The right sign is a matter of luck.
Wolfgang Pauli
The main goal of this notebook is to solve this integral:
∞
∫
0
log(x+1)
2
x

2
log
(x)+
2
π

x
​It will also be necessary to solve this integral to achieve the main goal:
∞
∫
0
1
(x+1-u)
2
log
(x)+
2
π

x
​As a secondary objective this integral is solved for
n∈
:
∞
∫
0
1
n
(1+x)

2
log
(x)+
2
π

x
To get these and more integrals and proofs, you can look here:
​https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R(x,log)
Some of these integrals can be solved directly with Mathematica (LightBlue), some can be solved but the answer ist not the simplest possible (LightPurple), some can not be solved (LightOrange) but can be checked numerically and some can not be solved and not even be checked numerically although a solution is known (LightRed).
This Notebook is about the LightOrange and LightRed cases:
Out[]=
(0.1)
1
∫
0
log(2)-log(1+x)
1+
2
x
x
1
8
πlog(2)≈0.272198
Out[]=
(0.2)
1
∫
0
log(2)-log(1+x)
1-
2
x
x
2
π
24
≈0.411234
Out[]=
(0.3)
∞
∫
1
log(1+x)-log(2)
1+
2
x
xC-
1
8
πlog(2)​≈0.643767
Out[]=
(0.4)
∞
∫
1
log(2)-log(1+x)
1-
2
x
x
2
π
12
≈0.822467
Out[]=
(0.5)
1
∫
0
-
log(x)
1+
2
x
xC≈0.822467
Out[]=
(0.6)
1
∫
0
-
log(x)
1-
2
x
x
2
π
8
≈0.822467
Out[]=
(0.7)
∞
∫
0
log(1+x+
2
x
)
1+
2
x
x
4C
3
+
1
3
πlog(2+
3
)≈2.6004
Mathematica gives the same answer in disguise:
Out[]=
-
1
24
πArcCosh[2]+6Log[7-4
3
]+
1
2
PolyLog2,1-
1/6
(-1)
-PolyLog2,
1
2
((2+)-
3
)-PolyLog2,
2
(2-)+
3
+PolyLog2,
2
(2+)+
3
≈2.6004
Out[]=
(0.8)
1
∫
0
-log(-log(x))x≈0.577216
Out[]=
(0.9)
1
∫
0
2log(x)
2
x
-4x+8
-
3log(x)
2
x
+2x+2
xC≈0.915966
Mathematica gives the same answer in disguise:
Out[]=
1
2
3PolyLog2,-
1
2
-

2
-3PolyLog2,-
1
2
+

2
-PolyLog2,
1
4
-

4
+PolyLog2,
1
4
+

4
≈0.915966
This integral is to hard to solve (but still the answer is known), and it is even to hard to be computed numerically:
Out[]=
(0.10)
∞
∫
0
log(x+1)
2
x

2
log
(x)+
2
π

x≈0.577216
This integral is to hard to solve (but still the answer is known).
​
F
n
are Fontana numbers (Gregory’s coefficients)
n∈
:
Out[]=
(1.4)
∞
∫
0
1
n
(1+x)

2
log
(x)+
2
π

x
F
n
Mathematica can solve the case
n=1
:
Out[]=
(​
*
1.4
​)
∞
∫
0
1
(1+x)
2
log
(x)+
2
π

x
1
2
≈0.5
… and for example
n=5
gives:
Out[]=
(​
*
1.4
​)
∞
∫
0
1
5
(1+x)

2
log
(x)+
2
π

x
3
160
≈0.01875
This can be checked numerically:
In[]:=
NIntegrate
1
5
(1+x)

2
Log[x]
+
2
π

,{x,0,∞}
Out[]=
0.01875
It’s all about the sum of the residues (there is only one residue here):
Out[]=
∑ Res =
3
160
The Fontana numbers (Gregory’s coefficients) are discussed in the Addendum:
In[]:=
Fontana[5]
Out[]=
3
160
The integration path is like that:
Out[]=
https://en.wikipedia.org/wiki/Branch_point​​The typical example of a branch cut is the complex logarithm. If a complex number is represented in polar form
z=r
ϑ

, then the logarithm of
z
is
Out[]=
log(z)log(r)+ϑlog(z)+ϑ
Let
z=x+
+
0
be a point on
γ
1
, then
1
log(-z)
=
1
log(-x-
+
0
)
=
1
log(x)-π
.Let
z=x-
+
0
be a point on
γ
3
, then
1
log(-z)
=
1
log(-x+
+
0
)
=
1
log(x)+π
.
And we use this property of log for
x>0
, i.e.
x∈
+

:
In[]:=
With[{n=3},{Limit[1/Log[-(x+ε)],{x->n,ε->0},Direction->"FromAbove"],Limit[1/Log[-(x-ε)],{x->n,ε->0},Direction->"FromAbove"]}]
Out[]=

1
-π+Log[3]
,
1
π+Log[3]

And the minus in the next expression is because
γ
3
is integrated from ∞ to 0: ​
∫
γ
3
f(z)z=
0
∫
∞
f(x)x=-
∞
∫
0
f(x)x
And amazingly
x∈
+

:
Out[]=
1
log(x)-π
-
1
log(x)+π

2π
2
log
(x)+
2
π
The integral over the path
γ
2
and
γ
4
go to zero, and the rest of the proof is as follows:
Out[]=
∫
γ
1
f(z)z+
∫
γ
3
f(z)z =
∫
γ
1
1
n
(1+z)
log(-z)
z+
∫
γ
3
1
n
(1+z)
log(-z)
z =
∞
∫
0
1
n
(1+x)
(log(x)-π)
x +
0
∫
∞
1
n
(1+x)
(log(x)+π)
x =
∞
∫
0
1
n
(1+x)
(log(x)-π)
x -
∞
∫
0
1
n
(1+x)
(log(x)+π)
x =
∞
∫
0
2π
n
(1+x)

2
log
(x)+
2
π

x = 2πres
-n
(z+1)
log(-z)
,{z,-1} = 2π
F
n
This integral is to hard (but still the answer is known):
Out[]=
(1.5)
∞
∫
0
1
(x+1-u)
2
log
(x)+
2
π

x
1
u
+
1
log(1-u)
; u∈
×

​
≥1

What does
×

mean?
​https://math.stackexchange.com/questions/4625250/difference-between-mathbbc-times-and-mathbbc
The integration path is the same as above:
Out[]=
It’s all about the sum of the residues (there are only two here):

Proof of (0.10)

First observe that:
Than replace the integrand by (1.5):
Than swap the integrals:
And observe that:
And inserting (4) into (3) is the final result:

CITE THIS NOTEBOOK

A hard integral to solve part two​
by Oliver Seipel​
Wolfram Community, STAFF PICKS, January 2, 2023
​https://community.wolfram.com/groups/-/m/t/3094086