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Evaluate the integrals \begin{align} I_{1} &= \int_{0}^{1} \ln^{2n}(x) \ \ln\left(\ln\left(\frac{1}{x}\right)\right) dx \end{align} and \begin{align} I_{2} = \int_{0}^{1} \ln^{2n}(x) \ \ln^{2}\left(\ln\left(\frac{1}{x}\right)\right) dx. \end{align} From the resulting values of the integrals above is it possible to evaluate the integral \begin{align} I_{3} = \int_{0}^{1} \left( x^{a} + \frac{1}{x^{a}} \right) \ \ln^{p}\left(\ln\left(\frac{1}{x}\right)\right) dx \end{align} where $p=1,2$ in a compact form? Please show all work in in the solutions.

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  • $\begingroup$ Is it known to have a closed form for general $n$? $\endgroup$ – user111187 May 14 '14 at 16:48
  • $\begingroup$ @user111187 Integrals $I_{1}$ and $I_{2}$ should whereas $I_{3}$ may not. $\endgroup$ – Leucippus May 14 '14 at 17:34
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$I_j$ with $j = 1,2$ can be calculated as follows:

Substitute $u = -\ln x$, $\,dx = e^{-u} \,du$ to obtain \begin{align} I_1 &= \int_0^{\infty} x^{2n}e^{-x} \ln x \,dx = \partial_{\mu}|_0\int_0^{\infty} x^{2n+\mu}e^{-x}\,dx \\ &=\partial_{\mu}|_0 \,\Gamma(2n+\mu+1) \\ &= \Gamma(2n+1) \ \psi(2n+1) \end{align} and similarly, by differentiating twice before setting $\mu=0$, we obtain \begin{align} I_{2} = \Gamma(2n+1) \ \left[\psi(2n+1)^2+\psi_1(2n+1)\right]. \end{align}

Now we will use these results (with $n = 0$) to prove some lemmas. First notice that for $b>0$, we have \begin{align} \int_0^{\infty} e^{-bx} \ln x \,dx &= \frac{1}{b}\int_0^{\infty} e^{-x} (\ln x - \ln b)\,dx \\ &= - \frac{\ln b}{b}+ \frac{1}{b}\int_0^{\infty} x^0e^{-x} \ln x \,dx \\ &= - \frac{\ln b}{b}+ \frac{1}{b}\psi(1)\Gamma(1) \\ &= - \frac{\ln b+\gamma}{b} \end{align} where I used the known value $\psi(1) = -\gamma$. Also note that \begin{align} \int_0^{\infty} e^{-bx} \ln^2 x \,dx &= \frac{1}{b}\int_0^{\infty} e^{-x} \left(\ln^2 x - 2 \ln b \ln x + \ln^2 b\right)\,dx \\ &= \frac{1}{b}\int_0^{\infty} e^{-x} (\ln^2 x - 2 \ln b \ln x + \ln^2 b)\,dx \\ &= \frac{1}{b} \left(\pi^2/6+\gamma^2 + 2 \gamma \ln b + \ln^2 b \right) \\ &= \frac{\pi^2/6+(\gamma+\ln b)^2}{b} \end{align} where I used that $\psi_1(1) = \zeta(2) = \pi^2/6$.

Now substitute $u = -\ln x$ to obtain for $p = 1$ \begin{align} \int_0^{1} \left(x^a + x^{-a} \right) \ln\left( \ln \frac{1}{x} \right) \,dx &= \int_0^{\infty} \left(e^{-(1+a)x} + e^{-(1-a)x} \right) \ln x \,dx \\ &= \frac{2 \gamma}{1-a^{2}} - \frac{\ln (1-a)}{1-a}- \frac{\ln (1+a)}{1+a}, \end{align} where $a \neq 1$, and for $p = 2$ \begin{align} \int_0^{1} \left(x^a + x^{-a} \right) \ln^2\left( \ln \frac{1}{x} \right) \,dx &= \int_0^{\infty} \left(e^{-(1+a)x} + e^{-(1-a)x} \right) \ln^2 x \,dx \\ &= \frac{\pi^{2}}{3(1-a^{2})} + \frac{(\gamma+\ln (1-a))^2}{1-a} + \frac{(\gamma+\ln (1+a))^2}{1+a}, \end{align} where $a \neq 1$.

