CITE THIS NOTEBOOK: Wolfram R&D LIVE: Ask Integration Questions to Oleg Marichev by Oleg Marichev. Wolfram Community JAN 31 2023.
Sometimes users ask how to evaluate “simple looking” integrals, for example from x^x. I am ready to talk about do-ability different integrals from mathematical functions.
https://youtu.be/Ye7ZORKPFak

Remarks about the do-ability of integrals

Operations with
{
α
z
,
z
a
,
z
z
}

In[]:=
D[{
α
z
,
z
a
,
z
z
},z]
Out[]=
{
-1+α
z
α,
z
a
Log[a],
z
z
(1+Log[z])}
In[]:=
Integrate[{
α
z
,
z
a
,
z
z
},z]
Out[]=

1+α
z
1+α
,
z
a
Log[a]
,
z
z
z
∫f[z]zIntegrate[f[z],z]
In[]:=

-
2
z

z,
z

z
z,
α-1
z
-z

z
Out[]=

1
2
π
Erf[z],ExpIntegralEi[z],-Gamma[α,z]
In[]:=
Daz,
z
a
,
a
z
,z
Out[]=
a,
1
a
,-
a
2
z

In[]:=
Integrateaz,
z
a
,
a
z
,z
Out[]=

a
2
z
2
,
2
z
2a
,aLog[z]

Integrals from the literature

An example integral used for calculating “Neutrino forces”
​https://arxiv.org/abs/2209.07082v1

Example usage for “Neutrino forces” with
1
∫
0
StruveH0,
1-
2
z
zz

Recently we found article https://arxiv.org/abs/2209.07082v1 about “Neutrino forces in neutrino backgrounds”, which on the page 14 mentioned that Mathematica can not evaluate one integral with the Struve H function. ​​
"We note that Mathematica contains some unidentified bugs leading to incorrect results of integrals involvingthe Struve H function,e.g.
1
∫
0
StruveH0,​
1-
2
z
zz should be nonzero while Integrate in Mathematica onlyproduces a vanishing result."​​
​In reality this integral after rather complicated procedures can be converted to value of StruveTransform from case of MeijerG function. It allows one to apply algorithm developed in my book of 1978 (translated in 1983, see https://en.wikipedia.org/wiki/Oleg_Marichev). As result we got formula
1
∫
0
StruveH0,
1-
2
z
zz
2
3π
HypergeometricPFQ{1,1},
5
4
,
3
2
,
7
4
,-
1
16

In[]:=
NIntegrateStruveH0,
1-
2
z
z,{z,0,1},
2
3π
HypergeometricPFQ{1,1},
5
4
,
3
2
,
7
4
,-
1
16
//N
Out[]=
{0.208197,0.208197}
In[]:=
Solve
1-
2
z
z==t,z
Solve
:There may be values of the parameters for which some or all solutions are not valid.
Out[]=
z-
1-
1-4
2
t
2
,z
1-
1-4
2
t
2
,z-
1+
1-4
2
t
2
,z
1+
1-4
2
t
2

In[]:=
D
1-
1-4
2
t
2
,tStruveH0,
1-
2
z
z/.z
1-
1-4
2
t
2
//Simplify
Out[]=
tStruveH0,
2
t

1
2
-2
2
t
1-
1-4
2
t

Action-angle integrals
https://mathoverflow.net/questions/96182/action-integral

Kepler potential integral

For the Kepler potential, this integral is
In[]:=
Integrate
(z-a)(b-z)
z
,{z,a,b}
Out[]=
-
(a-b)
1
2
(a+b)π-
a
b
Log-

