Thursday, December 2, 2021

Integral of the day XI

Integral from Twitter

Integral involving the trigamma constant ψ(1)(13)


Today we show the proof of the following result proposed by @Ali3934213 0ln(x+1)x3+1dx=19[π2+π3ln(3)+ψ(1)(13)]
The proof relies on some properties of the polylogarithm and the trigamma function.

Proof

I=0ln(x+1)x3+1dx=10ln(x+1)x3+1dx+1ln(x+1)x3+1dxJ
J=1ln(x+1)x3+1dxw1x=10ln(w+1w)ww3+1dw=10ln(w+1)ww3+1dw10ln(w)ww3+1dw
Hence I=10ln(x+1)x3+1dx+10ln(w+1)ww3+1dw10ln(w)ww3+1dw=10ln(w+1)(w+1)1+w3dw10ln(w)ww3+1dw=10ln(w+1)w2w+1dwK10ln(w)ww3+1dwL
Note that (231w12i32)=1w2w+1
Hence K=10ln(w+1)w2w+1dw=(2310ln(w+1)w12i32dw)=(2312i32(12+i32)ln(t+32+i32)tdt)=(2312i32(12+i32)ln((32+i32)(t32+i32+1))tdt)=(2ln(32+i32)312i32(12+i32)1tdt+2312i32(12+i32)ln(t32+i32+1)tdt)=(2ln(32+i32)3(ln(12i32))ln(12i32))+(2312i32(12+i32)ln(t32+i32+1)tdt)=πln(3)33+(23i312+i23ln(1s)sds)(st32+i32)=πln(3)3323(Li2(i3)Li2(12+i23))
The polylogarithm function satisfies: (Li2(i3)Li2(12+i23))=π2183+1123ψ(1)(13)
Hence K=πln(3)33+π227118ψ(1)(13)
Now, for L; L=10ln(w)w1+w3dw=10(ddt|t=0+wt)w1+w3dw=ddt|t=0+10wt+11+w3dw=ddt|t=0+10n=0(1)nw3nwt+1dw=ddt|t=0+n=0(1)n10wt+1+3ndw=ddt|t=0+n=0(1)n1(3n+2+t)=n=0(1)n+1(3n+2)2=n=01(6n+5)2n=01(6n+2)2=136n=01(n+56)2136n=01(n+13)2
Recall the series representation of the polygamma function: ψ(m)(z)=(1)m+1m!k=01(z+k)m+1
Hence L=136[ψ(1)(56)ψ(1)(13)]
Therefore I=KL=πln(3)33+π227136ψ(1)(13)136ψ(1)(56)
Now using the identity ψ(1)(13)+ψ(1)(56)=4ψ(1)(23)
and the reflection formula ψ(1)(1z)+ψ(1)(z)=π2sin2(πz)
with z=23 we have ψ(1)(13)+ψ(1)(23)=43π2
Therefore 0ln(x+1)x3+1dx=19[π2+π3ln(3)+ψ(1)(13)]

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