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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 $\int_{-\infty }^{\infty }\!{{\rm e}^{b-{{\rm e}^{b}}}}\,{\rm d}b=1$

crow
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  • I don't see it in Gradshteyn and Ryzhik. Why do you think it is true? – Will Jagy Feb 07 '17 at 01:39
  • Because I evaluated the integral numerically over the range of -100 to 100 and put that number into the inverse symbol calculator which said thats what it is. http://wayback.cecm.sfu.ca/cgi-bin/isc/lookup?number=.57721566490153286061&lookup_type=simple then I evaluated gamma in Maple with evalf, and showed its the same for at least 16 digits – crow Feb 07 '17 at 01:52

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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

Community
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Will Jagy
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  • what does the abbreviation FI stand for? – crow Feb 07 '17 at 17:54
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    @crow G. M. Fikhtengol'ts author. The integral is intimately tied with the Gamma function and its derivative; surely all known by the time of Weierstrass, probably long before that (Legendre) – Will Jagy Feb 07 '17 at 18:23