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+π√3ln(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+1dx⏟J
J=∫∞1ln(x+1)x3+1dxw↦1x=∫10ln(w+1w)ww3+1dw=∫10ln(w+1)ww3+1dw−∫10ln(w)ww3+1dw
Hence
I=∫10ln(x+1)x3+1dx+∫10ln(w+1)ww3+1dw−∫10ln(w)ww3+1dw=∫10ln(w+1)(w+1)1+w3dw−∫10ln(w)ww3+1dw=∫10ln(w+1)w2−w+1dw⏟K−∫10ln(w)ww3+1dw⏟L
Note that
ℑ(2√31w−12−i√32)=1w2−w+1
Hence
K=∫10ln(w+1)w2−w+1dw=ℑ(2√3∫10ln(w+1)w−12−i√32dw)=ℑ(2√3∫12−i√32−(12+i√32)ln(t+32+i√32)tdt)=ℑ(2√3∫12−i√32−(12+i√32)ln((32+i√32)(t32+i√32+1))tdt)=ℑ(2ln(32+i√32)√3∫12−i√32−(12+i√32)1tdt+2√3∫12−i√32−(12+i√32)ln(t32+i√32+1)tdt)=ℑ(2ln(32+i√32)√3(ln(12−i√32))−ln(−12−i√32))+ℑ(2√3∫12−i√32−(12+i√32)ln(t32+i√32+1)tdt)=πln(3)3√3+ℑ(2√3∫i√312+i2√3ln(1−s)sds)(s↦−t32+i√32)=πln(3)3√3−2√3ℑ(Li2(i√3)−Li2(12+i2√3))
The polylogarithm function satisfies:
ℑ(Li2(i√3)−Li2(12+i2√3))=−π218√3+112√3ψ(1)(13)
Hence
K=πln(3)3√3+π227−118ψ(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+∫10∞∑n=0(−1)nw3nwt+1dw=ddt|t=0+∞∑n=0(−1)n∫10wt+1+3ndw=ddt|t=0+∞∑n=0(−1)n1(3n+2+t)=∞∑n=0(−1)n+1(3n+2)2=∞∑n=01(6n+5)2−∞∑n=01(6n+2)2=136∞∑n=01(n+56)2−136∞∑n=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=K−L=πln(3)3√3+π227−136ψ(1)(13)−136ψ(1)(56)
Now using the identity
ψ(1)(13)+ψ(1)(56)=4ψ(1)(23)
and the reflection formula
ψ(1)(1−z)+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+π√3ln(3)+ψ(1)(13)]
No comments:
Post a Comment