 David Lewis

2022-01-06

Evaluate ${\int }_{0}^{1}{\left(\frac{1}{\mathrm{ln}x}+\frac{1}{1-x}\right)}^{2}dx$ Jeffery Autrey

Here is an another approach:
Let I denote the integral. By the substitution $x={e}^{-t}$, we have
$I={\int }_{0}^{\mathrm{\infty }}\left[\frac{1}{{\left(1-{e}^{-t}\right)}^{2}}-\frac{2}{t\left(1-{e}^{-t}\right)}+\frac{1}{{t}^{2}}\right]{e}^{-t}dt$
$={\int }_{0}^{\mathrm{\infty }}\left[\frac{{e}^{t}}{{\left({e}^{t}-1\right)}^{2}}-\frac{1}{{t}^{2}}\right]dt+{\int }_{0}^{\mathrm{\infty }}\left[\frac{1+{e}^{-t}}{{t}^{2}}-\frac{2}{t\left({e}^{t}-1\right)}\right]dt$
It is easy to observe that the first integral is
${\int }_{0}^{\mathrm{\infty }}\left[\frac{{e}^{t}}{{\left({e}^{t}-1\right)}^{2}}-\frac{1}{{t}^{2}}\right]dt={\left[\frac{1}{t}-\frac{1}{{e}^{t}-1}\right]}_{0}^{\mathrm{\infty }}=-\frac{12}{}$
We thus focus on the second integral. Associated to it, we introduce
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}\left[\frac{1+{e}^{-t}}{{t}^{2}}-\frac{2}{t\left({e}^{t}-1\right)}\right]{e}^{-st}dt$
By the twice differentiation, we have
$F{}^{″}\left(s\right)={\int }_{0}^{\mathrm{\infty }}\left[1+{e}^{-t}-\frac{2t}{{e}^{t}-1}\right]{e}^{-st}dt$
$=\frac{1}{s}+\frac{1}{s+1}-2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(n+s\right)}^{2}}$
$=\frac{1}{s}+\frac{1}{s+1}-2{\psi }_{1}\left(s+1\right)$
Integrating and using the condition ${F}^{\prime }\left(+\mathrm{\infty }\right)=0$, we have
${F}^{\prime }\left(s\right)=\mathrm{log}s+\mathrm{log}\left(s+1\right)-2{\psi }_{0}\left(s+1\right)$
Here we used the estimate ${\psi }_{0}\left(s\right)\sim \mathrm{log}s$ as $s\to \mathrm{\infty }$. Integrating again, we have A bit late to the party, but here is a slightly different approach that doesn't use any series, Stirling's approximations or anything, though does make use of some values of the zeta function.
First substitute $x={e}^{-y}$ to get
${\int }_{0}^{\mathrm{\infty }}\left(\frac{1}{1-{e}^{-y}}\right)-\frac{1}{y}{\right)}^{2}{e}^{-y}dy$
The integrand is awkward because when you multiply it out, the pieces separately diverge. Multiplying it by ${y}^{z}$ regulates the singularity at 0 but also gives standard integrals. So for $Re\left(z\right)>1$ we have (by parts)
${\int }_{0}^{\mathrm{\infty }}\frac{1}{{\left(1-{e}^{-y}\right)}^{2}}{e}^{-y}{y}^{z}dy=z{\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-y}}{1-{e}^{-y}}{y}^{z-1}dy$
$=z\zeta \left(z\right)\mathrm{\Gamma }\left(z\right)$
and
${\int }_{0}^{\mathrm{\infty }}\frac{1}{y\left(1-{e}^{-y}\right\}{e}^{-y}{y}^{z}dy=\zeta \left(z\right)\mathrm{\Gamma }\left(z\right)}$
${\int }_{0}^{\mathrm{\infty }}\frac{1}{{y}^{2}}{e}^{-y}{y}^{z}dy=\mathrm{\Gamma }\left(z-1\right)$
Putting the bits together:
${\int }_{0}^{\mathrm{\infty }}{\left(\frac{1}{1-{e}^{-y}}-\frac{1}{y}\right)}^{2}{e}^{-y}{y}^{z}dy$
$=\mathrm{\Gamma }\left(z\right)\left(\frac{1}{z-1}+\left(z-2\right)\zeta \left(z\right)\right)$
The derivation was valid for $Re\left(z\right)>1$, but both sides of the above equation are analytic for $Re\left(z\right)\succ 1$ (RHS has a removable singularity at $z=0$), so by analytic continuation we may take the limit $z\to 0$. Near $z=0,\mathrm{\Gamma }\left(z\right)=\frac{1}{z}+O\left(1\right)$, and the final answer is the derivative of $\frac{1}{z-1}+\left(z-2\right)\zeta \left(z\right)$ at o, i.e.,
$-1+\zeta \left(0\right)-2{\zeta }^{\prime }\left(0\right)=\mathrm{log}\left(2\pi \right)-\frac{3}{2}$ nick1337

