On Mathoverflow, the question about a closed form for the integral
∞
∫
0
3
tanh
(x)
2
x
x
was raised. Currently, Mathematica can’t give a closed form answer for this integral.
In[]:=
Integrate[Tanh[x]^3/x^2,{x,0,∞}]
Out[]=
∞
∫
0
3
Tanh[x]
2
x
x
But we can easily get a numerical value.
In[]:=
num=NIntegrate[Tanh[x]^3/x^2,{x,0,∞},WorkingPrecision->100,PrecisionGoal->30]//N[#,30]&
Out[]=
1.15478531332317626405907045194
How could one find a closed form result? With Integrate, Sum, SeriesCoefficient, ... we have a lot of tools at our hands to do some experimental mathematics. A simple manual change of variables does not help.
In[]:=
Integrate[y^3/(ArcTanh[y]^2)Dt[ArcTanh[y],y],{y,0,1}]
Out[]=
1
∫
0
3
y
(1-
2
y
)
2
ArcTanh[y]
y
The next typical trick is expanding the integrand in a series and integrating term by term:
In[]:=
TrigToExp[Tanh[x]^3]
Out[]=
3
(-
-x

+
x

)
3
(
-x

+
x

)
In[]:=
SeriesCoefficient[%/.{Exp[-x]->g,Exp[x]->1/g},{g,0,n}]//FullSimplify[#,n∈Integers]&
Out[]=
(2+
2
n
)Cos
nπ
2

n>0
1
n0
0
True
The resulting integrals do diverge:
In[]:=
Integrate[Exp[-nx]/x^2,{x,0,∞}]
Integrate
:Integral of
-nx

2
x
does not converge on {0,∞}.
Out[]=
∞
∫
0
-nx

2
x
x
We can make the integral convergent by adding a factor. For the case
α=0
that we need, the integral diverges:
In[]:=
Integrate[Exp[-nx]/x^2x^α,{x,0,∞}]
Out[]=
1-α
n
Gamma[-1+α] if Re[n]>0&&Re[α]>1
And the sum can be done in closed form for general α:
In[]:=
Sum[n^(1-α)Gamma[-1+α]Cos[(nπ)/2](2+n^2),{n,∞}]
Out[]=
-
2-2α
2
Gamma[-1+α](-32Zeta[-3+α]+
1+α
2
Zeta[-3+α]-4Zeta[-1+α]+
α
2
Zeta[-1+α])
Expanding the result around
α=0
gives a finite leading term:
In[]:=
Series[%,{α,0,1}]
Out[]=
1
15
(-15-4Log[2]+180Log[Glaisher]-1800
′
Zeta
[-3])+-
1
60
(4(-1+EulerGamma)+8Log[2])(-15-4Log[2]+180Log[Glaisher]-1800
′
Zeta
[-3])+
1
15
5Log[2]-2
2
Log[2]
-60Log[2]Log[Glaisher]+120Log[2]
′
Zeta
[-3]-900
′′
Zeta
[-3]-90
′′
Zeta
[-1]α+
2
O[α]
And indeed this is the result we were looking for:
In[]:=
N[(-1-(4Log[2])/15+12Log[Glaisher]-120Derivative[1][Zeta][-3]),30]
Out[]=
1.15478531332317626405907045194
Alternatively, we can split of the divergent point from the integral:
In[]:=
Integrate[Exp[-nx]/x^2,{x,ϵ,∞}]
Out[]=
-nϵ

ϵ
-nGamma[0,nϵ] if Re[ϵ]>0&&Im[ϵ]0&&Re[n]>0
The leading terms are:
In[]:=
Series[(E^(-nϵ)/ϵ-nGamma[0,nϵ]),{ϵ,0,2},Assumptions->ϵ>0]
Out[]=
1
ϵ
+(-n+EulerGamman+nLog[n]+nLog[ϵ])-
2
n
ϵ
2
+
3
n
2
ϵ
12
+
3
O[ϵ]
The n term does give no sum contribution:
In[]:=
Sum[nCos[(nπ)/2](2+n^2),{n,∞},Regularization->"Dirichlet"]
Out[]=
0
And the
nln(n)
term gives the same result as above:
In[]:=
Sum[nLog[n]Cos[(nπ)/2](2+n^2)n^α,{n,∞}]
Out[]=
-
-4-α
π
-2
2
π
Cos
πα
2
Gamma[2+α]Log[2]Zeta[2+α]+
4+α
2
2
π
Cos
πα
2
Gamma[2+α]Log[2]Zeta[2+α]+Cos
πα
2
Gamma[4+α]Log[2]Zeta[4+α]-
5+α
2
Cos
πα
2
Gamma[4+α]Log[2]Zeta[4+α]-
3+α
2
4+α
π
′
Zeta
[-3-α]+
7+2α
2
4+α
π
′
Zeta
[-3-α]-
2+α
2
4+α
π
′
Zeta
[-1-α]+
4+2α
2
4+α
π
′
Zeta
[-1-α]
In[]:=
Series[%,{α,0,1}]
Out[]=
1
15
(-15-4Log[2]+180Log[Glaisher]-1800
′
Zeta
[-3])+
1
90
4
π
-79
4
π
Log[2]+24EulerGamma
4
π
Log[2]-48
4
π
2
Log[2]
+2520
4
π
Log[2]Log[Glaisher]+24
4
π
Log[2]Log[π]-22320
4
π
Log[2]
′
Zeta
[-3]-1260
2
π
Log[2]
′
Zeta
[2]+16740Log[2]
′
Zeta
[4]+10800
4
π
′′
Zeta
[-3]+1080
4
π
′′
Zeta
[-1]α+
2
O[α]
By conjecturing the constants involved in the integral, we could have used FindIntegerNullVector to get a conjecture for a closed form:
In[]:=
FindIntegerNullVector[{1.15478531332317626405907045194,1,Log[2],EulerGamma,Log[Glaisher],Derivative[1][Zeta][-3]}]
Out[]=
{-15,-15,-4,0,180,-1800}

CITE THIS NOTEBOOK

Don't fear divergencies:
∞
∫
0
3
tanh
(x)
2
x
x
​by Michael Trott​Wolfram Community, STAFF PICKS, 2017​https://community.wolfram.com/groups/-/m/t/1170493