# Integral with $\ln$ and rational function

I saw the following integral on Facebook: $$\int_0^\infty \frac{x \ln(1+x)}{(1+x)(2x^2+2x+1)} dx$$

I thought at first that it would be easy to solve using a contour integral in the complex plane, but I didn't manage to find a useful contour.

My idea was to make a cut along $(-1, \infty)$ and follow $(0, \infty)$ on both sides and connect them at infinity with a hugh circle since the integral over that should be $0$, but how do I close the path close to origin so that it can be easily calculated or estimated?

• $\frac {x}{(1+x)(2 x^2 + 2 x +1) } = \frac{ -1}{(1+x))} + \frac{(2x+1)}{(2x^2+2x+1)}$. Now I think supposing $x=e^{it}$ may be helpful. x →∞ ⇒ t → π/2 and x →0 ⇒ $t →- π/2$. Aug 18, 2017 at 16:59
• @sirous. With $x=e^{it}$ it's not correct that $x \to \infty \implies t \to \pi/2$ and $x \to 0 \implies t \to -\pi/2$. Aug 18, 2017 at 21:07
• You are right, that was a silly mistake. Aug 19, 2017 at 9:06

\begin{align*} I=\int_0^\infty \frac{x \ln(1+x)}{(1+x)(2x^2+2x+1)} dx=\frac{5\pi^2}{96}-\frac{\ln^2 2} 8. \end{align*}

Here are some alternative ways to evaluate it. (I prefer method 2)

Method 1: Series + Residue

Substitute $$x=\frac 1t$$,

\begin{align*} I &= \int_0^\infty \frac{\ln(1+t)}{(1+t)(t^2+2t+2)} dt-\int_0^\infty \frac{\ln t}{(1+t)(t^2+2t+2)} dt \\ &= I_1-I_2. \end{align*}

Claim 1: \begin{align*} I_1=\int_0^\infty \frac{\ln(1+t)}{(1+t)(t^2+2t+2)} dt=\frac{\pi^2}{48}. \end{align*}

Put $$t=-1+\frac 1u$$ in $$I_1$$, \begin{align*} I_1 &= \int_0^1 -\frac{u\ln u}{1+u^2}du=- \int_0^1 \sum_{k=0}^\infty (-1)^k u^{2k+1}\ln u\,du\\ &=- \sum_{k=0}^\infty (-1)^k \left[\frac{u^{2k+2}\ln u}{2k+2}-\frac{u^{2k+2}}{(2k+2)^2}\right]_0^1\\ &=\frac 14 \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^2} \tag {1.1}\\ &=\frac 14\cdot\frac{\pi^2}{12}=\frac{\pi^2}{48}. \end{align*}

Comment:

Claim 2: \begin{align*} I_2=\int_0^\infty \frac{\ln t}{(1+t)(t^2+2t+2)} dt=\frac{\ln^2 2}{8}-\frac{\pi^2}{32}. \end{align*}

Consider the key-hole contour as shown:

Because $$\frac{\partial}{\partial\alpha}t^\alpha=t^\alpha\ln t$$, one uses the function

\begin{align*} f(z)=\frac{z^\alpha}{(1+z)(z^2+2z+2)},\qquad\forall -1<\alpha<1. \end{align*}

The poles of $$f$$ are at $$-1$$ and $$-1\pm i$$.

Set

\begin{align*} I_2(\alpha) =\int_0^\infty f(t)\,dt=\int_0^\infty \frac{t^\alpha}{(1+t)(t^2+2t+2)}\,dt. \end{align*}

Then,

\begin{align*} 2\pi i\sum \text{Res } f&=\oint f(z) dz\\ &=\int_\gamma+\int_\epsilon^R\frac{z^\alpha}{(1+z)(z^2+2z+2)}\,dz+\int_\Gamma+\int_R^\epsilon\frac{\left(ze^{2\pi i}\right)^\alpha}{(1+z)(z^2+2z+2)}\,dz\\ &=\left(1-e^{2\pi\alpha i}\right)\int_\epsilon^R\frac{z^\alpha}{(1+z)(z^2+2z+2)}\,dz+\left(\int_\gamma+\int_\Gamma\right)\\ &=\left(1-e^{2\pi\alpha i}\right)I_2(\alpha). \end{align*}

And,

\begin{align*} I_2&=\left.\frac{\partial}{\partial \alpha}\right|_{\alpha=0}I_2(\alpha)\\ &=\left.\frac{\partial}{\partial \alpha}\right|_{\alpha=0}\frac{\pi\left(2^{\frac {\alpha}2}\cos\left(\frac{\pi \alpha}4\right)-1\right)}{\sin(\pi \alpha)}\\ &=\frac{\ln^2 2}{8}-\frac{\pi^2}{32}. \end{align*}

OR

Use the same contour and consider function $$g$$:

\begin{align*} g(z)=\frac{\ln^2 z}{(1+z)(z^2+2z+2)}. \end{align*}

The poles of $$g$$ are at $$-1$$ and $$-1\pm i$$.

