David Lewis

Answered

2022-01-06

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

Answer & Explanation

Jeffery Autrey

Expert

2022-01-07Added 35 answers

Here is an another approach:

Let I denote the integral. By the substitution$x={e}^{-t}$ , we have

$I={\int}_{0}^{\mathrm{\infty}}[\frac{1}{{(1-{e}^{-t})}^{2}}-\frac{2}{t(1-{e}^{-t})}+\frac{1}{{t}^{2}}]{e}^{-t}dt$

$={\int}_{0}^{\mathrm{\infty}}[\frac{{e}^{t}}{{({e}^{t}-1)}^{2}}-\frac{1}{{t}^{2}}]dt+{\int}_{0}^{\mathrm{\infty}}[\frac{1+{e}^{-t}}{{t}^{2}}-\frac{2}{t({e}^{t}-1)}]dt$

It is easy to observe that the first integral is

$\int}_{0}^{\mathrm{\infty}}[\frac{{e}^{t}}{{({e}^{t}-1)}^{2}}-\frac{1}{{t}^{2}}]dt={[\frac{1}{t}-\frac{1}{{e}^{t}-1}]}_{0}^{\mathrm{\infty}}=-\frac{12}{$

We thus focus on the second integral. Associated to it, we introduce

$F\left(s\right)={\int}_{0}^{\mathrm{\infty}}[\frac{1+{e}^{-t}}{{t}^{2}}-\frac{2}{t({e}^{t}-1)}]{e}^{-st}dt$

By the twice differentiation, we have

$F{}^{\u2033}\left(s\right)={\int}_{0}^{\mathrm{\infty}}[1+{e}^{-t}-\frac{2t}{{e}^{t}-1}]{e}^{-st}dt$

$=\frac{1}{s}+\frac{1}{s+1}-2\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{(n+s)}^{2}}$

$=\frac{1}{s}+\frac{1}{s+1}-2{\psi}_{1}(s+1)$

Integrating and using the condition${F}^{\prime}(+\mathrm{\infty})=0$ , we have

${F}^{\prime}\left(s\right)=\mathrm{log}s+\mathrm{log}(s+1)-2{\psi}_{0}(s+1)$

Here we used the estimate${\psi}_{0}\left(s\right)\sim \mathrm{log}s$ as $s\to \mathrm{\infty}$ . Integrating again, we have

Let I denote the integral. By the substitution

It is easy to observe that the first integral is

We thus focus on the second integral. Associated to it, we introduce

By the twice differentiation, we have

Integrating and using the condition

Here we used the estimate

0

Suhadolahbb

Expert

2022-01-08Added 32 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$x={e}^{-y}$ to get

${\int}_{0}^{\mathrm{\infty}}\left(\frac{1}{1-{e}^{-y}}\right)-\frac{1}{y}{)}^{2}{e}^{-y}dy$

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)

${\int}_{0}^{\mathrm{\infty}}\frac{1}{{(1-{e}^{-y})}^{2}}{e}^{-y}{y}^{z}dy=z{\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-y}}{1-{e}^{-y}}{y}^{z-1}dy$

$=z\zeta \left(z\right)\mathrm{\Gamma}\left(z\right)$

and

$\int}_{0}^{\mathrm{\infty}}\frac{1}{y(1-{e}^{-y}\}{e}^{-y}{y}^{z}dy=\zeta \left(z\right)\mathrm{\Gamma}\left(z\right)$

${\int}_{0}^{\mathrm{\infty}}\frac{1}{{y}^{2}}{e}^{-y}{y}^{z}dy=\mathrm{\Gamma}(z-1)$

Putting the bits together:

${\int}_{0}^{\mathrm{\infty}}{(\frac{1}{1-{e}^{-y}}-\frac{1}{y})}^{2}{e}^{-y}{y}^{z}dy$

$=\mathrm{\Gamma}\left(z\right)(\frac{1}{z-1}+(z-2)\zeta \left(z\right))$

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 $\frac{1}{z-1}+(z-2)\zeta \left(z\right)$ at o, i.e.,

$-1+\zeta \left(0\right)-2{\zeta}^{\prime}\left(0\right)=\mathrm{log}\left(2\pi \right)-\frac{3}{2}$

First substitute

The integrand is awkward because when you multiply it out, the pieces separately diverge. Multiplying it by

and

Putting the bits together:

The derivation was valid for

nick1337

Expert

2022-01-11Added 573 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

Now Taylor expand the factor

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

You integrate this twice to recover

where

so the integral takes on the value

The sum may be simplified by Taylor expanding the

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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