Informally speaking, an "almost integer" is a real number very close to an integer.
There are some known ways to construct such examples in a systematic way. One is through the use of certain algebraic numbers called Pisot numbers. These numbers $\alpha$ have the property that their powers can get arbitrarly close to integers, that is:
$\lim_{n \to \infty} \alpha - [\alpha^n] = 0$
where $[ .]$ is the nearest integer function.
A well-known example is given by the golden ratio $\varphi = \frac{1 + \sqrt{5}}{2}$, whose powers are increasingly close to integers:
$\varphi^{19} = 9349.000107...$
$\varphi^{25} = 167761.00000596...$
Another example comes from numbers of the form $e^{\pi\sqrt{n}}$.
A well-known example is Ramanujan's constant:
$e^{\pi\sqrt{163}} = 262537412640768743.99999999999925007...$
There's another interesting way to generate almost integers by using the numbers $e$ and $\pi$. By using the identity
$$\sum_{n=-\infty}^\infty e^{-\pi n^2x}=x^{-1/2}\sum_{n=-\infty}^\infty e^{-\pi n^2/x}.$$
we can derive the approximate identity
$$ (*) \sum_{k=0}^{n-1}{e^{-\frac{k^2\pi}{n}}}\approx\frac{1+\sqrt{n}}{2}$$
which provides a way to construct almost integers with increasing precision:
$ e^{-\frac{\pi}{9}} + e^{-4\frac{\pi}{9}} + e^{-9\frac{\pi}{9}} + e^{-16\frac{\pi}{9}} + e^{-25\frac{\pi}{9}} + e^{-36\frac{\pi}{9}} + e^{-49\frac{\pi}{9}} + e^{-64\frac{\pi}{9}} = 1.0000000000010504... $
$\sum_{k=1}^{24} e^{-k^2\frac{\pi}{25}} = 2.000000000000000000000000000000000310793...$
$\sum_{k=1}^{48} e^{-k^2\frac{\pi}{49}} = 3.000000000000000000000000000000000000000000000000000000000000000000838654...$
So, the question is: is there another way to generate almost integers -with or without increasing precision- by using transcendental functions, as in the previous example?
(Note that there's a trivial way to do this: By taking a convergent series $\sum_{k = 1 }^\infty x_k$ and its limit $L$, the number $1/L\sum_{k = 1 }^n x_k$ will be an almost integer, namely close to $1$, but I'm looking for a an example like identity (*), or for a different, non trivial one). So, I am looking for an example that may be of the form $\sum_{k = 1 }^n f(x_k)$, where $f(x)$ is a transcendental function of $x$, that is able to generate a set of different almost integers (zero excluded).
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) $$\left(e^{\pi\sqrt{163}}-744\right)^{1/3}=640319.99999999999999999999999939031735...$$