Asymptotic antiderivatives involving x^(1/x)from M R Burns using the Wolfram Assistant
Asymptotic antiderivatives involving from M R Burns using the Wolfram Assistant
x^(1/x)
This notebook derives asymptotic antiderivative expansions for and for the oscillatory/exponential generalization
∫x
1/x
x
∫x.
ax
e
1/x
x
1. Asymptotic antiderivative of x^(1/x)
1. Asymptotic antiderivative of
x^(1/x)
We begin with
x^(1/x)==Exp[Log[x]/x]
Since 0 asx∞,we expand: , u.Therefore In Wolfram Language:
logx
x
u
e
∞
∑
k0
k
u
k!
logx
x
Series[x^(1/x),{x,Infinity,6}]
Formally integrating term by term gives
The first two terms are immediate:
For the remaining terms, a convenient closed form leads to
Wolfram Language definition
Wolfram Language definition
In[]:=
ClearAll[asymptoticAntiderivativexrootx];asymptoticAntiderivativexrootx[x_]:=x+Log[x]^2/2-Sum[x^(-(k-1))Sum[Log[x]^m/(m!(k-1)^(k-m+1)),{m,0,k}],{k,2,Infinity}]
A truncated version is often more practical:
In[]:=
asymptoticAntiderivativexrootx[x_,n_Integer?Positive]:=x+Log[x]^2/2-Sum[x^(-(k-1))Sum[Log[x]^m/(m!(k-1)^(k-m+1)),{m,0,k}],{k,2,n}]
Check by differentiation
Check by differentiation
In[]:=
lhs=Normal@Series[D[asymptoticAntiderivativexrootx[x,4],x],{x,Infinity,4}];rhs=Normal@Series[x^(1/x),{x,Infinity,4}];{lhs,rhs,FullSimplify[lhs==rhs]}
Out[]=
1++++,1++++,True
Log[x]
x
2
Log[x]
2
2
x
3
Log[x]
6
3
x
4
Log[x]
24
4
x
Log[x]
x
2
Log[x]
2
2
x
3
Log[x]
6
3
x
4
Log[x]
24
4
x
Built-in derivation with
Built-in derivation with
In[]:=
AsymptoticIntegrate[x^(1/x),x,{x,Infinity,6}]
Out[]=
x+++++(-15-60Log[x]-120-160-160-128)
2
Log[x]
2
-2-2Log[x]-
2
Log[x]
2x
-3-6Log[x]-6-4
2
Log[x]
3
Log[x]
48
2
x
-8-24Log[x]-36-36-27
2
Log[x]
3
Log[x]
4
Log[x]
1944
3
x
1
61440
4
x
2
Log[x]
3
Log[x]
4
Log[x]
5
Log[x]
Since
x^(1/x)==Exp[Log[x]/x]
you can also use
In[]:=
AsymptoticIntegrate[Exp[Log[x]/x],x,{x,Infinity,6}]
Out[]=
x+++++(-15-60Log[x]-120-160-160-128)
2
Log[x]
2
-2-2Log[x]-
2
Log[x]
2x
-3-6Log[x]-6-4
2
Log[x]
3
Log[x]
48
2
x
-8-24Log[x]-36-36-27
2
Log[x]
3
Log[x]
4
Log[x]
1944
3
x
1
61440
4
x
2
Log[x]
3
Log[x]
4
Log[x]
5
Log[x]
2. Asymptotic antiderivative of Exp[Pi I x] x^(1/x)
2. Asymptotic antiderivative of
Exp[Pi I x] x^(1/x)
We now derive an asymptotic expansion for
Let
f[x_]:=x^(1/x)
and look for an antiderivative of the form Because we have So to match the integrand , we want This suggests the formal asymptotic series
Therefore Equivalently,
Wolfram Language setup
Wolfram Language setup
In[]:=
ClearAll[x,f,gN,candidate,n];f[x_]:=x^(1/x);gN[x_,n_]:=Sum[(-1)^kD[f[x],{x,k}]/(PiI)^(k+1),{k,0,n}];candidate[x_,n_]:=Exp[PiIx]gN[x,n];
3. Robust Wolfram Language verification
3. Robust Wolfram Language verification
A direct symbolic with a symbolic truncation index may leave the result as finite sums rather than telescoping them automatically. A more reliable computational check is to verify the identity for several explicit truncation orders.
For the truncated asymptotic ansatz we expect This can be checked directly for the first several values of m:
In[]:=
ClearAll[a,x,f];f[x_]:=x^(1/x);Table[FullSimplify[aSum[(-1)^kD[f[x],{x,k}]/a^(k+1),{k,0,m}]+D[Sum[(-1)^kD[f[x],{x,k}]/a^(k+1),{k,0,m}],x]-(f[x]+(-1)^mD[f[x],{x,m+1}]/a^(m+1)),Assumptions->a!=0],{m,0,6}]
Out[]=
{0,0,0,0,0,0,0}
This returns
{0,0,0,0,0,0,0}
confirming the identity for these truncation orders.
Packaged function
Packaged function
5. Notes
5. Notes
◼
These are asymptotic expansions, not convergent power series in the usual sense.