Find the value of $$\int_0^1{\frac{nx^{n-1}}{x+1}}dx.$$
I had no luck while integrating it. I also tried differentiating w.r.t n but still couldn't reach anywhere. Need help.
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For what it's worth, Mathematica returns a result in terms of the digamma function. This suggests to me that it may not have a "nice" closed form. – Michael Seifert May 05 '17 at 17:23
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@MichaelSeifert It has closed forms for $n\in\mathbb N$. – Simply Beautiful Art May 05 '17 at 21:51
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@SimplyBeautifulArt There seems to be a different closed form in elementary functions for each positive integer $n,$ but not a single closed form in elementary functions that is good for every positive integer $n.$ – David K May 06 '17 at 01:07
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@DavidK How is Chapper's answer not closed form? – Simply Beautiful Art May 06 '17 at 01:15
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@SimplyBeautifulArt As soon as you see $\sum_{k=1}^n,$ it's not a closed form. – David K May 06 '17 at 01:18
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@DavidK As far as I care, finite summations involving elementary functions are closed form solutions of integrals. But of course, that is up to debate. – Simply Beautiful Art May 06 '17 at 01:22
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@SimplyBeautifulArt If the summation is $\sum_{k=1}^{1017}$ or some other sum with a fixed number of terms, then yes. But there's a whole class of problems along the lines of how to express $\sum_{k=1}^n k^3$ (for example) in closed, form, or how to write the $n$th term of a recursively defined sequence in closed form. – David K May 06 '17 at 01:28
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1@DavidK Well I think depending on context, finite sums with variable indexes could easily be considered closed form. – Simply Beautiful Art May 06 '17 at 01:31
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After trying to search using Approach0 if find a few posts about limit for $n\to\infty$, but not about actual value of the integral. Like Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ and other questions linked there. – Martin Sleziak May 06 '17 at 05:48
4 Answers
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Put $y=1+x$ and the integral becomes $$ \int_{1}^2 \frac{n(y-1)^{n-1}}{y} \, dy = \int_1^2 \sum_{k=0}^n n\binom{n-1}{k} (-1)^{n-k-1} y^{k-1} \, dy = \left[ n(-1)^{n-1}\log{y} + \sum_{k=1}^n \frac{n}{k} \binom{n-1}{k} (-1)^{n-k-1} y^k \right]_1^2 \\ = n(-1)^{n-1}\log{2} + \sum_{k=1}^n \frac{n}{k} \binom{n-1}{k} (-1)^{n-k-1}(2^k-1). $$
Chappers
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HINT:
If $\displaystyle I_n=\int_0^1\dfrac{nx^{n-1}}{x+1}dx,$
$$I_{m+1}+I_m=\int_0^1\dfrac{(m+1)x^m+mx^{m-1}}{x+1}dx$$ $$=m\int_0^1x^{m-1}\ dx+\int_0^1\dfrac{x^m}{1+x}dx=1+\int_0^1\dfrac{x^m}{1+x}dx$$
Again if $\displaystyle J_m=\int_0^1\dfrac{x^m}{1+x}dx,$
$$J_{m+1}+J_m=?$$
lab bhattacharjee
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2Could we just say for $n \neq 0,$ $I_n = \frac1n J_{n-1},$ and skip directly to the $J_{m+1}+J_m$ part? – David K May 05 '17 at 13:30
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{n\,x^{n - 1} \over x + 1}\,\dd x & = n\int_{0}^{1}{x^{n - 1} - x^{n} \over 1 - x^{2}}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\,n\int_{0}^{1}{x^{n/2 - 1} - x^{n/2 - 1/2} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 2}\,n\bracks{% \int_{0}^{1}{1 - x^{n/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{n/2 - 1} \over 1 - x}\,\dd x} \\[5mm] & = \bbx{{1 \over 2}\,n\pars{H_{n/2 - 1/2} - H_{n/2 - 1}}} \end{align}
where $\ds{H_{z}}$ is the Harmonic Number.
Felix Marin
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We can write the integral as $$ F(n) = \int_0^{\;1} {\frac{{nx^{n - 1} }}{{1 + x}}dx} = \int_0^{\;1} {\frac{1}{{1 + x}}dx^n } = \int_0^{\;1} {\left( {1 - \frac{x}{{1 + x}}} \right)dx^n } = 1 - \int_0^{\;1} {\frac{x}{{1 + x}}dx^n } $$
$F(n+1)$ will be $$ F(n + 1) = \int_0^{\;1} {\frac{{\left( {n + 1} \right)x^n }}{{1 + x}}dx} = \int_0^{\;1} {\frac{x}{{1 + x}}dx^n } + \int_0^1 {\frac{{x^n }}{{1 + x}}dx} = 1 - F(n) + \frac{1}{{n + 1}}F(n + 1) $$ i.e.: $$ \begin{array}{l} \left( {1 - \frac{1}{{n + 1}}} \right)F(n + 1) = 1 - F(n) \\ \frac{{F(n + 1)}}{{n + 1}} = \frac{1}{n} - \frac{{F(n)}}{n} \\ \left( { - 1} \right)^{\,n} \frac{{F(n + 1)}}{{n + 1}} = \frac{{\left( { - 1} \right)^{\,n} }}{n} + \left( { - 1} \right)^{\,n - 1} \frac{{F(n)}}{n} \\ \end{array} $$
So we have $$ \left\{ \begin{array}{l} G(n) = \left( { - 1} \right)^{\,n} \frac{{F(n + 1)}}{{n + 1}} \\ G(0) = F(1) = \ln 2 \\ G(n + 1) - G(n) = \frac{{\left( { - 1} \right)^{\,n + 1} }}{{n + 1}} \\ \end{array} \right. $$
which leads to $$ \bbox[lightyellow] { \begin{array}{l} G(n) = \ln (2) + \sum\limits_{k = 0}^{n - 1} {\frac{{\left( { - 1} \right)^{\,k + 1} }}{{k + 1}}} = \left( { - 1} \right)^{\,n} \Phi \left( { - 1,1,n + 1} \right) \\ F(n) = n\;\Phi \left( { - 1,1,n} \right) \\ \end{array} }$$ where $\Phi$ denotes the Lerch Transcendent.
In fact, by the integral representation of $\Phi$ $$ \Phi \left( {z,s,a} \right)\mathop \equiv \limits^{def} \frac{1} {{\Gamma (s)}}\int_0^\infty {\frac{{t^{\,s - 1} e^{\, - a\,t} }} {{1 - z\,e^{\, - \,t} }}dt} $$ and $$ \Phi \left( { - 1,1,n} \right) = \int_{t\, = \,0}^\infty {\frac{{e^{\, - n\,t} }} {{1 + \,e^{\, - \,t} }}dt} \quad \xrightarrow{{e^{\, - \,t} = x}}\quad - \int_{x = 1}^0 {\frac{{x^{\,n - 1} }} {{1 + \,x}}dt} $$
G Cab
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