33
$\begingroup$

There is a curious known integral: $$\int_0^1\frac{\ln\left(1+x^{2+\sqrt{3\vphantom{\large3}}}\right)}{1+x}dx=\frac{\pi^2}{12}\left(1-\sqrt{3\vphantom{\large3}}\right)+\ln \left(1+\sqrt{3\vphantom{\large3}}\right)\ln2.$$ If we consider $\alpha=2+\sqrt{3\vphantom{\large3}}$ as a parameter and take a derivative w.r.t. $\alpha$ at this point, we get the following: $$I=\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt{3\vphantom{\large3}}\right)}\right)}dx.$$ Is it possible to express the integral $I$ in a closed form?

$\endgroup$
3
  • $\begingroup$ mathematica 7.0 and 9.0 both have no closed form solutions. $\endgroup$ Nov 14 '13 at 0:21
  • 9
    $\begingroup$ @SeyhmusGüngören It cannot evaluate the first integral either, but a closed form exists for it. $\endgroup$ Nov 14 '13 at 0:41
  • $\begingroup$ I'm getting approximations of about $-0.0093$ish $\endgroup$ Nov 17 '13 at 23:26
17
+100
$\begingroup$

Here is a partial progress report. I am basically repeating Jim Belk's analysis from the previous answer.

Set $F(a) = \int_{x=0}^1 \frac{\log(1+x^a)}{1+x} dx$. Then $$F(a) = \int_{x=0}^1 \int_{y=0}^{x^a} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)}$$ so $$F(a) + F(a^{-1}) = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y \leq x^{1/a} \leq 1} \frac{dx dy}{(1+x)(1+y)}$$ $$= \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y^a \leq x \leq 1} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq x,y \leq 1} \frac{dx dy}{(1+x)(1+y)} = (\log 2)^2.$$ (In order to combine the integrals, first switch the names of $x$ and $y$ in the second one.)

So $$F'(a) - a^{-2} F'(a^{-1})=0.$$ This gives a linear relation between $F'(2 + \sqrt{3})$ and $F'(2-\sqrt{3})$. If we find a second one, we can solve the linear equations and be done.


Notice that $$F'(a) = \int_{x=0}^1 \frac{x^a \log x dx}{(1+x)(1+x^a)} = \sum_{m,n=0}^{\infty} \int_{x=0}^1 (-1)^{m+n} x^{m+(n+1) a} \log x dx.$$ Integrating by parts, $\int_{x=0}^1 x^b \log x dx = \frac{-1}{(b+1)^2}$. So, ignoring issues of convergence, we should have $$F'(a) = \sum_{m,n=0}^{\infty} \frac{(-1)^{m+n+1}}{(m+(n+1) a + 1)^2} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{(m+n a)^2}$$ In the last step, we turned $m+1$ and $n+1$ into $m$ and $n$ to make things pretty. My guess is that the convergence issues can be dealt with for any $a>0$, but I haven't thought much about it.

So $$F'(a) + F'(a^{-1}) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \frac{(-1)^{m+n+1}}{(m+n a)^2} +\frac{(-1)^{m+n+1}}{(m+n a^{-1})^2} \right).$$ Putting $a=2 + \sqrt{3}$, this is $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n+1} \frac{2 (m^2+4mn+7n^2)}{(m^2+4mn+n^2)^2} $$ $$= 2 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{m^2+4mn+n^2} +12 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1} n^2}{(m^2+4mn+n^2)^2}.$$

Here is where I run out of ideas. The first sum is basically the one at the end of Jim Belk's post, but I have no ideas for the second one.

$\endgroup$
4
$\begingroup$

Jim Belk's analysis is very impressive. But I'm afraid there is an assumption that $\alpha\ge 0$ being implicitly made in his analysis. Here presents the other half of the answer. \begin{eqnarray} F(\alpha)&=&\int_0^1 {\frac{\ln (1+x^\alpha)}{1+x}dx} \\ I(\alpha)&=&\frac{dF}{d\alpha}=\int_0^1 {\frac{\ln x}{(1+x)(1+x^{-\alpha})}dx} \\ &=&\int_0^1 {\frac{\ln x}{1+x}\frac{x^\alpha}{1+x^\alpha}dx} \\ &=&\int_0^1 {\frac{\ln x}{1+x} \left( 1 - \frac{1}{1+x^\alpha} \right) dx } \\ &=&\int_0^1 {\frac{\ln x}{1+x}dx}-\int_0^1{ \frac{1}{1+x^\alpha} \frac{\ln x}{1+x}dx } \\ &=&-\frac{\pi^2}{12}-I(-\alpha) \\ \end{eqnarray}

