# Integrate $\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln{x}} \mathop{dx}$

Insane integral $$\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln x} \mathop{dx}$$ I know $$x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}$$ but does it help? I think $$u=\ln{x}$$ might be necessary some point.

• WA doesn't give a closed form. Please try to add more context: from where the integral is, what did you try, etc. Thanks. Jul 4, 2020 at 18:44
• Divide numerator and denominator by $x^2$ and set $$x+\dfrac1x=y$$ Jul 4, 2020 at 18:51
• @labbhattacharjee What about the $\ln x$? Jul 4, 2020 at 18:55
• You can use the same method in this math.stackexchange.com/questions/285130/…, both the series way and digamma function way is possible, denominator would not be a big deal, as long as it is a regular polynomial. Jul 4, 2020 at 19:12
• @AlexeyBurdin. It does actually have a closed form solution: $\ln{\phi}$. I think Nikunj's solution is the way to go. According to google, that infinite product (after using log rules) is equal to $\phi$, but Wolfram spits out a value slightly less than $\ln{\phi}$ for that summation.
– Ty.
Jul 4, 2020 at 19:33

(edit for a little glitch in writing) so, basically you want to solve $$\int_{0}^{1} \frac{(x^2-1)(x-1)}{(x^5-1)\ln x} \mathrm{d}x$$ let $$I(s) = \int_{0}^{1} \frac{(x^2-1)(x-1)x^s}{(x^5-1)\ln x} \mathrm{d}x$$ take derivative of which \begin{aligned} I'(s) & = \int_{0}^{1} \frac{(x^2-1)(x-1)x^s}{x^5-1} \mathrm{d}x = \int_{0}^{1} \frac{x^{s+3}-x^{s+2}-x^{s+1}+x^{s}}{x^5-1} \mathrm{d}x\\ & = \frac1{5} \int_{0}^{1} \frac{x^{\tfrac{s-1}{5}}-x^{\tfrac{s-2}{5}}-x^{\tfrac{s-3}{5}}+x^{\tfrac{s-4}{5}}}{x-1} \mathrm{d}x \end{aligned} where you take $$x^5\to x$$. by recalling the representation of digamma function $$\psi(s+1) = -\gamma + \int_{0}^{1} {\frac{x^{s}-1}{x-1} \mathrm{d}x}$$ we have $$I'(s)=\frac1{5}\left(\psi\left(\frac{s+1}{5}\right)+\psi\left(\frac{s+4}{5}\right)-\psi\left(\frac{s+2}{5}\right)-\psi\left(\frac{s+3}{5}\right)\right)$$ with $$\lim_{s\to\infty}I(s)=0$$ and asymptotic expansion $$\ln\Gamma(z) = (z-\tfrac1{2})\ln z - z + \ln2\pi + \tfrac1{12z} + o(z^{-3})$$ to anti-derivative $$I(0) = -\int_{0}^{\infty}I'(s)\,\mathrm{d}s = -\ln\left.\left(\frac{\Gamma\left(\frac{s+1}{5}\right)\Gamma\left(\frac{s+4}{5}\right)}{\Gamma\left(\frac{s+2}{5}\right)\Gamma\left(\frac{s+3}{5}\right)}\right)\right|_{s=0}^{\infty} = \ln\left(\frac{\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{4}{5}\right)}{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}\right)$$ where using the reflection formula to obtain final answer $$\int_{0}^{1} \frac{x^2-1}{(x^4+x^3+x^2+x+1)\ln x} \mathrm{d}x = \ln\left(\frac{\sqrt5+1}{2}\right)$$

• very very very-good answer +1
– user808985
Jul 18, 2020 at 0:51

Write this as:

$$\int_{0}^1\frac{(x^2-1)(x-1)\,dx}{(x^5-1)\ln x}$$

Note that $$\int_0^1 x^y\,dy = (x-1)/\ln x$$

Replacing this and converting the problem into a double integral gives

$$\int_{0}^1\int_0^1\frac{(x^2-1)}{(x^5-1)}x^y\,dx\, dy$$

Writing $$\frac{1}{1-x^5} = \sum_{k=0}^\infty x^{5k}$$

This becomes

$$\int_{0}^1\int_0^1(1-x^2)x^y\sum_{k=0}^\infty x^{5k}\, dx\, dy$$

Interchanging the summation and integration

$$= \sum_{k=0}^\infty \int_{0}^1\int_0^1(1-x^2)x^yx^{5k}\,dx\,dy$$ Integrating with respect to $$x$$,

$$= \sum_{k=0}^\infty \int_{0}^1 \left(\frac{1}{5k+y+1} - \frac{1}{5k+y+3}\right)\,dy$$

