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_29 327∫π40xarctanh(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+z1−z)
Therefore
I=∫π40xarctanh(tanx)dx=12∫π40xln(1+tanx1−tanx)dx=12∫∞1arctan(w−1w+1)ln(w)1+w2dw(w↦1+tanx1−tanx)
Recall the following addition formula for arctan(x):
arctan(w)−π4=arctan(w−1w+1)
Hence
I=∫π40xarctanh(tanx)dx=12∫∞1arctan(w−1w+1)ln(w)1+w2dw=12∫∞1(arctan(w)−π4)ln(w)1+w2dw=12∫∞1arctan(w)ln(w)1+w2dw⏟J−π8∫∞1ln(w)1+w2dw⏟K
K=∫∞1ln(w)1+w2dw=−∫10ln(s)1+s2ds(s↦1w)=−∫10∞∑n=0(−1)ns2nln(s)ds=−∞∑n=0(−1)n∫10s2n(ddt|t=0+st)ds=ddt|t=0+−∞∑n=0(−1)n∫10s2n+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(w↦1s)=−∫10(−arctan(s)+π2)ln(s)1+s2(arctan(1b)=−arctan(b)−sign(b)π2)=∫10arctan(s)ln(s)1+s2ds−π2∫10ln(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((2k−1)x)2k−1=12ln(cotx2)
Hence
J=−∫π40tln(cot(t))dt+π2β(2)=−2∫π40t∞∑k=1cos((2k−1)2t)2k−1dt+π2β(2)=−∞∑k=122k−1∫π40tcos((2k−1)2t)dt+π2β(2)=−12∞∑k=11(2k−1)3∫π(2k−1)20ycos(y)dy+π2β(2)IBP=−12∞∑k=11(2k−1)3[sin(y)y|π(2k−1)20−∫π8(2k−1)0sin(y)dy]+π2β(2)=−12∞∑k=11(2k−1)3[sin(π(2k−1)2)π(2k−1)2+cos(π(2k−1)2)−1]+π2β(2)=−π4∞∑k=1sin(π(2k−1)2)(2k−1)2−12∞∑k=1cos(π(2k−1)2)(2k−1)3+12∞∑k=11(2k−1)3+π2β(2)=π4∞∑k=1cos(πk)(2k−1)2−12∞∑k=1sin(πk)(2k−1)3+12λ(3)+π2β(2)
Note that
(sin(πk))k∈N=(0,0,0,0,...)
(cos(πk))k∈N=(−1,1,−1,1,−1,..)
Therefore
J=π4∞∑k=1(−1)k(2k−1)2+12λ(3)+π2β(2)=π4β(2)+12λ(3)
From the identity
λ(v)=(1−2−v)ζ(v)
we have
J=∫∞1arctan(w)ln(w)1+w2dw=716ζ(3)+π4β(2)
Hence
I=∫π40xarctanh(tanx)dx=12J−π8K=732ζ(3)−π8β(2)+π8β(2)
Therefore, we can conclude
327∫π40xarctanh(tanx)dx=ζ(3)
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