how to show $-\int_{-\infty }^{\infty }\!b{{\rm e}^{b-{{\rm e}^{b}}}}\,{\rm d}b= \gamma$?

Question

how can one go about showing that $\int_{-\infty }^{\infty }\!b{{\rm e}^{b-{{\rm e}^{b}}}}\,{\rm d}b= -\gamma$ where $\gamma=.57721566490153286061...$ is Euler's constant? It can be thought of as the expectation of the distribution $b{{\rm e}^{b-{{\rm e}^{b}}}} $ since $\int_{-\infty }^{\infty }\!{{\rm e}^{b-{{\rm e}^{b}}}}\,{\rm d}b=1$

Answer 1

It works The first integral in G+R, 8.367.4, is $$ - \gamma = \int_0^\infty e^{-t} \log t \, dt $$

This is also the first example in https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant#Integrals

Substitute $$ t = e^x $$

The one above is pretty famous. I will see if I can decipher their attribution, it is a sort of bibliographic code. alright, FI was a three volume calculus book in Russian, 1947-1949

ADDED some good proofs of the "famous" item here, with further references: Integral representation of Euler's constant