Examples of transcendental functions giving almost integers
Solution 1:
1) If $a,b\in\textbf{R}$ and $a<x<b$, then if $f(x)$ is continuous: $$ \frac{b-a}{N}\sum^{N}_{n=0}f\left(a+\frac{b-a}{N}n\right)=\int^{b}_{a}f(t)dt+o(1)\textrm{, }N\rightarrow\infty. $$ 2) If $\left[x\right]$ denotes largest integer $\leq x$ and $a$ is a non-rational real number, then $$ \lim_{n\rightarrow\infty}\frac{\left[na\right]}{n}=a $$ i.e. every real number can be approximated by rational numbers arbitrarily well.
3) If $\beta_{1}=1/2$,
$$
\beta_{r+1}=\frac{1-\sqrt{1-\beta_{r/2}}}{2},
$$
then
$$
\sin\left(\frac{\pi}{2(r+1)}\right)=\sqrt{\beta_r}
$$
4) This one is quite involved.
$$
e^{-25\pi}=\frac{1}{3}-\frac{1}{6}\sqrt{4+75\pi-12\log 2+6\log k}-6.2619875...\times 10^{-103},
$$
where
$$
k=\sqrt{\frac{1}{2}-\frac{1}{2}\sqrt{1-(51841-23184\sqrt{5})Y^{12}}},
$$
where $Y$ is a root of
$$
Y^3+5s^{-1}Y^2-sY-1=0,
$$
where $s$ is such
$$
s=\sqrt[3]{\frac{(t-1)^5}{11+6t+6t^2+t^3+t^4}},
$$
where
$$
t=2\sinh\left(\frac{1}{4}\textrm{arcsinh}\left(\frac{9+\sqrt{5}}{2}\right)\right).
$$
Note that $t$ is an algebraic number.
5) Set $$ p=\sqrt{2+216\cdot 5^{1/4}-96\cdot 5^{3/4}} $$ and $$ k=1-\frac{2}{1+t}\textrm{, }t=\frac{\left(\sqrt{2}+\sqrt{p}\right)^2}{2\cdot 2^{3/4}p^{1/4}\sqrt{2+p}}. $$ Also $$ l=\left(1+\frac{2^{3/4}p^{1/4}}{\sqrt{2+p}}\right)^2\frac{4+2\sqrt{5}+\sqrt{2}(3+2\cdot 5^{1/4})}{160}. $$ Then $$ \frac{\Gamma\left(\frac{1}{4}\right)^2}{\pi^{3/2}}=\frac{4+k^2-6k^4}{4l}-7.01743379...\times 10^{-107}. $$ 6) (Ramanujan) For $|x|<1$, $$ \prod_{k=1}^{\infty}\left(1-x^{p_k}\right)^{-1}=1+\sum^{\infty}_{k=1}\frac{x^{p_1+p_2+\ldots+p_k}}{(1-x)(1-x^2)\ldots(1-x^k)},\tag 1 $$ where $p_1,p_2,\ldots,$ denote the primes in ascending order. The above formula $(1)$ is canceled i.e. the Taylor series on both sides of $(1)$ agree only to the first 22 terms. (see Bruce C. Berndt. "Ramanujan's Notebooks I." Springer-Verlag, New York Inc. (1985) page 130).
7) This one is inspired from a formula of Ramanujan
Let $a,b$ be positive reals with $ab=2\pi$ and $\Psi(x)$ analytic on $\textbf{R}$. Let also
$$
M\Psi(s)=\int^{\infty}_{0}\Psi(x)x^{s-1}dx,
$$
be the Mellin transform of $\Psi$. If also
$$
\phi(x)=Re\left(M\Psi(ix)n^{-ix}\right).
$$
Then
$$
a\sum^{\infty}_{k=0}\Psi\left(ne^{ak}\right)=a\left(1/2-\sum^{\infty}_{k=1}\frac{\Psi^{(k)}(0)}{k!}\frac{n^k}{e^{ak}-1}\right)+c+2\sum^{\infty}_{k=1}\phi(bk),\tag 2
$$
where $c=\lim_{h\rightarrow 0}\phi(h)=:\phi(0)$.
Example
For $\Psi(x)=e^{-x}$, we get $$ a\sum^{\infty}_{k=0}e^{-ne^{ak}}=a\left(1/2-\sum^{\infty}_{k=1}\frac{(-1)^kn^k}{k!(e^{ak}-1)}\right)-\gamma-\log n+2\sum^{\infty}_{k=1}\phi(bk),\tag 3 $$ where $\phi(x)=Re\left(\Gamma(ix)n^{-ix}\right)$ and $c=\phi(0)=-\gamma-\log n$.
For $n=1$ in (3), we get the formula of Ramanujan and a good approximation of $\gamma$ constant (Euler's constant).
If $a=1/N$ then $b=2\pi N$ and we get as $N\rightarrow\infty$ $$ \gamma=-\frac{1}{N}\sum^{\infty}_{k=0}\exp\left(-e^{k/N}\right)+\frac{1}{N}\left(\frac{1}{2}-\sum^{\infty}_{k=1}\frac{(-1)^k}{k!\left(e^{k/N}-1\right)}\right)+O\left(e^{-\pi^2N}\right)\tag 4 $$ Also for $a=\frac{\log A}{N}$, then $b=\frac{2\pi N}{\log A}$ and holds $$ \gamma=-\frac{\log A}{N}\sum^{\infty}_{k=0}\exp\left(-A^{k/N}\right)+\frac{\log A}{N}\left(\frac{1}{2}-\sum^{\infty}_{k=1}\frac{(-1)^k}{k!(A^{k/N}-1)}\right)+O\left(e^{-\pi^2N/\log A}\right).\tag 5 $$ Set now $$ k_{10}(N)=\left[\frac{N}{\log A}\log\left(\frac{N\pi^2}{\log A}\right)\right]+1 $$ and $$ k_{20}(N)=\left[\frac{N\pi^2}{C_N\log A}\right]+1\textrm{, }C_N=P_L\left(\frac{A^{1/N}N\pi^2}{e\log A}\right), $$ where $P_L(x)$ is the product log function i.e. $e^{P_L(x)}P_L(x)=x$. Then $$ \frac{\gamma}{\log A}=-\frac{1}{N}\sum^{k_{10}(N)}_{k=0}\exp\left(-A^{k/N}\right)+\frac{1}{N}\left(\frac{1}{2}-\sum^{k_{20}(N)}_{k=1}\frac{(-1)^k}{k!(A^{k/N}-1)}\right)+O\left(e^{-\pi^2N/\log A}\right).\tag 6 $$
8) $$\left(e^{\pi\sqrt{163}}-744\right)^{1/3}=640319.99999999999999999999999939031735...$$