stuck on this integral
$$\int_0^1\dfrac{(x^2+x+1)}{(x^4+x^3+x^2+x+1)}\ dx$$
I was attempting to evaluate the infinity sum S = $ 1- \frac{1}{4} + \frac {1}{6} - \frac {1}{9} + \frac {1}{11} -\frac {1}{14}+ ........ $
what I then did was define the S to be equal to $$ \int_0^1 (1-x^3 +x^5-x^8+x^{10}-x^{13}+........) dx $$
I simplified this and got the above integral I tried to do partial fraction but did not succeed.
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Michael Rozenberg
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This is equivalent to, for $x\neq1$, $$\int\frac{x^3-1}{x^5-1}dx$$ – Rushabh Mehta Oct 22 '18 at 05:23
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Did you try partial fraction decomposition? It is a coarse method, but it might work. Since the degree in the numerator is less than the degree of denominator. – Bertrand Wittgenstein's Ghost Oct 22 '18 at 10:13
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Related https://math.stackexchange.com/questions/1784472 – Nosrati Oct 22 '18 at 11:09
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@RushabhMehta What next? – Frank W Oct 22 '18 at 15:33
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@FrankW. It's a hint, not a solution. Partial Fraction decomposition leads to a nice solution. – Rushabh Mehta Oct 22 '18 at 15:37
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Quite a deadly integral to solve. – abhishek ravindra Oct 22 '18 at 17:39
3 Answers
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If your purpose is to evaluate
$$ S=\sum_{k\geq 0}\left[\frac{1}{10k+1}-\frac{1}{10k+4}+\frac{1}{10k+6}-\frac{1}{10k+9} \right]$$
you do not need an indefinite integral, just an integral over $(0,1)$: that's a huge difference, in some cases.
Actually, since $1+9=4+6=10$, you may just invoke the identities
$$ \sum_{k\geq 0}\left[\frac{1}{10k+1}-\frac{1}{10k+9}\right]=\frac{\pi}{10}\cot\frac{\pi}{10} $$
$$ \sum_{k\geq 0}\left[\frac{1}{10k+4}-\frac{1}{10k+6}\right]=\frac{\pi}{10}\cot\frac{4\pi}{10} $$
which follow from Herglotz' trick / the reflection formula for the digamma function.
In particular
$$ S = \frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}.$$
Jack D'Aurizio
- 353,855
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Yeah it's definite integral I forgot to write it in the upper one I will edit that but yeah herglotz trick did worked fantastic for this question. – abhishek ravindra Oct 22 '18 at 16:19
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Math stackexhange is just so awesome it has community of just amazing people those who prepare for olympiad do self learning it is a great tool and kind of like friend giving just amazing insights from all sorts of mathematical areas – abhishek ravindra Oct 22 '18 at 16:33
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The hint:
Use $$x^4+x^3+x^2+x+1=\left(x^2+\frac{1}{2}x+1\right)^2-\left(\frac{\sqrt5}{2}x\right)^2=$$ $$=\left(x^2+\frac{1-\sqrt5}{2}x+1\right)\left(x^2+\frac{1+\sqrt5}{2}x+1\right).$$ Can you end it now?
Michael Rozenberg
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{x^{2} + x + 1 \over x^{4} + x^{3} + x^{2} + x + 1}\,\dd x} = \int_{0}^{1}{\pars{x^{3} - 1}/\pars{x - 1} \over \pars{x^{5} - 1}/\pars{x - 1}}\,\dd x \\[5mm] = &\ \int_{0}^{1}{1 - x^{3} \over 1 - x^{5}}\,\dd x \,\,\,\stackrel{x^{5}\ \mapsto\ x}{=}\,\,\, {1 \over 5}\int_{0}^{1}{x^{-4/5} - x^{-1/5} \over 1 - x}\,\dd x \\[5mm] = &\ {1 \over 5}\pars{\int_{0}^{1}{1 - x^{-1/5} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-4/5} \over 1 - x}\,\dd x} = {1 \over 5}\bracks{\Psi\pars{4 \over 5} - \Psi\pars{1 \over 5}} \\[5mm] = &\ {1 \over 5}\bracks{\pi\cot\pars{\pi{1 \over 5}}} = \bbx{{1 \over 5}\root{1 + {2 \over \root{5}}}\,\pi} \approx 0.8648 \end{align}
Felix Marin
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