Few minutes ago I did experiments with Wolfram Alpha. Check my codes for my calculations with this online calculator
integrate $e^{-x}\log((e^x)/(1+x))dx$
and do a click to get More digits in the infinite integral that showed the online calculator $$\int_0^\infty e^{-x}\log\left(\frac{e^x}{1+x}\right)dx,$$ to get from
0.403652637
when you type it as the input of the online calculator, a simple relationship with Gompertz constant.
I tried also a comparison from the reference from previous article of MathWorld to The On-Line Encyclopedia of Integer Sequences, I say the sequence A073003.
My
Question. Can yoy help me to get this nice integral $$\int_0^\infty e^{-x}\log\left(\frac{e^x}{1+x}\right)dx?$$ Many thanks.
I don't know if it is obvious (then what's the easy change of variable, or integration by parts, to get my relationship from the identity defining Gompertz constant?) or well known.
We can simplify the integral as $$ I= \int_{0}^{\infty} e^{-x} \ln(\frac{e^x}{1+x})\, dx = -\int_{0}^{\infty} e^{-x}(\ln(x+1)-x) \,dx$$ $$=\int_{0}^{\infty} -e^{-x}\ln(x+1) \,dx +\int_{0}^{\infty} xe^{-x} \,dx $$ $$= I_1 + I_2$$ Now, $$I_1 = \int_{0}^{\infty}- e^{-x}\ln(x+1) dx = \int_{0}^{\infty} \frac{e^{-x}}{-x-1} dx +e^{-x}\ln(x+1)$$ Substituting $u = x+1$, we can get the first part of $I_1$ as $-e\operatorname{Ei}(-1)$ which is the Gompertz Constant.
We can easily integrate $I_2$ to get the answer as $1$. Thus, $$ I = 1-e\operatorname{Ei}(-1) = 1-\text{Gompertz Constant}$$
Hope it helps.