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Integrate

$$I=\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)\tag1$$

$${x^{2n}-x\over 1+x}=\sum_{k=0}^{\infty}(-1)^k(x^{2n}-x)x^k\tag2$$

Sub $(2)$ into $(1)\rightarrow (3)$

$$I=\sum_{n=0}^{\infty}\int_{0}^{1}(x^{2n+k}-x^{k+1})\cdot{dx\over \ln{x}}\tag3$$ Frullani's theorem

$$\int_{0}^{1}(x^m-x^n)\cdot{dx\over \ln{x}}=\ln{m+1 \over n+1}\tag4$$

Apply $(4)$ to $(3)\rightarrow (5)$

$$I=\sum_{n=0}^{\infty}(-1)^n\ln\left({2n+k+1\over k+2}\right)\tag5$$

$${2\over \pi}={(2n-1)!!\over (2n)!}\prod_{k=0}^{\infty}\left({2n+k+1\over k+2}\right)^{(-1)^k}\tag6$$

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  • $\begingroup$ Abstract duplicate of a question you already asked: math.stackexchange.com/questions/1818436/… $\endgroup$ – Jack D'Aurizio Jun 10 '16 at 16:42
  • $\begingroup$ What is abstract duplicate? Have I already asked this question? Why are you the only person marking down my question with so called duplicate? When I see more than a finger I start to believe in you. $\endgroup$ – gymbvghjkgkjkhgfkl Jun 10 '16 at 18:11
  • $\begingroup$ Abstract duplicate because it is not an exact duplicate, but the techniques shown in the answers to your previous question clearly solve this problem, too, so this actual question is redundant, in my opinion. In any case, I am just A finger, not many fingers. $\endgroup$ – Jack D'Aurizio Jun 10 '16 at 18:15
  • $\begingroup$ It is easy to say when you have knowledge in maths. How is the other question be also duplicate? $\endgroup$ – gymbvghjkgkjkhgfkl Jun 10 '16 at 18:16
  • $\begingroup$ For instance: take the product in $(6)$ for $0\leq k\leq 2N$. Split the even $k$s and the odd ones, collect pieces, simplify, take the limit as $N\to +\infty$. $\endgroup$ – Jack D'Aurizio Jun 10 '16 at 18:16
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Enforce the substitution $x\to e^{-x}$ to write

$$\begin{align} I(n)&=\int_0^1 \frac{x^{2n}-x}{1+x}\frac{1}{\log(x)}\,dx\\\\ &=\int_0^\infty \frac{e^{-x}-e^{-2nx}}{x}\frac{e^{-x}}{1+e^{-x}}\,dx\\\\ &=\sum_{k=0}^\infty (-1)^k \int_0^\infty \frac{e^{-(k+2)x}-e^{-(k+2n+1)x}}{x}\\\\ &=\sum_{k=0}^\infty (-1)^k \log\left(\frac{k+2n+1}{k+2}\right)\\\\ &=\sum_{k=1}^\infty (-1)^{k-1} \log\left(\frac{k+2n}{k+1}\right)\\\\ \end{align}$$

Now, note that we can write the partial sum

$$\begin{align} \sum_{k=1}^{2N} (-1)^{k-1} \log\left(\frac{k+2n}{k+1}\right)&=\sum_{k=1}^N \log\left(\frac{2k-1+2n}{2k}\right)-\sum_{k=1}^N \log\left(\frac{2k+2n}{2k+1}\right)\\\\ &=\sum_{k=1}^N \log\left(\frac{2k-1+2n}{2k+2n}\right)+\sum_{k=1}^N \log\left(\frac{2k+1}{2k}\right)\\\\ &=\sum_{k=n+1}^{n+N} \log\left(\frac{2k-1}{2k}\right)+\sum_{k=1}^N \log\left(\frac{2k+1}{2k}\right)\\\\ &=\sum_{k=1}^{n+N} \log\left(\frac{2k-1}{2k}\right)+\sum_{k=1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\ &-\sum_{k=1}^{n} \log\left(\frac{2k-1}{2k}\right)-\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\ &=\sum_{k=1}^{n+N} \log\left(\frac{(2k-1)(2k+1)}{(2k)(2k)}\right)\\\\ &+\log\left(\frac{(2n)!!}{(2n-1)!!}\right)-\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\ \end{align}$$

Recalling Wallis' Product, we see that

$$\lim_{N\to \infty}\sum_{k=1}^{n+N} \log\left(\frac{(2k-1)(2k+1)}{(2k)(2k)}\right)=-\log(\pi/2)$$

And since $\lim_{N\to \infty}\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)=0$, we find

$$I(n)=\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)$$

as was to be shown!!


