How to evaluate $$\int_{0}^{\infty }\frac{\ln(2x+1)}{x(x+1)}\,\mathrm dx?$$ I tried $$\frac{\ln(2x+1)}{x(x+1)}=\ln(2x+1)\left (\frac{1}{x}-\frac{1}{1+x} \right)$$ but I don't know how to go on.


do the substitution $2x+1\rightarrow x$ and the let $x\rightarrow x^{-1}$ , hence $$\int_{0}^{\infty }\frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\, \mathrm{d}x=2\int_{1}^{\infty }\frac{\ln x}{x^{2}-1}\, \mathrm{d}x=-2\int_{0}^{1}\frac{\ln x}{1-x^{2}}\, \mathrm{d}x$$ then use the geometric series $$\frac{1}{1-x^{2}}=\sum_{n=0}^{\infty }x^{2n}$$ we get $$\int_{0}^{\infty }\frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\, \mathrm{d}x=2\sum_{n=0}^{\infty }\frac{1}{\left ( 2n+1 \right )^{2}}$$ and the answer will follow.

  • $\begingroup$ thx.nice approach! i'm doing the wrong way just now. $\endgroup$ – user308493 Jan 9 '17 at 3:07
  • $\begingroup$ Shouldn't you specify something about the uniform convergence of your series etc? I don't think you can exchange sums and integral that easily, although it is correct of course. $\endgroup$ – user8469759 Jan 26 '17 at 14:44

Alternatively we can use Feynman's Trick if we let, $$I(a)=\int_0^{\infty} \frac{\ln(ax+1)}{x(x+1)}\mathrm dx,$$ then if we take the derivative of $I(a)$ and expand using partial fractions we have $$I'(a)=\int_0^{\infty} \frac{1}{(ax+1)(x+1)}\mathrm dx =\int_0^{\infty} \frac{a}{(a-1)(ax+1)} - \frac{1}{(a-1)(x+1)}\mathrm dx.$$ Integrating this gives, $$I'(a)=\bigg(\frac{1}{a-1}\ln|ax+1|-\frac{1}{a-1}\ln|x+1| \bigg)|_0^{\infty}=\frac{1}{a-1}\ln|\frac{ax+1}{x+1}| |_0^{\infty}.$$ Noticing the lower bound is zero we're left with an easy limit$$I'(a)=\lim_{x\to\infty}\frac{1}{a-1}\ln|\frac{ax+1}{x+1}|=\frac{\ln(a)}{a-1}.$$ Now if we integrate we can get $I(a)$, this is easy because our integral closely resembles the definition of the dilogarithm, so we have $$I(a)=\int \frac{\ln(a)}{a-1} da=-Li_2(1-a)+C.$$ Now if we notice $I(0)=0$ and note that $Li_2(1)=\frac{\pi^2}{6}$ we see that $C=\frac{\pi^2}{6}.$ So finally if we note that for your problem $a=2$ have that, $$I(2)=-Li_2(-1)+\frac{\pi^2}{6}=\frac{\pi^2}{12}+\frac{\pi^2}{6}=\frac{\pi^2}{4}.$$

This is admittedly not as nice as the approach using the geometric series and makes heavy use of the dilogarithm and it's particular values but I think it's kinda fun.


Too long for a comment.

Making the problem more general, it is possible to compute $$I=\int \frac{\log (a x+b)}{(x+c) (x+d)}\,\mathrm dx$$ Using one integration by parts $$u=\log (a x+b)\implies u'=\frac{a}{a x+b}\,\mathrm dx$$ $$v'=\frac{\mathrm dx}{(c+x) (d+x)}\implies v=\frac{\log (x+d)-\log (x+c)}{c-d}$$ So, if $c\neq d$,$$(c-d)I=\log (a x+b) \log \left(\frac{(x+d) (a c-b)}{(x+c) (a d-b)}\right)-\text{Li}_2\left(\frac{a x+b}{b-a c}\right)+\text{Li}_2\left(\frac{a x+b}{b-a d}\right) $$ and, if $c=d$, $$I=\frac{a (x+c) \log (x+c)-(a x+b) \log (a x+b)}{(x+c) (b-a c)}$$


