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Hi I am trying to prove this result $$ I:=\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c},\quad c>1. $$

Thanks. Since $x\in[0,1] $ we can write$$ I=\sum_{n=0}^\infty \int_0^1 \log x (1+x^2) x^{c-2+2cn}dx. $$ since $\sum_{n=0}^\infty x^n= (1-x)^{-1} , |x| < 1.$ Simplifying $$ I=\sum_n \int_0^1 \log x\, x^{c-2+2n} dx+\sum_n \int_0^1 \log x \, x^{c+2cn}\, dx. $$ Now we can write $$ I=-\frac{1}{4}\psi_1 \left(\frac{c+1}{2}\right)-\frac{1}{4} \psi_1\left( \frac{3+c}{2}\right) $$ where I summed the results of the integrals using general result of $\int_0^1 \log x \, x^n \, dx=-\frac{1}{(n+1)^2}$. Howeever this is not the result $\sec^2...$. Thanks

The function $\psi$ is the polygamma function which is defined in general by $$ \psi_m(z)=\frac{d^{m+1}}{dz^{m+1}} \log \Gamma(z) $$ and $\Gamma(z)=(z-1)!$, for this case $m=1$.

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Another derivation that uses only very elementary complex analysis and no special functions: (all integrals are to be regarded as Cauchy PV integrals if necessary)

By substituting $x\mapsto 1/x$, we find that the original integral equals $$\int_1^{\infty} (-\log x) \frac{(1+1/x^2)x^{2-c}}{1-x^{-2c}}\frac{dx}{x^2} = \int_1^{\infty} \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}} \,dx$$ Adding both copies of the integral, the problem is reduced to showing that $$\int_0^{\infty} \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}} \,dx = -\frac{\pi^2}{2c^2}\sec^2\left(\frac{\pi}{2c}\right)$$ We will do this by the method that is by now well-known: let $I(\mu) = \int_0^{\infty} \frac{(1+x^2)x^{c-2+\mu}}{1-x^{2c}} \,dx$, then we are looking for $I'(0)$.

We have (remember we are taking PV everywhere)$$\int_0^{\infty} \frac{x^a}{1-x^b} \,dx = \frac{\pi}{b}\cot\left(\pi\frac{a+1}{b}\right)$$ which can be easily derived using residues. (Integrate around a pizza slice contour.)

This immediately gives $$I(\mu) = \frac{\pi}{2c} \left[\cot\left(\pi\frac{c-1+\mu}{2c}\right) + \cot\left(\pi\frac{c+1+\mu}{2c}\right) \right]$$ and hence $$ I'(\mu) = -\frac{\pi^2}{4c^2}\left[\csc^2\left(\frac{\pi}{2c} (c-1+\mu)\right) +\csc^2\left(\frac{\pi}{2c}(c+1+\mu)\right)\right]$$ $$ \implies I'(0) = -\frac{\pi^2}{2c^2}\sec^2\left(\frac{\pi}{2c}\right)$$

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  • $\begingroup$ Thanks again for your solutions to my problems...+1 $\endgroup$ – Jeff Faraci May 13 '14 at 0:18
  • $\begingroup$ @Integrals It is my pleasure. This one was a bit easy though, post some harder ones ;) $\endgroup$ – user111187 May 13 '14 at 6:22
  • $\begingroup$ Okay :) The reason I do not post real hard ones as much is because a lot of people seem to complain when I do so. So I have been making the integrals cleaner and simpler...But I will now for you. $\endgroup$ – Jeff Faraci May 13 '14 at 14:22
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    $\begingroup$ @Integrals People complain only when you post integrals that are literally unsolvable in terms of known functions. Please make sure they do have a known solution, or conjecture a solution e.g. by numerical calculations. $\endgroup$ – user111187 May 13 '14 at 14:27
  • $\begingroup$ all Integrals I post have closed form solutions and I have proofs of every single one them. They are not "unsolvable" $\endgroup$ – Jeff Faraci May 13 '14 at 14:29
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Using the identity \begin{align} \psi(1-x) - \psi(x) = \pi \cot(\pi x) \end{align} then the derivative with respect to $x$ yields \begin{align} \psi_{1}(1-x) + \psi_{1}(x) = \pi^{2} \csc^{2}(\pi x). \end{align} Now the integral \begin{align} I &= \int_{0}^{1} \frac{ x^{c-2} (1+x^{2}) \ln(x) }{ 1 - x^{2c} } \, dx \end{align} is evaluate as follows. \begin{align} I &= \sum_{n=0}^{\infty} \, \int_{0}^{1} (1+x^{2}) \, x^{2cn+c-2} \ln(x) \, dx \\ &= - \sum_{n=0}^{\infty} \left[ \frac{1}{(2cn+c-1)^{2}} + \frac{1}{(2cn+c+1)^{2}} \right] \\ &= - \frac{1}{4 c^{2}} \left[ \psi_{1}\left(\frac{c-1}{2c}\right) + \psi_{1}\left( \frac{c+1}{2c}\right) \right] \\ &= - \left(\frac{\pi}{2c}\right)^{2} \csc^{2}\left(\frac{\pi}{2} + \frac{\pi}{2c}\right) \\ &= - \left(\frac{\pi}{2c}\right)^{2} \sec^{2}\left(\frac{\pi}{2c}\right). \end{align} It can now be stated that \begin{align} \int_{0}^{1} \frac{ x^{c-2} (1+x^{2}) \ln(x) }{ 1 - x^{2c} } \, dx = - \left(\frac{\pi}{2c}\right)^{2} \sec^{2}\left(\frac{\pi}{2c}\right). \end{align}

