Is the function defined by :
$$f(x) = \int^x_0 \frac{1}{1+e^t}dt$$
convergent ? When graphing it, it seems that it converges to approximately $0.6931$ : $log(2)$.
The primitive of the function : $$f(x) = \frac{1}{1+e^x}$$ is $$F(x) = x-log(e^x+1)$$
Is this function convergent ?
As you showed $$f(x) = \int^x_0 \frac{1}{1+e^t}dt=x-\log \left(1+e^x\right)+\log (2)$$ Now, write $$\log \left(1+e^x\right)=\log(e^x)+\log(1+e^{-x})=x+\log(1+e^{-x})$$ So, $$f(x)=\log(2)-\log(1+e^{-x})$$ and $x\to \infty$.
Using Taylor series of $\log(1+a)$ around $a=0$ and replacing $a$ by $e^{-x}$, you would have $$f(x)=\log(2)-\left(e^{-x}-\frac{e^{-2 x}}{2}+\frac{e^{-3 x}}{3}+\cdots\right)$$
Using $x=10$, the last formula gives $$f(10)\approx 0.69310178166072844371$$ while the exact calculation would give $$f(10)\approx 0.69310178166072844477$$