Integrating this, we get

$$I = \sum_{k=0}^\infty \ln \frac{(5k+2)(5k+3)}{(5k+1)(5k+4)}$$

Using the identity provided by @Nanayajitzuki in the comments:

$$\ln \prod_{k=0}^\infty \frac{(5k+2)(5k+3)}{(5k+1)(5k+4)} = \ln\left(\frac{\Gamma(1/5)\Gamma(4/5)}{\Gamma(2/5)\Gamma(3/5)}\right)$$

Using Euler's reflection formula, $$\Gamma(z)\Gamma(1-z) = \pi /\sin (\pi z)$$

$$= \ln \frac{\sin (2\pi/5)}{\sin (\pi/5)}$$ $$= \ln (2\cos (\pi/5))$$ $$= \ln (\phi)$$

• And you may use this generalized formula for the final products. $$\prod_{k=0}^{\infty} \frac{(k+a_1) \cdots (k+a_j)}{(k + b_1) \cdots (k+b_j)} = \frac{\Gamma(b_1) \cdots \Gamma(b_k)}{\Gamma(a_1) \cdots \Gamma(a_k)}$$ where $a_1+a_2+⋯+a_j=b_1+b_2+⋯+b_j$ and no $b_j$ is zero or a negative integer. Jul 4, 2020 at 19:18
• @Nanayajitzuki How do I use it with $5k$? Is it possible to create some sort of generating function? Jul 4, 2020 at 19:39
• @Nikunj. Wolfram gives your final product as slightly less than $\ln{\phi}$, but it should be exactly equal to. If you use log properties and find the infinite product, that is equal to $\phi$. Wolfram also confirms that the closed form of the integral of \$\ln{\phi}, which confuses me why the product isn't equal according to wolf.
– Ty.
Jul 4, 2020 at 19:44
• @Nikunj. Just divide 5 both in the numerator and denominator, I may write another method to show how it deal with the Gamma function. Jul 4, 2020 at 19:47
• @Nanayajitzuki Oh, silly me. Jul 4, 2020 at 19:49

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[#ffd,15px]{\int_{0}^{1}{x^{2} - 1 \over \pars{x^{4} + x^{3} + x^{2} + x + 1}\ln\pars{x}}\,\dd x} \\ = &\ \int_{0}^{1}{x^{2} - 1 \over \pars{x^{5} - 1}/\pars{x - 1}}\ \overbrace{\pars{{1 \over x - 1}\int_{0}^{1}x^{y}\,\dd y}} ^{\ds{1 \over \ln\pars{x}}}\ \,\dd x = \int_{0}^{1}\int_{0}^{1}{x^{y} - x^{y + 2} \over 1 - x^{5}}\,\dd x\,\dd y \\[5mm] = &\ {1 \over 5}\int_{0}^{1}\int_{0}^{1}{x^{y/5 - 4/5} - x^{y/5 - 2/5} \over 1 - x}\,\dd x\,\dd y \\[5mm] = &\ {1 \over 5}\int_{0}^{1}\pars{\int_{0}^{1}{1 - x^{y/5 - 2/5} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{y/5 - 4/5} \over 1 - x}\,\dd x}\dd y \\[5mm] = &\ {1 \over 5}\int_{0}^{1}\bracks{\Psi\pars{{y \over 5} + {3 \over 5}} - \Psi\pars{{y \over 5} + {1 \over 5}}}\dd y \\[5mm] = &\ \bracks{\ln\pars{\Gamma\pars{{y \over 5} + {3 \over 5}}} - \ln\pars{\Gamma\pars{{y \over 5} + {1 \over 5}}}}_{\ 0}^{\ 1} \\[5mm] = &\ \ln\pars{\ln\pars{\Gamma\pars{4/5}}\ln\pars{\Gamma\pars{1/5}} \over \ln\pars{\Gamma\pars{2/5}}\ln\pars{\Gamma\pars{3/5}}} = \ln\pars{\sin\pars{2\pi/5} \over \sin\pars{\pi/5}} \\[5mm] = &\ \bbx{\ln\pars{2\cos\pars{\pi \over 5}}} = \bbx{\ln\pars{1 + \root{5} \over 2}}\ \approx\ 0.4812 \end{align}