Exploring a zero of the MRB Constant Integral Analog

This Assistant assisted notebook investigates the integral analog
of the MRB (Marvin Ray Burns) constant
∞
∑
n=1
n
(-1)
(
1/n
n
-1)
through the examination of complex integrals. ​​We focus on the function
f(x)
x
(-1)

1
x
x
-1
and its integration path from 1 to
i∞
. By utilizing the properties of entire functions and the Cauchy Residue Theorem, we aim to understand the behavior and evaluation of this integral. The result
a+b
, where
a
and
b
are the real and imaginary parts of the integral, is an analog to the MRB constant, providing insights into its mathematical structure.
---
This introduction sets the stage for the exploration of the integral and its connection to the MRB constant, providing a clear context for the computations and theoretical discussions that follow.
Given
9
Out[]=
9

a=Re
∞
∫
1
x
(-1)
(
1/x
x
-1)x

In[]:=

b=Im
∞
∫
1
x
(-1)
(
1/x
x
-1)x

we have

∞
∫
1
x
(-1)
(-πa+πb+
1/x
x
)x+

π
0.

Analytically shown here.
In[]:=
Integrate[(-1)^x(
1/x
x
-IaPi+bPi),{x,1,InfinityI}]+I/Pi
Out[]=

π
+
∞
∫
1
x
(-1)
-aπ+bπ+
1
x
x
x
In[]:=
a=Re[NIntegrate[(-1)^x(x^(1/x)-1),{x,1,InfinityI},WorkingPrecision->30]];​​b=Im[NIntegrate[(-1)^x(x^(1/x)-1),{x,1,InfinityI},WorkingPrecision->30]];​​​​result=NIntegrate[(-1)^x(x^(1/x)-IaPi+bPi),{x,1,InfinityI},WorkingPrecision->30]+I/Pi;​​​​result//N
Out[]=
2.65334×
-34
10
+0.
In[]:=
​​result=NIntegrate[(-1)^x(x^(1/x)),{x,1,InfinityI},WorkingPrecision->30]+I/Pi;​​​​result//N
Out[]=
0.070776-0.0473806
In[]:=
​​result=NIntegrate[(-1)^x(-IaPi+bPi),{x,1,InfinityI},WorkingPrecision->30];​​​​result//N
Out[]=
-0.070776+0.0473806
Proof
The values for
a
and
b
are chosen such that the integral evaluates to zero, specifically: ​​
•
aℜ
i∞
∫
1
x
(-1)
(
1/x
x
-1)x
​​
•
bℑ
i∞
∫
1
x
(-1)
(
1/x
x
-1)x
​​These values are calculated using the properties of the function
f(x)
x
(-1)
(
1/x
x
-1)
along the specified path. When you alter these values arbitrarily, the integral expression: ​​
i∞
∫
1
x
(-1)
(
1/x
x
-Iaπ+bπ)x+
i
π
​​will likely yield a non-zero result. This is because
a
and
b
are derived specifically to cancel out the contributions of the integral, ensuring that the real and imaginary parts balance out and result in zero. ​​
Explanation:
​​1. **Integral Dependence on
a
and
b
**: ​​
• The integral includes terms
-Iaπ
and
bπ
which are specifically designed to balance the real and imaginary parts of the original integral expression
x
(-1)
(
1/x
x
-1)
.
​​ ​​2. Cancellation Mechanism: ​​
• The choice of
a
and
b
ensures that the oscillatory and decaying nature of the function
f(x)
leads to cancellation across the integration path. Altering these values disrupts this delicate balance, causing the parts to no longer cancel each other out.
​​3. Analytical Conditions: ​​
• The analytical condition provided ensures zero contribution under the assumption of the specified
a
and
b
, since there are no poles within the contour and the sum of the residues is zero. Modifying these values violates the conditions under which the theorem (such as Cauchy Residue Theorem) was applied.
​​If you change
a
and
b
, it would be equivalent to modifying the conditions under which the integral was shown to be zero, and the result would no longer be zero due to the imbalance in contributions from the real and imaginary parts.
The cancellation mechanism for the values of
a
and
b
in the integral expression is rooted in the properties of complex integration and the specific nature of the function being integrated. Let's break down how
a
and
b
are chosen to achieve this cancellation: ​​
Integral Expression
​​The integral we are considering is: ​​
i∞
∫
1
x
(-1)
(
1/x
x
-Iaπ+bπ)x+
i
π
​​
Role of
a
and
b
​​1. Complex Nature of the Integral:​​
• The function
f(x)
x
(-1)
(
1/x
x
-1)
is complex-valued, meaning it has both real and imaginary components.
​​
• The terms
-Iaπ
and
bπ
are introduced to adjust these components so that the overall integral evaluates to zero.
​​2. Balancing Real and Imaginary Parts:​​
• The real part of the integral is primarily affected by the term
x
(-1)
ℜ(
1/x
x
-1)
.
​​
• The imaginary part is influenced by
x
(-1)
ℑ(
1/x
x
-1)
.
​​
• By choosing
a
as the real part and
b
as the imaginary part of the integral of
x
(-1)
(
1/x
x
-1)
along the path from 1 to
i∞
, we ensure that these terms precisely cancel out the contributions of the oscillatory and decaying nature of the function.
​​3. Zero Result:​​
• The choice of
a
and
b
ensures that the integral becomes zero because the modifications
-Iaπ
and
bπ
are specifically calculated to offset the contributions from
1/x
x
-1
.
​​
• The additional term
i
π
is used to further ensure that any small residual parts are canceled.
​​
Mathematical Justification
​​
• The integral is computed using properties of complex integration, such as the Cauchy Residue Theorem, which relies on the absence of poles within the contour and the sum of residues being zero.
​​
• The specific values of
a
and
b
are derived from the computation of the integral along the path from 1 to
i∞
, ensuring that the conditions of the residue theorem hold.
​​In summary,
a
and
b
are not arbitrary constants but are carefully calculated to ensure that the integral evaluates to zero by balancing the real and imaginary parts of the complex function being integrated. If these values were altered, this balance would be disrupted, leading to a non-zero result.
Todemonstratethatthefunction
f(x)
x
(-1)
(
1/x
x
-1)
isentire(i.e.,withoutpoles)
, we need to analyze its components:​​​​1. **Exponential Component
x
(-1)
:**​​​​ The expression
x
(-1)
can be rewritten using Euler's formula as:​​
x
(-1)

xln(-1)
e

x(ln|1|+iπ)
e

iπx
e
​​ This represents a continuous and differentiable function across the complex plane, as it's simply a complex exponential function. Thus, it does not introduce any poles.​​​​2. **Fractional Power Component
1/x
x
:**​​​​ The expression
1/x
x
can be expressed as:​​
1/x
x

ln(x)
x
e
​​ This is also entire because the natural logarithm function
ln(x)
is analytic (holomorphic) except at
x0
, but since we are interested in integration from 1 to
i∞
, this point is not on our path. Furthermore, the division by
x
does not introduce any singularities along the path since
x≠0
.​​​​3. Combination of Components:​​​​ Since both
x
(-1)
and
1/x
x
are entire functions, their combination
x
(-1)
(
1/x
x
-1)
remains entire. Subtracting 1 from an entire function does not affect its analyticity.​​​​Therefore, the function
f(x)
is entire on the specified integration path from 1 to
i∞
, indicating that it is free of poles or singularities in this region. This property allows the use of the Cauchy Residue Theorem, which confirms that the integral over a closed contour that includes this path is zero if the contour does not enclose any poles.