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

With $\ds{x \equiv t^{1/\pars{2c}}\quad\imp\quad t = x^{2c}}$: \begin{align} &\color{#c00000}{% \int_{0}^{1}\ln\pars{x}\,{\pars{1 + x^{2}}x^{c - 2} \over 1 - x^{2c}}\,\dd x} =\int_{0}^{1} \ln\pars{t^{1/\bracks{2c}}}\,{\pars{1 + t^{1/c}}t^{1/2 - 1/c} \over 1 - t} \,{1 \over 2c}\,t^{1/\pars{2c} - 1}\,\dd t \\[3mm]&={1 \over 4c^{2}}\bracks{% \int_{0}^{1}{\ln\pars{t}t^{-\pars{1 + 1/c}/2} \over 1 - t}\,\dd t +\int_{0}^{1}{\ln\pars{t}t^{-\pars{1 - 1/c}/2} \over 1 - t}\,\dd t} \\[3mm]&=-\,{1 \over 4c^{2}} \lim_{\mu \to -\pars{1 + 1/c}/2} \partiald{}{\mu}\int_{0}^{1}{1 - t^{\mu} \over 1 - t}\,\dd t + \pars{~c \to -c~} \\[3mm]&=\color{#c00000}{-\,{1 \over 4c^{2}} \lim_{\mu \to -\pars{1 + 1/c}/2} \partiald{\bracks{\Psi\pars{\mu + 1} + \gamma}}{\mu} + \pars{~c \to -c~}} \end{align} where we used the A&S table identity ${\bf\mbox{6.3.22}}$. $\ds{\Psi\pars{z}}$ is the Digamma Function and $\ds{\gamma}$ is the Euler-Mascheroni Constant.

\begin{align} &\color{#c00000}{% \int_{0}^{1}\ln\pars{x}\,{\pars{1 + x^{2}}x^{c - 2} \over 1 - x^{2c}}\,\dd x} = -\,{1 \over 4c^{2}}\bracks{\Psi'\pars{\half + {1 \over 2c}} + \Psi'\pars{\half - {1 \over 2c}}} \end{align} With Euler Reflection Formula ${\bf\mbox{6.4.7}}$: \begin{align} &\color{#c00000}{% \int_{0}^{1}\ln\pars{x}\,{\pars{1 + x^{2}}x^{c - 2} \over 1 - x^{2c}}\,\dd x} = -\,{1 \over 4c^{2}}\bracks{-\pi\,\totald{\cot\pars{\pi z}}{z}} _{z\ =\ 1/2\ -\ 1/\pars{2c}} \\[3mm]&=-\,{\pi^{2} \over 4c^{2}}\,\csc^{2}\pars{\pi\bracks{\half - {1 \over 2c}}} =-\,{\pi^{2} \over 4c^{2}}\,\sec^{2}\pars{\pi \over 2c} \end{align}

$$ \color{#00f}{\large% \int_{0}^{1}\ln\pars{x}\,{\pars{1 + x^{2}}x^{c - 2} \over 1 - x^{2c}}\,\dd x} =\color{#00f}{\large-\,\pars{\pi \over 2c}^{2}\sec^{2}\pars{\pi \over 2c}}\,,\qquad \verts{c} > 1 $$

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  • $\begingroup$ @Integrals You're welcome. Thanks a lot. $\endgroup$ – Felix Marin May 13 '14 at 21:51
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Here is an approach. Follow the steps

1) Make the change of variables $\ln(x)=-t$

2) Follow it with another change of variables $y=2ct$

3) Use the Hurwitz Zeta function. See this problem.

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