How could we tell that we would end up with exactly 30 digits of convergence per almost every 22 i units in the upper limit?

Question

I noticed a remarkable fact about the rate of convergence of the following MRB constant (m) integral.

$$m=-\Im\left(\int_1^{i \infty } \frac{t^{1/t}-1}{\sin (\pi t)} \, dt\right).$$

Let m be an approximation of the MRB constant.

$$m=\sum _{n=1}^{\infty } \frac{n^{1/n}-1}{\cos(\pi n) }.$$

m = N[NSum[Exp[ I Pi n] (1 - (1 + n)^(1/(1 + n))), {n, 0, Infinity}, \
WorkingPrecision -> 1207, Method -> "AlternatingSigns"], 1200];

Let i n be an upper limit of

$$m_{in}=\Im\left(\int_1^{i n} \frac{t^{1/t}-1}{\sin (\pi t)} \, dt\right),n\in{\displaystyle \mathbb {N} }.$$ Which, as a n gets large, ends up yielding exactly 30 digits for most 22i unit increases of n.

Table[{22 n "as an upper bound gives around ", -(MantissaExponent[
       m + Im[NIntegrate[(t^(1/t) - 1)/Sin[Pi t], {t, 1, 22 n I}, 
          WorkingPrecision -> 1200, MaxRecursion -> 11]]][[2]]), 
   "accurate digits."}, {n, 1, 30}] // TableForm

Table[{22 n "as an upper bound gives around ", -(MantissaExponent[
       m + Im[NIntegrate[(t^(1/t) - 1)/Sin[Pi t], {t, 1, 22 n I}, 
          WorkingPrecision -> 1800, MaxRecursion -> 11]]][[2]]), 
   "accurate digits."}, {n, 31, 40}] // TableForm

I noticed the following:

m = N[NSum[Exp[I Pi n] (1 - (1 + n)^(1/(1 + n))), {n, 0, Infinity}, 
    WorkingPrecision -> 1807, Method -> "AlternatingSigns"], 1800];

Table[{22 n "as an upper bound gives around ", -(MantissaExponent[
       m + Im[NIntegrate[(t^(1/t) - 1)/Sin[Pi t], {t, 1, 22 n I}, 
          WorkingPrecision -> 1800, MaxRecursion -> 12]]][[2]]), 
   "accurate digits."}, {n, 40, 46}] // TableForm

My motivation is that it is not known whether m is rational. I am thinking that if new digits of m are always arising from higher and higher upper limits of $\frac{t^{1/t}-1}{\sin (\pi t)},$ I'm just guessing that m would be irrational.

Answer 1

Here is a heuristic explanation for the observed behavior.

Write the integral as an infinite series, $m= \sum_{k = 1}^\infty a_k$ with $a_k = \int_{i kM}^{i (k+1)M} \frac{t^{1/t}-1}{\sin (\pi t)} \, dt$ for $k \ge 2$ and the obvious modification for $k = 1$. You are computing the partial sums of these series with $M = 22$ and your question is why the series remainders decrease by a factor of $10^{-30}$ for each additional term.

The integrand is a quotient with numerator $t^{1/t} - 1 \approx \log t\, / t$ and denominator $1/\sin \pi t \approx e^{i \pi t}$ for large imaginary $t$. The absolute values of these terms therefore are $|a_k| \approx \log |kM|/|kM| \cdot e^{-\pi kM}$. This implies $$ \frac{|a_{k+1}|}{|a_k|} \to e^{-\pi M} $$ as $k \to \infty$. Consequently the remainders $\sum_{k = N}^\infty$ behave like $e^{- \pi N M}$. They decrease by a factor of $e^{-\pi M}$ for each additional term. And for $M = 22$, this is approximately $10^{-30}$, predicting an increase in accuracy of 30 digits whenever the upper integration bound in creased by $22i$.