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

With $\ds{x = \expo{-t}\quad\imp\quad t = -\ln\pars{x}}$: \begin{align} {\cal I}_{n}\pars{\mu}& =\int_{\infty}^{0}\ln^{2n}\pars{\expo{-t}}\ln^{\mu}\pars{\ln\pars{\expo{t}}} \,\bracks{\expo{-t}\pars{-\dd t}} =\int_{0}^{\infty}t^{2n}\ln^{\mu}\pars{t}\expo{-t}\,\dd t \\[3mm]&=\lim_{\alpha \to 2n}\partiald[\mu]{}{\alpha} \int_{0}^{\infty}t^{\alpha}\expo{-t}\,\dd t =\lim_{\alpha \to 2n}\partiald[\mu]{\Gamma\pars{\alpha + 1}}{\alpha} \end{align} where $\ds{\Gamma\pars{z}}$ is the Gamma Function.

\begin{align} {\cal I}_{n}\pars{\mu}& =\lim_{\alpha \to 2n}\left\lbrace% \begin{array}{ll} \Gamma\pars{\alpha + 1}\Psi\pars{\alpha + 1}\,, & \quad\mu = 1 \\[2mm] \Gamma\pars{\alpha + 1} \bracks{\Psi^{2}\pars{\alpha + 1} + \Psi'\pars{\alpha + 1}}\,, & \quad\mu = 2 \end{array}\right. \end{align} $\ds{\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}}$ is the Digamma Function.

\begin{align} \color{#00f}{\large{\cal I}_{n}\pars{\mu}}& =\color{#00f}{\large\left\lbrace% \begin{array}{ll} \pars{2n}!\,\Psi\pars{2n + 1}\,, & \quad\mu = 1 \\[2mm] \pars{2n}!\,\bracks{\Psi^{2}\pars{2n + 1} + \Psi'\pars{2n + 1}}\,, & \quad\mu = 2 \end{array}\right.} \end{align} where we used Recurrence Formula ${\bf\mbox{6.1.15}}$: $\ds{\Gamma\pars{m + 1} = m!\,,\quad m \in {\mathbb N}}$.

$\color{#f00}{\ds{I_{3}\equiv\int_{0}^{1}\pars{x^{a} + {1 \over x^{a}}} \ln^{p}\pars{\ln\pars{1 \over x}}\,\dd x:\ {\large ?}}}$ \begin{align} I_{3}&=\int_{\infty}^{0}\pars{\expo{-at} + \expo{at}} \ln^{p}\pars{\ln\pars{\expo{t}}}\,\pars{-\expo{-t}\,\dd t} \\[3mm]&=\int_{0}^{\infty}\ln^{p}\pars{t}\expo{-\pars{1 + a}t}\,\dd t \int_{0}^{\infty}\ln^{p}\pars{t}\expo{-\pars{1 - a}t}\,\dd t \\[3mm]&=\lim_{\alpha \to 0}\partiald[p]{}{\alpha}\bracks{% \int_{0}^{\infty}t^{\alpha}\expo{-\pars{1 + a}t}\,\dd t \int_{0}^{\infty}t^{\alpha}\expo{-\pars{1 - a}t}\,\dd t} \end{align} Both integrals converge simultaneously when $\ds{\Re\pars{\alpha} > - 1}$ and $\ds{\verts{\Re\pars{a}} < 1}$. In that case \begin{align} I_{3}&=\lim_{\alpha \to 0}\partiald[p]{}{\alpha}\bracks{% \pars{1 + a}^{-\alpha - 1} + \pars{1 - a}^{-\alpha - 1}}\Gamma\pars{\alpha + 1} \end{align}

$$\color{#00f}{\large% \left.I_{3}\right\vert_{p\ =\ 1} =-\,{\pars{1 - a}\ln\pars{1 + a} + \pars{1 + a}\ln\pars{1 - a} + 2\gamma \over 1 - a^{2}}}\,,\quad \verts{\Re\pars{a}} < 1 $$

$$\color{#00f}{\large% \left.I_{3}\right\vert_{p\ =\ 2} = {Num \over 3\pars{1 - a^{2}}}} $$

\begin{align} \color{#00f}{Num}&=\color{#00f}{3 (a+1) \ln^2(1-a)-3 (a-1) \ln^2(a+1)+6 \gamma (a+1) \ln(1-a)} \\[3mm]&\color{#00f}{-6 \gamma(a-1)\ln(a+1)+6 \gamma ^2+\pi^{2}} \,,\quad \verts{\Re\pars{a}} < 1 \end{align} $\ds{\gamma}$ is the Euler-Mascheroni Constant ${\bf\mbox{6.1.3}}$.

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  • $\begingroup$ This solutions is missing the evaluation of $I_{3}$ as proposed in the problem, but does contain a nice method of calculating the integrals of $I_{1}$ and $I_{2}$. It is evident that the same result for $I_{3}$ would be obtained as given by @user111187's solution. $\endgroup$ – Leucippus May 14 '14 at 20:29
  • $\begingroup$ @Leucippus I went to have lunch. I just came back. Be patient. $\endgroup$ – Felix Marin May 14 '14 at 20:36
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    $\begingroup$ Lunch is important; Also nice solution $\endgroup$ – Leucippus May 14 '14 at 21:53

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