$\lim\limits_{N\to\infty}\sum\limits_{n=1}^{2N}(-1)^ng(n)=ℑ\lim\limits_{N\to\infty}\int\limits_{0}^{2N}\frac{g(1+it)}{\sinh(\pi t)}dt.$

Question

@Dark Malthorp got me started on Abel-Plana integral-equations for the MRB constant, here, and by trial and error, I got them concluded in a way that looks beautiful to me, and that includes equations for its integrated analog. But I would like to be able to present proof with them. Any help here?

Let $g(x) = x^{1/x}$.

$$\mathrm{CMRB} = \lim\limits_{N\to\infty}\sum\limits_{n=1}^{2N}(-1)^ng(n)=ℑ\lim\limits_{N\to\infty}\int\limits_{0}^{2N}\frac{g(1+it)}{\sinh(\pi t)}dt.$$

$$MKB=\lim\limits_{N\to\infty}\int\limits_1^{2N}e^{i\pi t}g(t)dt=(-i)\lim\limits_{N\to\infty}\int\limits_{0}^{2N}\frac{g(1+it)}{\exp(\pi t)}dt. $$

Here is some Mathematica code with results that justify them to me:

Clear[g]; g[x_] = x^(1/x); CMRB = N[NSum[(-1)^n (g[n] - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 57], 30]

 (* 0.187859642462067120248517934054*)

  g[x_] = x^(1/x); CMRB -  NIntegrate[Im[g[1 + I t]/(Sinh[Pi t])], {t, 0, Infinity}, 
  WorkingPrecision -> 30]

 (* 0.*10^-31*)

  g[x_] = x^(1/x); Timing[ MKB = N[NIntegrate[Exp[I Pi t] (g[t]), {t, 1, Infinity}, 
     WorkingPrecision -> 57], 30] - I/Pi]

(*{0.03125, 
 0.070776039311528803539528021830 - 0.684000389437932129182744459993 I}*)

  g[x_] = x^(1/x); Timing[MKB - (N[NIntegrate[(-1)^t (g[t]), {t, 1, Infinity}, 
       WorkingPrecision -> 57, MaxRecursion -> 3500], 30] - I/Pi)] // Quiet

 (* {30.8438, 6.3*10^-29 - 6.00*10^-28 I}*)

  g[x_] = x^(1/x); Timing[ MKB -
    (-N[ I NIntegrate[(g[(1 + t I)])/( Exp[Pi t]), {t, 0, Infinity},  WorkingPrecision -> 57] + I/Pi, 30])]

(* {0.046875, 0.*10^-31 + 0.*10^-31 I}*)

g[x_] = x^(1/x); Timing[ MKB - (-N[
 I NIntegrate[Exp[I^2 Pi t] (g[1 + t I]), {t, 0, Infinity},  WorkingPrecision -> 57] + I/Pi, 30])]

(* {0.03125, 0.*10^-31 + 0.*10^-31 I}*)

Here in Wolfram Community is where I would like to post the proofs after I understand them.

I tried variations of the following Wikipedia article.

I looked at $$ g(x) = x^{1/x} \text{ and }f(x)=e^{i \pi x}(x^{1/x}-1). $$$$\text{Re(MKB)}=i\lim\limits_{N\to\infty}\int _0^{2 N}\frac{g (1-i t)-g (1+i t)}{2 e^{\text{$\pi $t}}}$$$$=ℑ \int_0^{\infty } \frac{f(1-i t)-f(1+i t)}{e^{2 \pi t}+1} \, dt=ℑ \int_0^{\infty } \frac{f(1+i t)+f(1-i t)}{e^{2 \pi t}-1} \, dt.$$$$CMRB=Re(MKB)+ℑ \int_0^{\infty } \frac{f (1-i t)-f (1+i t)}{\exp (2 \pi t)-1} \, dt.$$$$\text{ℑ} \text{(MKB)}=(-i)\lim\limits_{N\to\infty}\int\limits_0^{2N}\frac{g (1+i t)+g (1-i t)}{2 e^{\text{$\pi $t}}}.$$ The (often missing in the denominator and appearing in the numerator)"1" can come from the convergent $CMRB=\sum\limits_{n=1}^{\infty}(-1)^n(g(n)-1).$ Nonetheless, I can't quite make the proof work.

Here on the Wolfram cloud is where I justified and summarized all my findings like I just posted above.

Answer 1

Here's the details on how I used the Abel-Plana formula: Let $f(x) = 1-(1+x)^{\frac1{1+x}}$. Then we have $$ C_{MRB} = \sum_{n=1}^\infty (-1)^n \left(n^{\frac1n}-1\right) =\sum_{n=0}^\infty (-1)^n \left(1-(n+1)^{\frac1{n+1}}\right)= \sum_{n=0}^\infty (-1)^nf(n) $$ To apply Abel-Plana, we need bounds on $f(z)$. It doesn't quite satisfy Wikipedia's assumptions, but actually it's pretty nice anyway, as $\lim_{x\rightarrow \infty} f(x+yi) = 0$ for all fixed $y$, and it is bounded in the right half-plane. Then we use the alternating series formulation of Abel-Plana:\begin{eqnarray} C_{MRB}&=&\sum_{n=0}^n(-1)^nf(n) = \frac12f(0)+i\int_0^\infty\frac{f(it)-f(-it)}{2\sinh(\pi t)}dt\\ &=&\frac12 \cdot 0 + i\int_0^\infty \frac{ 1-(1+i t)^{\frac1{1+it}} -1+(1-i t)^{\frac1{1-it}} }{2\sinh(\pi t)}dt\\ &=&i\int_0^\infty \frac{ -(1+i t)^{\frac1{1+it}} +(1-i t)^{\frac1{1-it}} }{2\sinh(\pi t)}dt \end{eqnarray} Because $(1+z)^{\frac1{1+z}}$ is holomorphic for $\Re z\ge 0$ and real valued for real $z$ we know that $f(\overline z) = \overline {f(z)}$. Because $\overline z - z= -2i(\Im z)$, this implies $$ C_{MRB} = \int_0^\infty \frac{\Im(1+it)^{\frac1{1+it}}}{\sinh(\pi t)}dt $$ Strictly speaking, we cannot pull the imaginary part out of the integral, as the function $\frac{(1+it)^{\frac1{1+it}}}{\sinh(\pi t)}$ has a pole at $0$ and thus its integral doesn't converge. The imaginary part, however, is bounded for $t\in(0,\infty)$.