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\ln\pars{2x + 1} \over x\pars{x + 1}}\,\dd x & \,\,\,\stackrel{2x + 1\ \mapsto\ x}{=}\,\,\, \int_{1}^{\infty}{\ln\pars{x} \over \bracks{\pars{x - 1}/2}\bracks{\pars{x - 1}/2 + 1}}\,{\dd x \over 2} = 2\int_{1}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x \\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\, -2\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, -\,{1 \over 2}\int_{0}^{1}{x^{-1/2}\,\ln\pars{x} \over 1 - x}\,\dd x \\[5mm] & = \left.{1 \over 2}\,\totald{}{\mu}\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x\, \right\vert_{\ \mu\ =\ -1/2} = \left.{1 \over 2}\,\totald{\bracks{\Psi\pars{\mu + 1} + \gamma}}{\mu} \right\vert_{\ \mu\ =\ -1/2}\label{1}\tag{1} \end{align}

where $\ds{\Psi}$ is the $Digamma\ Function$ and $\gamma$ is the $Euler\mbox{-}Mascheroni\ Constant$: See $\mathbf{6.3.22}$ in A&S Table.

From expression \eqref{1}: $$ \int_{0}^{\infty}{\ln\pars{2x + 1} \over x\pars{x + 1}}\,\dd x = {1 \over 2}\,\Psi\, '\pars{1 \over 2} = \bbx{\ds{\pi^{2} \over 4}} $$

See $\mathbf{6.4.4}$ in A&S Table.


This might be a late response, but consider the double integral

$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \frac{x}{1+x^2} \ dy \ dx.$$

We will evaluate $I$ in two ways. Integrating with respect to $y$ first, we see that

$$\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \ dy= \lim_{b \rightarrow \infty}\frac{\arctan((2b+1)x)}{2x}-\frac{\arctan(x)}{2x}=\frac{\pi}{4x}-\frac{\arctan(x)}{2x}$$ and get

$$I=\int_{0}^{\infty} \left(\frac{\pi}{4x}-\frac{\arctan(x)}{2x} \right) \frac{x}{1+x^2} \ dx=\int_{0}^{\infty} \frac{\pi}{4(1+x^2)}-\frac{\arctan(x)}{2(1+x^2)} \ dx=\frac{\pi^2}{8}-\frac{\pi^2}{16}=\frac{\pi^2}{16}.$$

Now we reverse the order of integration as such:

$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \frac{x}{1+x^2} \ dx \ dy.$$

To clean things up a little bit, let $u=x^2, \ du = 2 x \ dx.$ So

$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+u(2y+1)^2} \frac{u}{2(1+u)} \ du \ dy.$$

Integrating with respect to $u,$ we can use partial fractions to get that

$$\frac{1}{1+u(2y+1)^2} \frac{1}{2(1+u)}=\frac{(2y+1)^2}{2(1+u(2y+1)^2)((2y+1)^2-1)}-\frac{1}{2(u+1)((2y+1)^2-1)}.$$

We see

$$\int_{0}^{\infty} \frac{(2y+1)^2}{2(1+u(2y+1)^2)((2y+1)^2-1)}-\frac{1}{2(u+1)((2y+1)^2-1)} \ du $$ $$= \lim_{b \rightarrow \infty}\frac{\ln(b(2y+1)^2+1)}{2(2y+1)^2-2} - \frac{\ln(b+1)}{2(2y+1)^2-2}= \frac{\ln(2y+1)}{(2y+1)^2-1}=\frac{\ln(2y+1)}{4y(y+1)}.$$


$$I=\int_{0}^{\infty}\frac{\ln(2y+1)}{4y(y+1)} \ dy=\frac{\pi^2}{16},$$

So $$I=\int_{0}^{\infty}\frac{\ln(2y+1)}{y(y+1)} \ dy=\frac{\pi^2}{4}.$$


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