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

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

yotaniwc · Accepted Answer

This is by no means a complete solution but a possible route.  Letting t=1x note that  I=∫01(ln⁡t1−t)2dt  =∫1∞(ln⁡t1−t)2dt  =∫0∞(ln⁡(1+t)t)2dt  Setting 1+t=ex, we get  I=∫0∞x2(ex−1)2exdx  =∫0∞x2(ex2−e−x2)2dx  =2∫−∞∞x2sinh2(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 · Accepted Answer

∫01(log⁡t1−t)2=∫0∞x2ex(ex−1)2dx  Heres

star233 · Accepted Answer

Consider ∑k=0∞tk=11−t (1) Differentiating (1) with respect to t yields ∑k=1∞ktk−1=1(1−t)2 (2) Multiplying both sides of (2) by ln2⁡t and integrating from t=0 to t=1 yields ∫01(ln⁡t1−t)2dt=∫01∑k=1∞ktk−1ln2⁡tdt =∑k=1∞k∫01tk−1ln2⁡tdt (3) Using formula ∫01xαlnn⁡xdx=(−1)nn!(α+1)n+1, for n=0,1,2,... (4) then (3) becomes ∫01(ln⁡t1−t)2dt=2∑k=1∞1k2 =π23 where ∑k=1∞1k2=ζ(2)=π26

What is the proof of the following: \int_0^1(\frac{\ln t}{1-t})^2dt=\frac{\pi^2}{3} ?

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New Questions in Calculus 1