Hello Brilliants, here I wish to share my solution for the proposed problem.
Before we start with the problem we shall consider the integral of the form for all n≥1I(n)=∫01xn−1ln(1−x)dx=n1∫01dxd(xn−1)ln(1−x)dx and by integration by parts we yield
I(n)=n1⎣⎡0(xn−1)ln(1−x)⎦⎤01−n1∫011−x1−xndx=−n1∫01j=1∑nxj−1=−n1j=1∑n∫01xj−1dx=−n1j=1∑nj1=−nHn⋯(1)
Note that D(k)=n2m+m2n+kmn1=mn1∫01xm+n+k−1dx=mnxk−1∫01xm+ndx and therefore 1≤n≤m≤∞∑D(k)=∫01xk−1n=0∑∞m=0∑∞nxn⋅mxmdx=∫01xk−1ln2(1−x)dx Now we consider the latter integral I(k)=∫01xkln2(1−x)dx for k≥0 which further can be written as k+11∫01dxd(xk+1−1)ln2(1−x)dx and hence on Integration by parts we see that I(k)=k+11⎣⎡0(xk+1−1)ln2(1−x)⎦⎤01−k+12∫011−x1−xk+1ln(1−x)dx=−k+12∫01n=1∑k+1xn−1ln(1−x)=−k+12n=1∑k+1∫01xn−1ln(1−x)dx Plugging the result from (1) to last integral we have k+12n=1∑k+1nHn=k+12(2Hk+12+Hk+1(2))=k+1Hk+12+Hk+1(2) Further note that nth partial sum of Hk+1(2)=ζ(2)−ψ1(k+2) giving us required result of D(k)=k+1Hk+12−ψ1(k+2)+6(k+1)π2
Now we notice that
(q+b)q+b+1+(q+b+1)q+b1=(q+b)q+b+1(−1)(q+b)q+b+1−(q+b+1)q+b=q+b1−q+b+11 and hence n=0∑∞q=0∑nxn⎝⎜⎜⎜⎛Telescopingq+b1−q+b+11⎠⎟⎟⎟⎞=n=0∑∞xn(b1−n+b+11) and by hence by algebra of sum and for for ∣x∣<1n=0∑∞bxn−n=0∑∞n+b+1xn=b(1−x)1−Φ(x,21,1) and we deduce that α=2,β=1,γ=6. Thus Φ(β−α,β,(α+2βγ)−1)=Φ(−1,1,101)=n=0∑∞(n+101)(−1)n=DD=n=0∑∞(2n+10−1)(2n+10−1+1)1=41n=0∑∞(n+20−1)(n+20−1+2−1)1=2−1(ψ(21+101)−ψ(201))=2−1(ψ(101)−2ψ(201)−2ln2)and finally we have D=ψ(101)−ψ(201)−ln2. Here to evaluate this expression D , we use the Gauss's-Diagamma theorem ie ψ(mr)=−γ−ln(2m)−2πcot(mrπ)+2q=1∑⌊2m−1⌋cos(m2πqr)lnsin(nπq) plugging r=1,m=10,20 we have
I=ψ(101)=−γ−ln(20)−2πcot(10π)+2q=1∑4cos(5πq)lnsin(10πq) and K=ψ(201)=−γ−ln(40)−2πcot(20π)+2q=1∑9cos(10πq)lnsin(20πq) and also D=I−k−ln2 and on further simplification of the D we obtain that D=4πsin(201)cos(201)1+A(logcot(20π))+B(logcot(203π)) with A=25+5=ϕ+2 and B=25−5=2−110−25 and 4πsin(20π)cos(20π)1=2πsin(18∘)1=5−12π since sin(18∘)=4−1(5−1).
Further we have cot(20π)=cot(9∘). To evaluate cot(9∘ we shall be using the fact that
cot(3x)=sin(3x)cos(3x)=3sinx−4sin3x4cos3x−3cosx=sinx(3−4sin2x)cosx(1−4sin2x)=cotx(1−3−4sin2x2) setting x=3 we have the expression for cot9∘ but then we will show that sin3∘=418−10−25−15−3
{Note}: This identity is proposed by Narendra Bhandari and is proved by Sergio Esteban beautiful geometry work) and Ahmed Hegazi(trigonometry work).
From half angle formula we have that sin2ϕ=21−cosϕ now set ϕ=6∘. Giving us sin3∘=21−cos6∘=21−cos(36∘−30∘) Now using compound angle formula we can get cos(36∘−30∘)=cos36∘cos30∘+sin36∘sin30∘=(45+1)23+(410−25)21=810−25+3+15putting in original equation sin3∘=21⎝⎛1−810−25+3+15⎠⎞=41(8−10−25−3−15)
and hence cos3∘=1−sin23∘=418+10−25+10+3 and hence we deduce easily that cot9∘=cot3∘(1−4+10−25+15+38) also cot(203π)=cot(27∘=cot(30−3) and using compound angle formula for cot(a−b) we have cot27∘=cot3∘+cot30∘cot3∘cot30∘−1=cot3∘+33cot3∘−1 and hence we have D=Φ(β−α,β,(α+2β+γ)−1=5−12π+ϕ+2log(θ−4+10−25+15+38θ)+2−110−25log(θ+33θ−1) and thus θ=8−10−25−15−38+10−25+15+3we are done
Other different approaches are highly appreciated. I wish to be corrected if any such typos or errors are spotted in the above solution.
Thank you!
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Comments
I don't know calculus but I like your solutions (both of them!)
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I'm glad to know, you like it.Thank you :)
Man I didn't understood the step where you pull out the x to the k-1 variable out from the integral.