Evaluate ∫01(1ln⁡x+11−x)2dx

Question

Jeffery Autrey · Accepted Answer

Here is an another approach:Let I denote the integral. By the substitution x=e−t, we haveI=∫0∞[1(1−e−t)2−2t(1−e−t)+1t2]e−tdt=∫0∞[et(et−1)2−1t2]dt+∫0∞[1+e−tt2−2t(et−1)]dtIt is easy to observe that the first integral is∫0∞[et(et−1)2−1t2]dt=[1t−1et−1]0∞=−12We thus focus on the second integral. Associated to it, we introduceF(s)=∫0∞[1+e−tt2−2t(et−1)]e−stdtBy the twice differentiation, we haveF″(s)=∫0∞[1+e−t−2tet−1]e−stdt=1s+1s+1−2∑n=1∞1(n+s)2=1s+1s+1−2ψ1(s+1)Integrating and using the condition F′(+∞)=0, we haveF′(s)=log⁡s+log⁡(s+1)−2ψ0(s+1)Here we used the estimate ψ0(s)∼log⁡s as s→∞. Integrating again, we have

Suhadolahbb · Accepted Answer

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}^{\infty} (\frac{1}{1 - e^{- y}}) - \frac{1}{y})^{2} e^{- y} d y

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

R e (z) > 1

we have (by parts)

\int_{0}^{\infty} \frac{1}{{(1 - e^{- y})}^{2}} e^{- y} y^{z} d y = z \int_{0}^{\infty} \frac{e^{- y}}{1 - e^{- y}} y^{z - 1} d y

= z ζ (z) Γ (z)

and

\int_{0}^{\infty} \frac{1}{y (1 - e^{- y}} e^{- y} y^{z} d y = ζ (z) Γ (z)}

\int_{0}^{\infty} \frac{1}{y^{2}} e^{- y} y^{z} d y = Γ (z - 1)

Putting the bits together:

\int_{0}^{\infty} {(\frac{1}{1 - e^{- y}} - \frac{1}{y})}^{2} e^{- y} y^{z} d y

= Γ (z) (\frac{1}{z - 1} + (z - 2) ζ (z))

The derivation was valid for

R e (z) > 1

, but both sides of the above equation are analytic for

R e (z) ≻ 1

(RHS has a removable singularity at

z = 0

), so by analytic continuation we may take the limit

z \to 0

. Near

z = 0, Γ (z) = \frac{1}{z} + O (1)

, and the final answer is the derivative of

\frac{1}{z - 1} + (z - 2) ζ (z)

at o, i.e.,

- 1 + ζ (0) - 2 ζ^{'} (0) = \log (2 π) - \frac{3}{2}

nick1337 · Accepted Answer

OK, I&#039;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 ∫0∞dye−y((e−y−(1−y))2y2(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: ∑k=1∞k∫0∞dy(e−y−(1−y))2y2)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 α and differentiate with respect to α inside the integral twice (to clear the pesky y2 in the denominator) to get a function I(α)=∫0∞dy((e−y−(1−y))2y2)e−αyd2Idα2=∫0∞dy(e−y−(1−y))2e−αy=1α+2−2α+1+2(α+1)2+1α−2α2+2α3 You integrate this twice to recover α; the constants of integration may be shown to vanish by considering the limit as a→∞. The original integral is then ∑k=1∞kI(k) where I(k)=(k+2)log⁡[k(k+2)(k+1)2]+1k so the integral takes on the value ∑k=1∞[1+[(k+1)2−1]log⁡(1−1(k+1)2)]=∑k=1∞[1+(k+1)2log⁡(1−1(k+1)2)]+log⁡2 The sum may be simplified by Taylor expanding the log term; note that the unit value cancels and we get that the integral equals log⁡2+∑k=2∞[1−k2∑m=1∞1m(1k2)m]=log⁡2−∑m=1∞1m+1[ζ(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.

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

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Answer & Explanation

New Questions in Calculus 1