OK, I'm going to lay this out up to a sum, which will likely evaluate into whatever answer was provided above. This integral is subject to the same sorts of tricks that I did for another integral involving a factor of $1/\mathrm{log}x$ in the integral. The first piece is to let $x={e}^{-y}$; the integral becomes
${\int }_{0}^{\mathrm{\infty }}dy{e}^{-y}\left(\frac{\left({e}^{-y}-\left(1-y\right){\right)}^{2}}{{y}^{2}\left(1-{e}^{-y}{\right)}^{2}}\right)$
Now Taylor expand the factor $\left(1-{e}^{-y}{\right)}^{-2}$, and if we can reverse the order of summation and integration, we get:
$\sum _{k=1}^{\mathrm{\infty }}k{\int }_{0}^{\mathrm{\infty }}dy\left(\frac{{e}^{-y}-\left(1-y\right){\right)}^{2}}{{y}^{2}}\right){e}^{-ky}$
The integral inside the sum is a bit difficult, although it is convergent. The way I see through it is to replace k with a continuous parameter $\alpha$ and differentiate with respect to $\alpha$ inside the integral twice (to clear the pesky ${y}^{2}$ in the denominator) to get a function
$I\left(\alpha \right)={\int }_{0}^{\mathrm{\infty }}dy\left(\frac{\left({e}^{-y}-\left(1-y\right){\right)}^{2}}{{y}^{2}}\right){e}^{-\alpha y}\phantom{\rule{0ex}{0ex}}\frac{{d}^{2}I}{d{\alpha }^{2}}={\int }_{0}^{\mathrm{\infty }}dy\left({e}^{-y}-\left(1-y\right){\right)}^{2}{e}^{-\alpha y}\phantom{\rule{0ex}{0ex}}=\frac{1}{\alpha +2}-\frac{2}{\alpha +1}+\frac{2}{\left(\alpha +1{\right)}^{2}}+\frac{1}{\alpha }-\frac{2}{{\alpha }^{2}}+\frac{2}{{\alpha }^{3}}$
You integrate this twice to recover $\alpha$; the constants of integration may be shown to vanish by considering the limit as $a\to \mathrm{\infty }$. The original integral is then
$\sum _{k=1}^{\mathrm{\infty }}kI\left(k\right)$
where
$I\left(k\right)=\left(k+2\right)\mathrm{log}\left[\frac{k\left(k+2\right)}{\left(k+1{\right)}^{2}}\right]+\frac{1}{k}$
so the integral takes on the value
$\sum _{k=1}^{\mathrm{\infty }}\left[1+\left[\left(k+1{\right)}^{2}-1\right]\mathrm{log}\left(1-\frac{1}{\left(k+1{\right)}^{2}}\right)\right]=\sum _{k=1}^{\mathrm{\infty }}\left[1+\left(k+1{\right)}^{2}\mathrm{log}\left(1-\frac{1}{\left(k+1{\right)}^{2}}\right)\right]+\mathrm{log}2$
The sum may be simplified by Taylor expanding the $\mathrm{log}$ term; note that the unit value cancels and we get that the integral equals
$\mathrm{log}2+\sum _{k=2}^{\mathrm{\infty }}\left[1-{k}^{2}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m}\left(\frac{1}{{k}^{2}}{\right)}^{m}\right]=\mathrm{log}2-\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m+1}\left[\zeta \left(2m\right)-1\right]$
I have not yet evaluated this sum yet, but unless someone else does it before me, I will figure it out and come back.

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