Evaluate \int_0^1(\frac{1}{\ln x}+\frac{1}{1-x})^2dx

David Lewis 2022-01-06 Answered
Evaluate \(\displaystyle{\int_{{0}}^{{1}}}{\left({\frac{{{1}}}{{{\ln{{x}}}}}}+{\frac{{{1}}}{{{1}-{x}}}}\right)}^{{2}}{\left.{d}{x}\right.}\)

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Expert Answer

Jeffery Autrey
Answered 2022-01-07 Author has 3447 answers
Here is an another approach:
Let I denote the integral. By the substitution \(\displaystyle{x}={e}^{{-{t}}}\), we have
\(\displaystyle{I}={\int_{{0}}^{\infty}}{\left[{\frac{{{1}}}{{{\left({1}-{e}^{{-{t}}}\right)}^{{2}}}}}-{\frac{{{2}}}{{{t}{\left({1}-{e}^{{-{t}}}\right)}}}}+{\frac{{{1}}}{{{t}^{{2}}}}}\right]}{e}^{{-{t}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{0}}^{\infty}}{\left[{\frac{{{e}^{{t}}}}{{{\left({e}^{{t}}-{1}\right)}^{{2}}}}}-{\frac{{{1}}}{{{t}^{{2}}}}}\right]}{\left.{d}{t}\right.}+{\int_{{0}}^{\infty}}{\left[{\frac{{{1}+{e}^{{-{t}}}}}{{{t}^{{2}}}}}-{\frac{{{2}}}{{{t}{\left({e}^{{t}}-{1}\right)}}}}\right]}{\left.{d}{t}\right.}\)
It is easy to observe that the first integral is
\(\displaystyle{\int_{{0}}^{\infty}}{\left[{\frac{{{e}^{{t}}}}{{{\left({e}^{{t}}-{1}\right)}^{{2}}}}}-{\frac{{{1}}}{{{t}^{{2}}}}}\right]}{\left.{d}{t}\right.}={{\left[{\frac{{{1}}}{{{t}}}}-{\frac{{{1}}}{{{e}^{{t}}-{1}}}}\right]}_{{0}}^{\infty}}=-{\frac{{12}}{}}\)
We thus focus on the second integral. Associated to it, we introduce
\(\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{\left[{\frac{{{1}+{e}^{{-{t}}}}}{{{t}^{{2}}}}}-{\frac{{{2}}}{{{t}{\left({e}^{{t}}-{1}\right)}}}}\right]}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}\)
By the twice differentiation, we have
\(\displaystyle{F}{''}{\left({s}\right)}={\int_{{0}}^{\infty}}{\left[{1}+{e}^{{-{t}}}-{\frac{{{2}{t}}}{{{e}^{{t}}-{1}}}}\right]}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{s}}}}+{\frac{{{1}}}{{{s}+{1}}}}-{2}{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{\left({n}+{s}\right)}^{{2}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{s}}}}+{\frac{{{1}}}{{{s}+{1}}}}-{2}\psi_{{1}}{\left({s}+{1}\right)}\)
Integrating and using the condition \(\displaystyle{F}'{\left(+\infty\right)}={0}\), we have
\(\displaystyle{F}'{\left({s}\right)}={\log{{s}}}+{\log{{\left({s}+{1}\right)}}}-{2}\psi_{{0}}{\left({s}+{1}\right)}\)
Here we used the estimate \(\displaystyle\psi_{{0}}{\left({s}\right)}\sim{\log{{s}}}\) as \(\displaystyle{s}\to\infty\). Integrating again, we have
\(\displaystyle{F}{\left({s}\right)}={s}{\log{{s}}}+{\left({s}+{1}\right)}{\log{{\left({s}+{1}\right)}}}-{2}{s}-{1}-{2}{\log{\Gamma}}{\left({s}+{1}\right)}+{C}\)
To determine the constant C, we rearrange the terms as
\(\displaystyle{F}{\left({s}\right)}={\left[{\left({s}+{1}\right)}{\log{{\left({\frac{{{s}+{1}}}{{{s}}}}\right)}}}-{1}\right]}+{2}{\left[{\left({s}+{\frac{{{1}}}{{{2}}}}\right)}{\log{{s}}}-{s}-{\log{\Gamma}}{\left({s}+{1}\right)}\right]}+{C}\)
Then by the Stirling's formula, we have
\(\displaystyle{0}={F}{\left(+\infty\right)}=-{\log{{\left({2}\pi\right)}}}+{C}\)
and thus \(\displaystyle{C}={\log{{\left({2}\pi\right)}}}\). Therefore
\(\displaystyle{I}=-{\frac{{{1}}}{{{2}}}}+{F}{\left({0}\right)}={\log{{\left({2}\pi\right)}}}-{\frac{{{3}}}{{{2}}}}\)
as desired.
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Suhadolahbb
Answered 2022-01-08 Author has 863 answers
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 \(\displaystyle{x}={e}^{{-{y}}}\) to get
\(\displaystyle{\int_{{0}}^{\infty}}{\left({\frac{{{1}}}{{{1}-{e}^{{-{y}}}}}}\right)}-{\frac{{{1}}}{{{y}}}}{)}^{{2}}{e}^{{-{y}}}{\left.{d}{y}\right.}\)
The integrand is awkward because when you multiply it out, the pieces separately diverge. Multiplying it by \(\displaystyle{y}^{{z}}\) regulates the singularity at 0 but also gives standard integrals. So for \(\displaystyle{R}{e}{\left({z}\right)}{>}{1}\) we have (by parts)
\(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{1}}}{{{\left({1}-{e}^{{-{y}}}\right)}^{{2}}}}}{e}^{{-{y}}}{y}^{{z}}{\left.{d}{y}\right.}={z}{\int_{{0}}^{\infty}}{\frac{{{e}^{{-{y}}}}}{{{1}-{e}^{{-{y}}}}}}{y}^{{{z}-{1}}}{\left.{d}{y}\right.}\)
\(\displaystyle={z}\zeta{\left({z}\right)}\Gamma{\left({z}\right)}\)
and
\(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{1}}}{{{y}{\left({1}-{e}^{{-{y}}}\right\rbrace}{e}^{{-{y}}}{y}^{{z}}{\left.{d}{y}\right.}=\zeta{\left({z}\right)}\Gamma{\left({z}\right)}}}}\)
\(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{1}}}{{{y}^{{2}}}}}{e}^{{-{y}}}{y}^{{z}}{\left.{d}{y}\right.}=\Gamma{\left({z}-{1}\right)}\)
Putting the bits together:
\(\displaystyle{\int_{{0}}^{\infty}}{\left({\frac{{{1}}}{{{1}-{e}^{{-{y}}}}}}-{\frac{{{1}}}{{{y}}}}\right)}^{{2}}{e}^{{-{y}}}{y}^{{z}}{\left.{d}{y}\right.}\)
\(\displaystyle=\Gamma{\left({z}\right)}{\left({\frac{{{1}}}{{{z}-{1}}}}+{\left({z}-{2}\right)}\zeta{\left({z}\right)}\right)}\)
The derivation was valid for \(\displaystyle{R}{e}{\left({z}\right)}{>}{1}\), but both sides of the above equation are analytic for \(\displaystyle{R}{e}{\left({z}\right)}\succ{1}\) (RHS has a removable singularity at \(\displaystyle{z}={0}\)), so by analytic continuation we may take the limit \(\displaystyle{z}\to{0}\). Near \(\displaystyle{z}={0},\Gamma{\left({z}\right)}=\frac{{1}}{{z}}+{O}{\left({1}\right)}\), and the final answer is the derivative of \(\displaystyle\frac{{1}}{{{z}-{1}}}+{\left({z}-{2}\right)}\zeta{\left({z}\right)}\) at o, i.e.,
\(\displaystyle-{1}+\zeta{\left({0}\right)}-{2}\zeta'{\left({0}\right)}={\log{{\left({2}\pi\right)}}}-{\frac{{{3}}}{{{2}}}}\)
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nick1337
Answered 2022-01-11 Author has 9672 answers

