Integral sigmoid to softplus function proof

Question

Do someone help me to prove the function below?

where

Answer 1

Note that

$\sigma(x) = \dfrac{e^x}{e^x + 1} = \dfrac{(e^x + 1)'}{e^x + 1}; \tag 1$

thus

$\displaystyle \int_{-\infty}^x \sigma(y) \; dy = \int_{-\infty}^x \dfrac{(e^y + 1)'}{e^y + 1} \; dy$ $= \lim_{L \to -\infty} \displaystyle \int_L^x \dfrac{(e^y + 1)'}{e^y + 1} \; dy = \lim_{L \to -\infty} \ln(e^y + 1\vert_L^x$ $= \displaystyle \lim_{L \to -\infty} (\ln(e^x + 1) - \ln(e^L + 1)) = \ln(e^x + 1) - \lim_{L \to -\infty}(\ln(e^L + 1))$ $= \ln(e^x + 1) - \ln 1 = \ln(e^x + 1) - 0 = \ln(e^x + 1) = \zeta(x). \tag 2$