What is the proof of the following: ∫01(ln⁡t1−t)2dt=π23 ?

Miguel Reynolds

Miguel Reynolds

Answered

2022-01-05

What is the proof of the following:
01(lnt1t)2dt=π23 ?

Answer & Explanation

yotaniwc

yotaniwc

Expert

2022-01-06Added 34 answers

This is by no means a complete solution but a possible route.
Letting t=1x note that
I=01(lnt1t)2dt
=1(lnt1t)2dt
=0(ln(1+t)t)2dt
Setting 1+t=ex, we get
I=0x2(ex1)2exdx
=0x2(ex2ex2)2dx
=2x2sinh2(x)dx
The last integral can be done by the method of residues to get
x2sinh2(x)dx=π26
I will fill this in once I get back home.
encolatgehu

encolatgehu

Expert

2022-01-07Added 27 answers

01(logt1t)2=0x2ex(ex1)2dx
Heres
star233

star233

Expert

2022-01-11Added 238 answers

Consider
k=0tk=11t (1)
Differentiating (1) with respect to t yields
k=1ktk1=1(1t)2 (2)
Multiplying both sides of (2) by ln2t and integrating from t=0 to t=1 yields
01(lnt1t)2dt=01k=1ktk1ln2tdt
=k=1k01tk1ln2tdt (3)
Using formula
01xαlnnxdx=(1)nn!(α+1)n+1, for n=0,1,2,... (4)
then (3) becomes
01(lnt1t)2dt=2k=11k2
=π23
where k=11k2=ζ(2)=π26

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