dx=\frac{\pi^{2}}{40}$$
Solution
We are first quoting a lemma:
\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 \\ &= \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*}
where we made use of the dilogarithmic value
$$ \operatorname{Li}_2 \left ( \frac{1}{\phi^2} \right )= \frac{\pi^2}{15}- \ln^2 \phi$$
where $\phi$ stands for the golden ration.
The exercise can also be found at mathimatikoi.org
$$\int_{0}^{\infty}\frac{(x+1)\ln^{2}(x+1)}{(4x^{2}+8x+5)^{3/2}}dx=\frac{\pi^{2}}{40}$$
Solution
We are first quoting a lemma:
Lemma: It holds that:Hence:
\begin{equation}\frac{x\cosh x}{\sinh x} = x + 2\sum_{n=1}^{\infty}xe^{-2nx} \end{equation}
Proof:$$\frac{x\cosh x}{\sinh x}= \frac{x\left ( e^x+e^{-x} \right )}{e^x-e^{-x}} = x + \frac{2xe^{-x}}{e^x-e^{-x}}= x + 2\sum_{n=1}^{\infty}xe^{-2nx}$$
\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 \\ &= \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*}
where we made use of the dilogarithmic value
$$ \operatorname{Li}_2 \left ( \frac{1}{\phi^2} \right )= \frac{\pi^2}{15}- \ln^2 \phi$$
where $\phi$ stands for the golden ration.
The exercise can also be found at mathimatikoi.org
The integrand has an antiderivative.
ReplyDelete$-\frac{\mathrm{asinh}\left( \frac{1}{2\cdot \left| x+1\right| }\right) }{4}-\frac{\mathrm{log}\left( x+1\right) }{4\cdot \sqrt{4\cdot {{x}^{2}}+8\cdot x+5}}$