$\endgroup$
4
$\begingroup$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\ln\pars{x} \over \pars{1 + x} \bracks{1 + x^{-\pars{2 + \root{3}}}}}\,\dd x:\ {\large ?}}$

\begin{align} \mbox{Let's consider}&\quad {\cal F}\pars{\mu}\equiv \int_{0}^{1}{\ln\pars{x} \over \pars{1 + x}\pars{1 + x^{-\mu}}}\,\dd x\quad \mbox{such that} \\[3mm]&\int_{0}^{1}{\ln\pars{x}\over \pars{1 + x} \bracks{1 + x^{-\pars{2 + \root{3}}}}}\,\dd x = {\cal F}\pars{2 + \root{3}}\tag{1} \end{align}

\begin{align} \color{#c00000}{{\cal F}\pars{\mu}} &=\totald{}{\mu}\int_{0}^{1}{\ln\pars{1 + x^{\mu}} \over 1 + x}\,\dd x =\totald{}{\mu}\int_{0}^{1}\sum_{m = 1}^{\infty}{\pars{-1}^{m + 1} \over m}x^{m\mu} \sum_{n = 0}^{\infty}\pars{-1}^{n}x^{n}\,\dd x \\[3mm]&=\totald{}{\mu}\sum_{n = 0}^{\infty}\pars{-1}^{n} \sum_{m = 1}^{\infty}{\pars{-1}^{m + 1} \over m}\int_{0}^{1}x^{m\mu + n}\,\dd x =\totald{}{\mu}\sum_{n = 0}^{\infty}\pars{-1}^{n} \sum_{m = 1}^{\infty}{\pars{-1}^{m + 1} \over m\pars{m\mu + n}} \\[3mm]&=\totald{}{\mu}\left\lbrace {1 \over \mu}\sum_{n = 0}^{\infty}\pars{-1}^{n}\times\right. \\[3mm]&\left.\phantom{\totald{}{\mu}\braces{\,\,\,}}\bracks{ \sum_{m = 0}^{\infty}{1 \over \pars{2m + 1}\pars{2m + 1 + n/\mu}} -\sum_{m = 0}^{\infty}{1 \over \pars{2m + 2}\pars{2m + 2 + n/\mu}}}\right\rbrace \\[3mm]&= {1 \over 4}\,\totald{}{\mu}\braces{{1 \over \mu} \sum_{n = 0}^{\infty}\pars{-1}^{n}\bracks{ {\Psi\pars{\bracks{1 + n/\mu}/2} - \Psi\pars{1/2} \over n/\bracks{2\mu}} - {\Psi\pars{1 + n/\bracks{2\mu}} - \Psi\pars{1} \over n/\bracks{2\mu}}}} \\[3mm]&=\half\,\totald{}{\mu} \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n}\bracks{ \Psi\pars{\half + {n \over 2\mu}} - \Psi\pars{1 + {n \over 2\mu}}} \\[3mm]&=\color{#c00000}{-\,{1 \over 8\mu^{2}} \sum_{n = 0}^{\infty}\pars{-1}^{n}\bracks{ -\Psi'\pars{\half + {n \over 2\mu}} + \Psi'\pars{1 + {n \over 2\mu}}}} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function.

Also, \begin{align} \Psi'\pars{\half + {n \over 2\mu}}&=4\Psi'\pars{n \over \mu} -\Psi'\pars{n \over 2\mu} \\[3mm] \Psi'\pars{1 + {n \over 2\mu}}&=\Psi'\pars{n \over 2\mu} - {4\mu^{2} \over n^{2}} \end{align}

\begin{align} \color{#00f}{\large{\cal F}\pars{\mu}}&= {\Psi'\pars{1/2} - \Psi'\pars{1} \over 8\mu^{2}} +{1 \over 4\mu^{2}} \sum_{n = 1}^{\infty}\pars{-1}^{n}\bracks{ 2\Psi'\pars{n \over \mu} - \Psi'\pars{n \over 2\mu}} -\half\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}} \\[3mm]&=\color{#00f}{\large{\pi^{2} \over 24}\pars{{1 \over \mu^{2}} + 1} +{1 \over 4\mu^{2}} \sum_{n = 1}^{\infty}\pars{-1}^{n}\bracks{ 2\Psi'\pars{n \over \mu} - \Psi'\pars{n \over 2\mu}}} \end{align}

So far !!!.

$\endgroup$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .