A definite integral involving the product of a rational function and a logarithm

Question

We have to do the following integral. $$\int_1^{\frac{1+\sqrt{5}}{2}}\frac{x^2+1}{x^4-x^2+1}\ln\left(1+x-\frac{1}{x}\right)dx$$ I tried it a lot. I substitute $t=1+x-(1/x)$, $dt=1+(1/x^2)$

But then I stuck at $$\int\limits_{1}^{2} \frac{\ln(t)}{(t-1)^{2} + 1} \mathrm{d}t$$

But now how to proceed.

Answer 1

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Let $I$ denote the integral

$$I:=\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln{\left(1+x-\frac{1}{x}\right)}\,\mathrm{d}x,$$

with $\phi$ of course being the golden ratio, $\phi=\frac{1+\sqrt{5}}{2}$.

As a numerical approximation, we find

$$I\approx0.272198.$$

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Try substituting instead

$$\frac12\left(x-\frac{1}{x}\right)=t,$$

$$\implies\frac12\left(1+\frac{1}{x^{2}}\right)\,\mathrm{d}x=\mathrm{d}t.$$

Taking the square, we have

$$\frac14\left(x^{2}-2+\frac{1}{x^{2}}\right)=t^{2}.$$

Then,

$$\begin{align} I &=\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln{\left(1+x-\frac{1}{x}\right)}\,\mathrm{d}x\\ &=\int_{1}^{\phi}\frac{2x^{2}}{x^{4}-x^{2}+1}\ln{\left(1+x-\frac{1}{x}\right)}\,\frac{x^{2}+1}{2x^{2}}\,\mathrm{d}x\\ &=\int_{1}^{\phi}\frac{2}{\left(x^{2}-2+x^{-2}\right)+1}\ln{\left(1+x-\frac{1}{x}\right)}\,\frac12\left(1+\frac{1}{x^{2}}\right)\,\mathrm{d}x\\ &=\int_{0}^{\frac12}\frac{2}{4t^{2}+1}\ln{\left(1+2t\right)}\,\mathrm{d}t;~~~\small{\left[\frac12\left(x-\frac{1}{x}\right)=t\right]}\\ &=\int_{0}^{1}\frac{\ln{\left(1+u\right)}}{u^{2}+1}\,\mathrm{d}u;~~~\small{\left[2t=u\right]}.\\ \end{align}$$

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Expressing the logarithmic term as an integral, $I$ becomes a double integral. Changing the order of integration, we obtain

$$\begin{align} I &=\int_{0}^{1}\frac{\ln{\left(1+x\right)}}{1+x^{2}}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{\mathrm{d}x}{1+x^{2}}\int_{0}^{1}\mathrm{d}t\,\frac{x}{1+xt}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{x}{\left(1+x^{2}\right)\left(1+xt\right)}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left(1+x^{2}\right)\left(1+tx\right)}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\left[\frac{t+x}{\left(1+t^{2}\right)\left(1+x^{2}\right)}-\frac{t}{\left(1+t^{2}\right)\left(1+tx\right)}\right];~~~\small{P.F.D.}\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\left(1+t^{2}\right)}\int_{0}^{1}\mathrm{d}x\,\left[\frac{t+x}{\left(1+x^{2}\right)}-\frac{t}{\left(1+tx\right)}\right]\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\left(1+t^{2}\right)}\left[\int_{0}^{1}\mathrm{d}x\,\frac{t}{\left(1+x^{2}\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left(1+x^{2}\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{t}{\left(1+tx\right)}\right]\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\left(1+t^{2}\right)}\left[\frac{\pi}{4}t+\frac12\ln{\left(2\right)}-\ln{\left(1+t\right)}\right]\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\left(1+t^{2}\right)}\left[\frac{\pi}{4}t+\frac12\ln{\left(2\right)}\right]-\int_{0}^{1}\frac{\ln{\left(1+t\right)}}{1+t^{2}}\,\mathrm{d}t\\ &=\frac{\pi}{4}\ln{\left(2\right)}-I.\\ \end{align}$$

Thus,

$$I=\frac{\pi}{8}\ln{\left(2\right)}.\blacksquare$$

Answer 2

Hint. We have that $$ \int_1^2\frac{\log(t)}{(t-1)^2+1}dt =\int_0^1\frac{\log(1+v)}{v^2+1}dv =\int_0^{\pi/4}\log(1+\tan(u))\,du\\ =\int_0^{\pi/4}\log(\cos(u)+\sin(u))\,du-\int_0^{\pi/4}\log(\cos(u))\,du\\ =\int_0^{\pi/4}\log(\sqrt{2}\cos(\pi/4-u)))\,du-\int_0^{\pi/4}\log(\cos(u))\,du.$$ Now work on the first integral and you will get the answer.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mc{J} & = \int_{1}^{\pars{1 + \root{5}}/2}{x^{2} + 1 \over x^{4} - x^{2} + 1}\, \ln\pars{1 + x - {1 \over x}}\,\dd x \\[5mm] & = \int_{1}^{\pars{1 + \root{5}}/2}{1 \over x^{2} - 1 + 1/x^{2}}\, \ln\pars{1 + x - {1 \over x}}\,\pars{1 + {1 \over x^{2}}}\dd x \\[5mm] & = \int_{1}^{\pars{1 + \root{5}}/2}{1 \over \pars{x - 1/x}^{2} + 1}\, \ln\pars{1 + x - {1 \over x}}\,\pars{1 + {1 \over x^{2}}}\dd x \\[5mm] \stackrel{t\ =\ x - 1/x}{=} &\ \int_{0}^{1}{\ln\pars{1 + t} \over t^{2} + 1}\,\dd t \,\,\,\stackrel{t\ =\ \tan\pars{\theta}}{=}\,\,\,\ \underbrace{\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta}_{\ds{=\ \mc{J}}}\ =\ \int_{-\pi/4}^{0}\ln\pars{1 + {\tan\pars{\theta} + 1 \over 1 - \tan\pars{\theta}}}\,\dd\theta \\[5mm] & = {1 \over 4}\,\pi\ln\pars{2} - \int_{-\pi/4}^{0}\ln\pars{1 - \tan\pars{\theta}}\,\dd\theta = \color{#f00}{{1 \over 4}\,\pi\ln\pars{2}} - \underbrace{\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta} _{\ds{=\ \mc{J}}} \end{align}

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$$ \mc{J} = \int_{1}^{\pars{1 + \root{5}}/2}{x^{2} + 1 \over x^{4} - x^{2} + 1}\, \ln\pars{1 + x - {1 \over x}}\,\dd x = {\mc{J} + \mc{J} \over 2} = {\pi\ln\pars{2}/4 \over 2} = \bbx{\ds{{1 \over 8}\,\pi\ln\pars{2}}} $$