Saturday, January 22, 2022

Integral of the day XX

Apéry's constant

Nice integral representation of the Apéry's constant ζ(3)


Today we show the proof of the following integral representation of the Aperys constant posted by @sounansya_29 327π40xarctanh(tanx)dx=ζ(3)
The proof relies on some properties of the arctanh and arctan functions and Fourier series.

Proof

Recall the definition in terms of logarithms of arctanh(z) arctanh(z)=12ln(1+z1z)
Therefore I=π40xarctanh(tanx)dx=12π40xln(1+tanx1tanx)dx=121arctan(w1w+1)ln(w)1+w2dw(w1+tanx1tanx)
Recall the following addition formula for arctan(x): arctan(w)π4=arctan(w1w+1)
Hence I=π40xarctanh(tanx)dx=121arctan(w1w+1)ln(w)1+w2dw=121(arctan(w)π4)ln(w)1+w2dw=121arctan(w)ln(w)1+w2dwJπ81ln(w)1+w2dwK
K=1ln(w)1+w2dw=10ln(s)1+s2ds(s1w)=10n=0(1)ns2nln(s)ds=n=0(1)n10s2n(ddt|t=0+st)ds=ddt|t=0+n=0(1)n10s2n+tds=ddt|t=0+n=0(1)n2n+t+1ds=n=0[ddt|t=0+(1)n2n+t+1]ds=n=0(1)n(2n+1)2=β(2)
J=1arctan(w)ln(w)1+w2dw=10arctan(1s)ln(1s)1+s2ds(w1s)=10(arctan(s)+π2)ln(s)1+s2(arctan(1b)=arctan(b)sign(b)π2)=10arctan(s)ln(s)1+s2dsπ210ln(s)1+s2ds=10arctan(s)ln(s)1+s2ds+π2β(2)=π40tln(tan(t))dt+π2β(2)=π40tln(cot(t))dt+π2β(2)
Recall the following Fourier series k=1cos((2k1)x)2k1=12ln(cotx2)
Hence J=π40tln(cot(t))dt+π2β(2)=2π40tk=1cos((2k1)2t)2k1dt+π2β(2)=k=122k1π40tcos((2k1)2t)dt+π2β(2)=12k=11(2k1)3π(2k1)20ycos(y)dy+π2β(2)IBP=12k=11(2k1)3[sin(y)y|π(2k1)20π8(2k1)0sin(y)dy]+π2β(2)=12k=11(2k1)3[sin(π(2k1)2)π(2k1)2+cos(π(2k1)2)1]+π2β(2)=π4k=1sin(π(2k1)2)(2k1)212k=1cos(π(2k1)2)(2k1)3+12k=11(2k1)3+π2β(2)=π4k=1cos(πk)(2k1)212k=1sin(πk)(2k1)3+12λ(3)+π2β(2)
Note that (sin(πk))kN=(0,0,0,0,...)
(cos(πk))kN=(1,1,1,1,1,..)
Therefore J=π4k=1(1)k(2k1)2+12λ(3)+π2β(2)=π4β(2)+12λ(3)
From the identity λ(v)=(12v)ζ(v)
we have J=1arctan(w)ln(w)1+w2dw=716ζ(3)+π4β(2)
Hence I=π40xarctanh(tanx)dx=12Jπ8K=732ζ(3)π8β(2)+π8β(2)
Therefore, we can conclude 327π40xarctanh(tanx)dx=ζ(3)

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