$\int_0^1\frac{1-t}{(t-2)\ln t}\,dt$ integral 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.
 A: 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,
$$I=\sum_{n=0}^\infty\frac{1}{2^{n+1}}\ln\frac{n+2}{n+1}=\sum_{n=1}^\infty\frac{1}{2^{n}}\ln\frac{n+1}{n}=2\sum_{n=1}^\infty\frac{\ln(n+1)}{2^{n+1}}-\sum_{n=1}^\infty\frac{\ln n}{2^n}=\sum_{n=1}^\infty\frac{\ln n}{2^n}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\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
$$\color{#66f}{\large%
\int_{0}^{1}{1 - t \over \pars{t - 2}\ln\pars{t}}\,\dd t
=-\lim_{{\rm s}\ \to\ 0}\partiald{{\rm Li_{s}}\pars{1/2}}{\rm s}}
\approx {\tt 0.5078}
$$
