# 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?

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.

• (1) The accuracy reported is only an estimate made by the software. (2) it isn't exactly 30 digits - there are two cases you've found where it increased by 31 digits, so it is a little over 30 digits on average. (3) This is called "linear" convergence, and is considered slow. Fast convergence would have the number of accurate digits doubling (or more) with each fixed-size increase in the input ("quadratic" convergence - Newton's methof for finding roots is an example). The speed here is determined by how fast the integrand decays. Aug 22 at 14:10
• @Paul Sinclair, I was wondering if there is a way to determine from the integrand alone that the convergence is linear on the order of being real close to 30 accurate digits per 22 input units. Aug 22 at 20:00
• @Dark Malthorp I hope you don't mind, but since you both helped find this integral $m=-\Im\left(\int_1^{i \infty } \frac{t^{1/t}-1}{\sin (\pi t)} \, dt\right),$ I was hoping you could help me determine how to find the rate of convergence from the integrand that I am looking for in this question. Aug 23 at 15:35
• @Diger I hope you don't mind, but since you both helped find this integral $m=-\Im\left(\int_1^{i \infty } \frac{t^{1/t}-1}{\sin (\pi t)} \, dt\right),$ I was hoping you could help me determine how to find the rate of convergence from the integrand that I am looking for in this question. Aug 23 at 15:35

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 $$22i$$.
• I knew it had something to do with logs and $\pi$. I just couldn't make it work. Aug 23 at 19:35
• Several years ago I found out that $m=\sum _{n=1}^{\infty } \frac{n^{1/n}-1}{\sec(\pi n) }.$ converges about n digits for every 10^(n+1) terms. I'm impressed with the increased rate of the integral! Aug 23 at 20:22