Prove that: ∫0π2ln2(cos⁡x)dx=π2ln22+π324

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Prove that:  ∫0π2ln2(cos⁡x)dx=π2ln22+π324

soanooooo40 · Accepted Answer

Let&#039;s get all powers of ln⁡(cos⁡(x)) at once, using an exponential generating function:G(z)=∫0π2∑j=0∞lnj(cos⁡(x))zjj!dx=∫0π2exp⁡(zln⁡(cos⁡(x)))dxChange variables: cos⁡(x)=t12, and using the Beta function:12∫01tz−121−tdt=12B(12,z+12)=Γ(12)Γ(z+12)2Γ(1+z2)=π2−πln⁡(2)2z+(π348+π(ln⁡(2))24)z2−π(4(ln⁡(2))3+π2ln⁡(2)+6ζ(3))48z3+…

lenkiklisg7 · Accepted Answer

Have a look at this other question. We have that log⁡(2cos⁡x) has a nice Fourier series:  log⁡(2cos⁡x)=∑n=1+∞(−1)n+1ncos⁡(2nx)  and since:  ∫0π2cos⁡(2nx)cos⁡(2mx)dx=π4δm,n  in follows that:  ∫0π2log2(2cos⁡x)dx=π4∑n=1+∞1n2=π4ζ(2)=π324 (1)  while  ∫0π2log⁡(cos⁡x)dx=−π2log⁡2 (2)  is a well-known result. (1) and (2) proves your claim:  ∫0π2ln2(cos⁡x)dx=π2ln22+π324  Since  log⁡(2sin⁡x)=−∑n=1+∞cos⁡(2nx)n  and ∫0π2cos⁡(2n1,x)cos⁡(2n2x)cos⁡(2n3x)dx  =π8δ2⋅max ni=(n1+n2+n3)  it follows that  ∫0π2log3(2sin⁡x)dx=−3π4∑n=1+∞Hn−1n2=−3π4ζ(3)

Vasquez · Accepted Answer

Let cos⁡x=t and dx=dt2t1−t, then ∫0π/2ln2⁡(cos⁡x)dx=18∫01ln2⁡tt1−tdt=18limx→12limy→12d2dx2∫01tx−1(1−t)y−1dt=18limx→12limy→12B(x,y)[(ψ(x)−ψ(x+y))2+ψ1(x)−ψ1(x+y)] where B(x,y) is beta function and ψk(z) is polygamma function. The same approach also works for int0π/2ln3⁡(cos⁡x)dx but for this one, we use third derivative of beta function.

Prove that: \int_0^{\pi/2}\ln^2(\cos x)dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}

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