APPENDIX:

In THIS ANSWER, I evaluated the integral

$$J(n)=\int_0^1 \frac{x^{2n+1}-x}{1+x}\frac{1}{\log(x)}\,dx=\log\left(\frac{(2n+1)!!}{(2n)!!}\right)$$

by making use of the integral evaluated herein.

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  • $\begingroup$ @JackD'Aurizio Thanks! Much appreciative. $\endgroup$ – Mark Viola Jun 11 '16 at 0:28
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\begin{align} &\color{#f00}{\int_{0}^{1}{x^{2n} - x \over 1 + x}\,{\dd x\over \ln\pars{x}}} = -\int_{0}^{1}{x^{2n} - x \over 1 + x}\ \overbrace{\int_{0}^{\infty}x^{y}\,\dd y}^{\ds{-\,{1 \over \ln\pars{x}}}}\ \,\dd x = -\int_{0}^{\infty}\int_{0}^{1}{x^{2n + y} - x^{1 + y} \over 1 + x}\,\dd x\,\dd y \\[3mm] = &\ \int_{0}^{\infty}\pars{\int_{0}^{1}{1 - x^{y + 2n} \over 1 + x}\,\dd x - \int_{0}^{1}{1 - x^{y + 1} \over 1 + x}\,\dd x}\,\dd y\tag{1} \end{align}


However, by using the well known digamma $\Psi$ function identity $\ds{\left.\int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t \,\right\vert_{\ \Re\pars{z}\ >\ 0} = \Psi\pars{z} + \gamma\quad}$ where $\gamma$ is the Euler-Mascheroni constant: \begin{align} \fbox{$\ds{\int_{0}^{1}{1 - x^{z} \over 1 + x}\,\dd x}$} &= 2\int_{0}^{1}{1 - x^{z} \over 1 - x^{2}}\,\dd x - \int_{0}^{1}{1 - x^{z} \over 1 - x}\,\dd x \\[3mm] & = \int_{0}^{1}{x^{-1/2} - x^{z/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{z} \over 1 - x}\,\dd x \\[3mm] & = \int_{0}^{1}{1 - x^{z/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{z} \over 1 - x}\,\dd x \\[3mm] & = \fbox{$\ds{% \Psi\pars{{z \over 2} + \half} - \Psi\pars{\half} - \Psi\pars{z + 1} - \gamma}$} \end{align}

we get, after replacing in $\pars{1}$, \begin{align} &\color{#f00}{\int_{0}^{1}{x^{2n} - x \over 1 + x}\,{\dd x\over \ln\pars{x}}} \\[3mm] = &\ \int_{0}^{\infty}\bracks{\Psi\pars{{y \over 2} + n + \half} - \Psi\pars{y + 2n + 1} - \Psi\pars{{y \over 2} + 1} + \Psi\pars{y + 2}}\,\dd y \end{align}

Since $\ds{\Psi\pars{z}\ \stackrel{\mbox{def.}}{=}\ \totald{\ln\pars{\Gamma\pars{z}}}{z}}$: \begin{align} &\color{#f00}{\int_{0}^{1}{x^{2n} - x \over 1 + x}\,{\dd x\over \ln\pars{x}}} = \left.\ln\pars{\Gamma^{2}\pars{y/2 + n + 1/2}\Gamma\pars{y + 2} \over \Gamma^{2}\pars{y/2 + 1}\Gamma\pars{y + 2n + 1}}\right\vert_{\ 0}^{\ \infty} \\[3mm] = &\ \color{#f00}{% \ln\pars{2^{1 - 2n}\,{\Gamma\pars{2n + 1} \over \Gamma^{2}\pars{n + 1/2}}}} \end{align}

Could you simplify it ?.

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