Some time ago I came across to the following integral: I=∫011−x1+xdxln⁡x What are the hints on how to compute this integral?

Question

Some time ago I came across to the following integral:  I=∫011−x1+xdxln⁡x  What are the hints on how to compute this integral?

esfloravaou · Accepted Answer

Make the substitution x=e−y and do a little algebra to get the value of the integral to be  ∫0∞dyye−y−e−2y1+e−y  Now Taylor expand the denominator and get  ∫0∞dyy(e−y−e−2y)∑k=0∞(−1)ke−ky  If we can reverse the order of sum and integral, we get  ∑k=0∞(−1)k∫0∞dyy(e−(k+1)y−e−(k+2)y)  The integral inside may be evaluated exactly, and the result is the sum  ∑k=0∞(−1)klog⁡k+1k+2  =limn→∞log⁡1234…2n−12n2356…2n+12n+2  =log⁡(2π)

jgardner33v4 · Accepted Answer

I tried with Differentiation under integration sign:  J(α)=∫011−xα1+xdxln⁡x α&amp;gt;0  Then as usual,  dJdα=−∫01xα1+xdx=−f(α)  Then, integrating  J(α)=−∫f(α)dα+C  I used Mathematica to evaluate f(α) (it gives difference of two Harmonic numbers) and its integral. The integral is just ln⁡Γ(2m+12)Γ(m+12). Note that J(0)=0 and putting α=1, the result follows.

nick1337 · Accepted Answer

Consider I(α)=∫011−xα(1+x)ln⁡xdx (1) Differentiating (1) with respect α yields dIdα=∫01ddα[1−xα(1+x)ln⁡x]dx=−∫01xα1+xdx=−∫01∑k=0∞(−1)kxα+kdx=−∑k=0∞(−1)k∫01xα+kdx=−∑k=0∞(−1)kα+k+1 (2) Now consider polygamma function ψn(z)=dn+1dzn+1ln⁡Γ(z) (3) and ψn(z)=(−1)n+1n!∑k=0∞1(z+k)n+1 (4) Hence by using (4) we obtain ∑k=0∞(−1)k(z+k)n+1=1(−2)n+1n![ψn(z2)−ψn(z+12)] (5) Using (3) and (5) then (2) becomes dIdα=12[ψ0(α+12)−ψ0(α+22)]I(α)=12∫[ψ0(α+12)−ψ0(α+22)]dα=ln⁡Γ(α+12)−ln⁡Γ(α+22)+C=ln⁡Γ(α+12)−ln⁡Γ(α+22)+12ln⁡πwhere I(0)=0 and C=−ln⁡Γ(12)Thus∫011−x(1+x)ln⁡xdx=I(1)=ln⁡Γ(1)−ln⁡Γ(32)−12ln⁡π=ln⁡(2π)

Some time ago I came across to the following integral: I=\int_0^1\frac{1-x}{1+x}\frac{dx}{\ln

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New Questions in Calculus 1