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 \(1/\log x\) in the integral. The first piece is to let \(x=e^{-y}\); the integral becomes
\(\int_0^\infty dye^{-y}(\frac{(e^{-y}-(1-y))^2}{y^2(1-e^{-y})^2})\)
Now Taylor expand the factor \((1-e^{-y})^{-2}\), and if we can reverse the order of summation and integration, we get:
\(\sum_{k=1}^\infty k\int_0^\infty dy(\frac{e^{-y}-(1-y))^2}{y^2})e^{-ky}\)
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
\(I(\alpha)=\int_0^\infty dy(\frac{(e^{-y}-(1-y))^2}{y^2})e^{-\alpha y} \\\frac{d^2 I}{d\alpha^2}=\int_0^\infty dy(e^{-y}-(1-y))^2e^{-\alpha y} \\=\frac{1}{\alpha+2}-\frac{2}{\alpha+1}+\frac{2}{(\alpha+1)^2}+\frac{1}{\alpha}-\frac{2}{\alpha^2}+\frac{2}{\alpha^3}\)
You integrate this twice to recover \(\alpha\); the constants of integration may be shown to vanish by considering the limit as \(a\to\infty\). The original integral is then
\(\sum_{k=1}^\infty kI(k)\)
where
\(I(k)=(k+2)\log[\frac{k(k+2)}{(k+1)^2}]+\frac{1}{k}\)
so the integral takes on the value
\(\sum_{k=1}^\infty[1+[(k+1)^2-1]\log(1-\frac{1}{(k+1)^2})]=\sum_{k=1}^\infty[1+(k+1)^2\log(1-\frac{1}{(k+1)^2})]+\log2\)
The sum may be simplified by Taylor expanding the \(\log\) term; note that the unit value cancels and we get that the integral equals
\(\log2+\sum_{k=2}^\infty[1-k^2\sum_{m=1}^\infty\frac{1}{m}(\frac{1}{k^2})^m]=\log2-\sum_{m=1}^\infty\frac{1}{m+1}[\zeta(2m)-1]\)
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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