a
b
-Log

a
b

2
(a-b)
if
condition
In[]:=
%//InputForm
Out[]//InputForm=
ConditionalExpression[-(((a - b)*(((a + b)*Pi)/2 - I*Sqrt[a]*Sqrt[b]*(Log[((-I)*Sqrt[a])/Sqrt[b]] - Log[(I*Sqrt[a])/Sqrt[b]])))/Sqrt[(a - b)^2]),
(a/(a - b) != 0 && Re[a/(-a + b)] >= 0) || Re[a/(a - b)] > 1 || NotElement[a/(a - b), Reals]]
In[]:=
With{a=Random[],b=Random[]+1},NIntegrate
(z-a)(b-z)
z
,{z,a,b},Inactive[ConditionalExpression][-(((a-b)*(((a+b)*Pi)/2-I*Sqrt[a]*Sqrt[b]*(Log[((-I)*Sqrt[a])/Sqrt[b]]-Log[(I*Sqrt[a])/Sqrt[b]])))/Sqrt[(a-b)^2]),(a/(a-b)!=0&&Re[a/(-a+b)]>=0)||Re[a/(a-b)]>1||NotElement[a/(a-b),Reals]],π
a+b
2
-
ab

In[]:=
NIntegrate
(z-a)(b-z)
z
,{z,a,b},Inactive[ConditionalExpression][-(((a-b)*(((a+b)*Pi)/2-I*Sqrt[a]*Sqrt[b]*(Log[((-I)*Sqrt[a])/Sqrt[b]]-Log[(I*Sqrt[a])/Sqrt[b]])))/Sqrt[(a-b)^2]),(a/(a-b)!=0&&Re[a/(-a+b)]>=0)||Re[a/(a-b)]>1||NotElement[a/(a-b),Reals]][[1]]-π
a+b
2
-
ab
//FullSimplify
In[]:=
%/.1
2
(a-b)
->1(b-a)//PowerExpand//Simplify
NIntegrate
:z = a is not a valid limit of integration.
General
:Further output of NIntegrate::nlim will be suppressed during this calculation.
Out[]=
NIntegrate
(z-a)(b-z)
z
,{z,a,b},0

Other potential integral

In[]:=
Integrate-
2ab
a
2
t
+b
+
2b
2
t
+1
-
2(b-a)
2
(
2
t
+1)
,{t,0,∞}
In[]:=
%//InputForm
In[]:=
With{a=Random[],b=Random[]+1},NIntegrate-
2ab
a
2
t
+b
+
2b
2
t
+1
-
2(b-a)
2
(
2
t
+1)
,{t,0,∞},ConditionalExpression
2
(a-b)
π
2
2

a
+
b

,Re
b
a
>0,-π
ab
+πb-
1
2
π(b-a)
Out[]=
{0.36206,ConditionalExpression[0.36206,True],0.36206}
In[]:=
NIntegrate-
2ab
a
2
t
+b
+
2b
2
t
+1
-
2(b-a)
2
(
2
t
+1)
,{t,0,∞},Inactive[ConditionalExpression][((a-b)^2*Pi)/(2*(Sqrt[a]+Sqrt[b])^2),Re[Sqrt[b]/Sqrt[a]]>0][[1]]--π
ab
+πb-π
b-a
2
//FullSimplify
NIntegrate
:The integrand -
2(-a+b)
2
(1+
2
t
)
+
2b
1+
2
t
-
2ab
b+a
2
t
has evaluated to non-numerical values for all sampling points in the region with boundaries {{∞,0.}}.
Out[]=
NIntegrate-
2ab
a
2
t
+b
+
2b
2
t
+1
-
2(b-a)
2
(
2
t
+1)
,{t,0,∞},-
a
b
+
ab
π

More potential integral

For the Kepler potential, this integral is
In[]:=
2
2
(b-a)
Integrate
2
t
(b
2
t
+a)
2
(1+
2
t
)
,{t,0,∞}
Out[]=
2
(-a+b)
π
2
2

a
+
b

if Re
a
b
>0
Out[]//InputForm=
ConditionalExpression[((-a + b)^2*Pi)/(2*(Sqrt[a] + Sqrt[b])^2), Re[Sqrt[a]/Sqrt[b]] > 0]

A difficult triple trigonometric integral
https://mathoverflow.net/questions/308834/difficult-trigonometric-integral

Mellin integrals

Euler integrals (Gamma & Beta functions)

MellinTransform (direct & inverse)

Questions