How can I find $\sum_{k=1}^{1233}f(\frac{k}{1234})$ 
Let $$f(x)=\frac{e^{2x-1}}{1+e^{2x-1}}$$ 
  Then find $$\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)$$ 

How I proceed: $$\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)=\int_{1}^{1233}\frac{e^{\frac{2x}{1234}-1}}{1+e^{\frac{2x}{1234}-1}}~dx$$ then how I solve this integral. please help.
 A: Note that:
\begin{align}
f(k/1234)+f((1234-k)/1234)=&
\frac{e^{\frac{k}{617}-1}}{e^{\frac{k}{617}-1}+1}+\frac{e^{\frac{1234-k}{617}-1}}{e^{\frac{1234-k}{617}-1}+1}&=&\\
=&\frac{e^{k/617}}{e^{k/617}+e}+\frac{e}{e^{k/617}+e}&=&1
\end{align}
So your sum is equal to $616+f(1/2)$.
A: Observe that $f(x)+f(1-x)=1$
Put $x=\frac k{1234}\implies  f\left(\frac k{1234}\right)+ f\left(\frac {1234-k}{1234}\right)=1$
Then put $k=1,2,,\cdots ,\frac{1234}2=617$ 
so $ f\left(\frac 1{1234}\right)+ f\left(\frac {1234-1}{1234}\right)=1,$
$ f\left(\frac 2{1234}\right)+ f\left(\frac {1234-2}{1234}\right)=1,$
**
$ f\left(\frac {617}{1234}\right)+ f\left(\frac {1234-617}{1234}\right)=1\implies  f\left(\frac 12\right)=\frac12$
Adding we get,
 $\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)=617(1)-f\left(\frac {617}{1234}\right)=617-\frac12$
A: I don't know if a closed form for your sum exists. However, we note that:
$$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{k=1}^{N}f(\frac{k}{N})=\int_0^1f(x)dx$$
Thus, for large values of $N$ you can estimate the sum using the integral.
To evaluate the integral, we note that $\frac{d}{dx}[e^{2x-1}+1]=2e^{2x-1}$, thus:
$$\int_0^1\frac{e^{2x-1}}{e^{2x-1}+1}dx=\frac{1}{2}\int_0^1\frac{2e^{2x-1}}{e^{2x-1}+1}dx=[\frac{1}{2}\log(1+e^{2x-1})]_{x=0}^{x=1}$$
