$\int_0^1\frac{1-t}{(t-2)\ln t}\,dt$ integral

Question

I have two related questions. The first is: Is there a closed form expression for: $$\int_0^1\frac{1-t}{(t-2)\ln t}\,dt\approx0.507834$$ I know that there are some very superb integrators on this site. I doubt that a closed form exists, but I may be wrong. If I'm wrong, I'm curious as to how you got your answer! (P.S. This page gives me hope that it might be expressable in terms of $\zeta(\cdot)$. If so, I'd consider that to be "closed form.")

Second question: Is there an elementary function $f(t)$ such that: $$\exp\left(\int_0^1\frac{1-t}{(t-2)\ln t}\,dt\right)=\int_0^1f(t)dt$$ That value is approximately $1.66169$. I am very interested in this value. The expression on the left is a great way to express this number. But, I feel like it would be "cleaner" if I could express it simply as an integral, like on the right.

Answer 1

The value of your integral could be expressed as $$I=\sum_{n=1}^\infty \frac{\ln n}{2^n}=-\frac{d}{ds}\Phi\left(\frac12,0,0\right)=-\frac12\frac{d}{ds}\Phi\left(\frac12,0,1\right),$$ where $$\Phi(z,s,a)=\sum_{n=1}^\infty\frac{z^n}{(n+a)^s}$$ is the Lerch transcendent. The second value you're interested in, surprisingly, has its own name: Somos' quadratic recurrence constant: $$\exp I=\sigma=\sqrt{1\sqrt{2\sqrt{3\ldots}}}=1^{1/2}2^{1/4}3^{1/8}\ldots$$

Here's the proof of the series representation. Using that $$\frac{1}{2-t}=\frac{1}{2}\sum_{n=0}^\infty\left(\frac{t}{2}\right)^n=\sum_{n=0}^\infty\frac{t^n}{2^{n+1}},$$ we can write $$I=\sum_{n=0}^\infty\frac{1}{2^{n+1}}\int_0^1\frac{t^{n+1}-t^n}{\ln t}\,dt=\sum_{n=0}^\infty\frac{1}{2^{n+1}}\ln\frac{n+2}{n+1},$$ since $$\int_0^1\frac{t^{n+1}-t^n}{\ln t}\,dt=\int_0^1\left(\int_{n}^{n+1}t^y\,dy\right)\,dt=\int_{n}^{n+1}\left(\int_0^1t^y\,dt\right)\,dy=$$$$=\int_{n}^{n+1}\frac{dy}{y+1}=\ln\frac{n+2}{n+1}.$$ Therefore, $$I=\sum_{n=0}^\infty\frac{1}{2^{n+1}}\ln\frac{n+2}{n+1}=\sum_{n=1}^\infty\frac{1}{2^{n}}\ln\frac{n+1}{n}=2\sum_{n=1}^\infty\frac{\ln(n+1)}{2^{n+1}}-\sum_{n=1}^\infty\frac{\ln n}{2^n}=\sum_{n=1}^\infty\frac{\ln n}{2^n}.$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{1 - t \over \pars{t - 2}\ln\pars{t}}\,\dd t \approx 0.507834: \ {\large ?}}$.

Lets consider $$ {\cal F}\pars{\mu}\equiv \int_{0}^{1}{1 - t^{\mu} \over \pars{t - 2}\ln\pars{t}}\,\dd t\,,\qquad\qquad \left\lbrace\begin{array}{rcl} {\cal F}\pars{1} & = & {\large ?} \\[1mm] {\cal F}\pars{0} & = & 0 \end{array}\right. $$

\begin{align} {\cal F}'\pars{\mu}& =\int_{0}^{1}{-t^{\mu}\ln\pars{t} \over \pars{t - 2}\ln\pars{t}}\,\dd t =\int_{0}^{1}{t^{\mu} \over 2 - t}\,\dd t =\half\sum_{n = 1}^{\infty}\int_{0}^{1/2}t^{\mu + n - 1}\pars{1/2}^{n - 1}\,\dd t \\[3mm]&=\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over \mu + n} \end{align}

\begin{align} \int_{0}^{1}{1 - t \over \pars{t - 2}\ln\pars{t}}\,\dd t&={\cal F}\pars{1} =\int_{0}^{1}{\cal F}'\pars{\mu}\,\dd\mu =\sum_{n = 1}^{\infty}\pars{\half}^{n}\bracks{\ln\pars{1 + n} - \ln\pars{n}} \\[3mm]&=\sum_{n = 2}^{\infty}\pars{\half}^{n - 1}\ln\pars{n} -\sum_{n = 1}^{\infty}\pars{\half}^{n}\ln\pars{n} =\sum_{n = 1}^{\infty}2^{-n}\ln\pars{n} \end{align}

Also $\ds{\pars{~\mbox{with}\ \verts{z} < 1~}}$, $$ \partiald{{\rm Li_{s}}\pars{z}}{\rm s} =\partiald{\sum_{n = 1}^{\infty}z^{n}/n^{\rm s}}{\rm s} =-\sum_{n = 1}^{\infty}{z^{n} \over n^{\rm s}}\,\ln\pars{n} $$ such that $$\color{#66f}{\large% \int_{0}^{1}{1 - t \over \pars{t - 2}\ln\pars{t}}\,\dd t =-\lim_{{\rm s}\ \to\ 0}\partiald{{\rm Li_{s}}\pars{1/2}}{\rm s}} \approx {\tt 0.5078} $$