Use $$0\le\arg z<2\pi$$ as the branch of the logarithm corresponding to.

\begin{align*} 2\pi i\sum \text{Res } g&=\oint g(z) dz\\ &=\int_\gamma+\int_\epsilon^R\frac{(\ln z)^2}{(1+z)(z^2+2z+2)}\,dz+\int_\Gamma+\int_R^\epsilon\frac{(\ln z+2\pi i)^2}{(1+z)(z^2+2z+2)}\,dz\\ &=\int_\epsilon^R\frac{(\ln z)^2-(\ln z+2\pi i)^2}{(1+z)(z^2+2z+2)}\,dz+\left(\int_\gamma+\int_\Gamma\right)\\ &=-4\pi i\int_0^\infty\frac{\ln z}{(1+z)(z^2+2z+2)}\,dz+\int_0^\infty\frac{4\pi^2}{(1+z)(z^2+2z+2)}\,dz\\ &=-4\pi iI_2+\int_0^\infty\frac{4\pi^2}{(1+z)(z^2+2z+2)}\,dz. \end{align*}

Method 2: Differentiation under $$\int$$

Substitute $$x=\frac{u}{1-u}$$,

\begin{align*} I&=-\int_0^1 \frac{u\ln(1-u)}{u^2+1}\,du. \end{align*}

Write

\begin{align*} J(\beta)&=-\int_0^1 \frac{u\ln(1-\beta u)}{u^2+1}\,du,\qquad \forall\beta\in[0,1]. \end{align*}

Then $$J(0)=0$$ and

\begin{align*} J_\beta (\beta)&=\frac{\partial}{\partial \beta}J(\beta)=\int_0^1 \frac{u^2}{(1-\beta u)\left(u^2+1\right)}\,du\\ &=\frac 1{1+\beta^2}\left(\color{red}{-\int_0^1\frac{1+\beta u}{1+u^2}\,du}\color{green}{+\int_0^1 \frac{1}{1-\beta u}\,du}\right)\\ &=\color{red}{-\frac{\pi+2\beta \ln 2}{4(1+\beta^2)}}\color{green}{-\frac{\ln(1-\beta)}{\beta}+\frac{\beta\ln(1-\beta)}{1+\beta^2}}. \end{align*}

\begin{align*} I&=J(1)=\int_0^1 J_\beta (\beta)\,d\beta\\ &=\color{red}{-\int_0^1\frac{\pi+2\beta \ln 2}{4(1+\beta^2)}\,d\beta}\color{blue}{-\int_0^1\frac{\ln(1-\beta)}{\beta}\,d\beta}+\int_0^1\frac{\beta\ln(1-\beta)}{1+\beta^2}\,d\beta\\ &=\color{red}{-\frac{\pi^2}{16}-\frac 14\ln^2 2}\color{blue}{+\frac{\pi^2}{6}}-I. \end{align*}

\begin{align*} \therefore 2I=\frac{5\pi^2}{48}-\frac 14\ln^2 2. \end{align*}

Method 3 : Series

Claim 3: \begin{align*} I=-\frac 12 \sum_{n=1}^\infty\frac{(-1)^n} n H_{2n},\qquad\qquad\text{where }H_m=\sum_{n=1}^m\frac 1 n,\quad m\in\mathbb N. \end{align*}

Substitute $$x=\frac{u}{1-u}$$,

\begin{align*} I&=-\int_0^1 \frac{u\ln(1-u)}{u^2+1}\,du. \end{align*}

Integrate by parts,

\begin{align*} I&=-\frac 12\int_0^1 \ln(1-u)\,d\left(\ln \frac {1+u^2}{2}\right)\\ &=\underbrace{\left.-\frac 12\ln(1-u)\left(\ln \frac {1+u^2}{2}\right)\right|_0^1}_0+\frac 12 \int_0^1\frac{\color{red}{\ln\left(1+u^2\right)}-\color{green}{\ln 2}}{u-1}\,du\\ &=\frac 12 \int_0^1 \frac 1{u-1}\left(\color{red}{\sum_{n=1}^\infty\frac{(-1)^{n-1}} n u^{2n}}-\color{green}{\sum_{n=1}^\infty\frac{(-1)^{n-1}} n }\right)\,du\\ &=-\frac 12 \sum_{n=1}^\infty\frac{(-1)^n} n\int_0^1\frac {u^{2n}-1}{u-1}\,du\\ &=-\frac 12 \sum_{n=1}^\infty\frac{(-1)^n} n\int_0^1\sum_{k=1}^{2n} u^{k-1}\,du\\ &=-\frac 12 \sum_{n=1}^\infty\frac{(-1)^n} n H_{2n}. \end{align*}

The claim is established.

