$g(x)=x^{1/x};\lim\limits_{N\to\infty}\int\limits_1^{2N}e^{i\pi t}g(t)dt=(-i)\lim\limits_{N\to\infty}\int\limits_{0}^{2N}e^{i^2\pi t}g(1+it)dt?$

Question

This linked arxiv article deals with a variant$\left(g(x)=x^{1/x};\lim\limits_{N\to\infty}\int\limits_1^{2N}e^{i\pi t}g(t)dt\right)$

of the MRB constant$\left(g(x)=x^{1/x}; \mathrm{CMRB} = \lim\limits_{N\to\infty}\sum\limits_{n=1}^{2N}(-1)^ng(n)\right).$

I found a faster-converging integral for the variant:

$\left(g(x)=x^{1/x};(-i)\lim\limits_{N\to\infty}\int\limits_{0}^{2N}e^{i^2\pi t}g(1+it)dt\text{ or }(-i)\lim\limits_{N\to\infty}\int\limits_{0}^{2N}e^{-\pi t}g(1+it)dt.\right)$

Here is some Mathematica code with results that justifies it to me:

  g[x_] = x^(1/x); Print[g[x_] = x^(1/x); 
 Timing[MKB = 
NIntegrate[Exp[I Pi t] (g[t]), {t, 1, Infinity}, 
  WorkingPrecision -> 200, PrecisionGoal -> 200] - 
 I/Pi][[1]]]; Print[g[x_] = x^(1/x); 
 Timing[N[MKB + (I NIntegrate[
    Exp[I^2 Pi t] (g[1 + t I]), {t, 0, Infinity}, 
    WorkingPrecision -> 200, PrecisionGoal -> 200] + I/Pi)]]]



 (* 1.25

  {0.890625,-1.0328794596386124*10^198+2.199848562388703*10^198 I}*)

I've shown the variant and my faster-converging integral to be equal to 35,000 digits at the end of this Wolfram community post.

It would be awesome if a scholar like you would prove that the variant and my faster-converging integral are really equal. Eternal fame awaits you if you can help prove this.

Answer 1

I got an answer on Quora. There is a lot of pretty math that might take me a while to transcribe.

I discovered that for

g(x)=x^(1/x),

Here it is proved by Ariel Gershon. Plugging in equations 5 and 6 into equation 2 gives us: Now take the limit as N→∞ and apply equations 3 and 4 : He went on to note that