Let $$S(N)=\sum_{n=1}^N\frac{(-1)^n} n H_{2n},\forall N\in\mathbb N$$.

\begin{align*} S(N)&=\sum_{n=1}^N\frac{(-1)^n} n \cdot \frac 1{2n}+\sum_{n=1}^N\frac{(-1)^n} n H_{2n-1}\tag{3.1}\\ &=\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}+\frac 12\sum_{n=1}^N\sum_{k=1}^{2n-1}\frac{(-1)^n} n\left(\frac 1k+\frac 1{2n-k}\right)\tag{3.2}\\ &=\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}+\sum_{n=1}^N\sum_{k=1}^{2n-1}\frac{(-1)^n}{k(2n-k)}\tag{3.3}\\ &=\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}+\Re\left\{\sum_{k=1}^{2N-1}\sum_{m=k+1}^{2N}\frac{(-1)^{m/2}}{k(m-k)}\right\}\tag{3.4}\\ &=\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}+\Re\left\{\sum_{k=1}^{2N-1}\sum_{l=1}^{2N-k}\frac{i^{k+l}}{kl}\right\}\tag{3.5}\\ &=\color{purple}{\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}}+\color{blue}{\Re\left\{\sum_{k=1}^{2N-1}\frac{i^k}k\sum_{l=1}^{2N-1}\frac{i^l}l\right\}}-\color{orange}{\Re\left\{\sum_{k=1}^{2N-1}\frac{i^k}k\sum_{l=2N-k-1}^{2N-1}\frac{i^l}l\right\}}\tag{3.6}\\ &=\color{purple}{S_1(N)}+\color{blue}{S_2(N)}-\color{orange}{S_3(N)}. \end{align*}

Comment:

• In $$(3.1)$$ we break $$H_{2n}$$ into $$\frac 1{2n}+H_{2n-1}$$.

• In $$(3.2)$$ we use $$\sum_{k=1}^{m-1}f(k)=\sum_{k=1}^{m-1}f(m-k)$$.

• In $$(3.4)$$ we swap two summation signs and set $$m=2n$$.

• In $$(3.5)$$ we set $$l=m-k$$.

• In $$(3.6)$$ we use $$\sum_{l=1}^{2N-k}=\sum_{l=1}^{2N-1}-\sum_{l=2N-k-1}^{2N-1}$$.

To prove $$\color{orange}{S_3(\infty)}=0$$, use the Alternating Series Test, one has

\begin{align*} \left|\color{orange}{S_3(N)}\right|&=\left|\Re\left\{\sum_{k=1}^{2N-1}\left(\frac{i^{2k}}{2k}+\frac{i^{2k-1}}{2k-1}\right)\sum_{l=2N-k-1}^{2N-1}\left(\frac{i^{2l}}{2l}+\frac{i^{2l-1}}{2l-1}\right)\right\}\right|\tag{3.7}\\ &=\left|\Re\left\{\sum_{k=1}^{2N-1}\left(\frac{(-1)^k}{2k}-i\frac{(-1)^k}{2k-1}\right)\sum_{l=2N-k-1}^{2N-1}\left(\frac{(-1)^l}{2l}-i\frac{(-1)^l}{2l-1}\right)\right\}\right|\tag{3.8}\\ &=\left|\sum_{k=1}^{2N-1}\frac{(-1)^k}{2k}\sum_{l=2N-k+1}^{2N-1}\frac{(-1)^l}{2l}+\sum_{k=1}^{2N-1}\frac{(-1)^k}{2k-1}\sum_{l=2N-k-1}^{2N-1}\frac{(-1)^l}{2l-1}\right|\tag{3.9}\\ &\le\left|\sum_{k=1}^{2N-1}\frac{(-1)^k}{k}\sum_{l=2N-k-1}^{2N-1}\frac{(-1)^l}{l}\right|\le\left|\sum_{k=1}^{2N-1}\frac{(-1)^k}{k}\sum_{l=N}^{2N-1}\frac{(-1)^l}{l}\right|\\ &\le\left|2\sum_{l=N}^{2N-1}\frac1{N(N-1)}\right|= \frac 2{N-1}. \end{align*}

Comment:

• In $$(3.7)$$ we break indices into odd and even parts.

Therefore, as $$N\to\infty$$, \begin{align*} \lim_{N\to\infty} S(N)&=\color{purple}{S_1(\infty)}+\color{blue}{S_2(\infty)}-\color{orange}{S_3(\infty)}\\ &=\color{purple}{\frac 12\left(-\frac{\pi^2}{12}\right)}+\color{blue}{\Re\left\{\ln(1-i)\ln(1-i)\right\}}-\color{orange}{0}\\ &=\color{purple}{-\frac{\pi^2}{24}}+\color{blue}{\left(\frac{\ln 2}{2}\right)^2+\left(\frac{-\pi i}{4}\right)^2}-\color{orange}{0}\\ &=-\frac{5\pi^2}{48}+\frac{\ln^2 2} 4. \end{align*}

• This looks very promising. I haven't had time to study your solution in detail yet, but I will do it in the coming days. Aug 20, 2017 at 19:29
• I marked your answer as accepted although I don't understand everything. Aug 22, 2017 at 20:45
• Also, how can $g(z)=\frac{\ln^2 z}{(1+z)(z^2+2z+2)}$ be used to calculate an integral of $\frac{\ln z}{(1+z)(z^2+2z+2)}$? Aug 22, 2017 at 20:50
• @md2perpe Answer is updated, I have found a new method (method 3). Aug 23, 2017 at 8:03