Proving Equivalence of Euler-Macheroni Constant

[BigKissLilKis]

Hi Everyone,

I just registered for PF today because this problem was driving me nuts and I was hoping to get some help. It comes from pg. 5 of Peter Miller's "Applied Asymptotic Analysis" and goes like this:

The Euler gamma constant has one definition as
$$\gamma := \int_0^\infty e^{-t}\log\left(\frac{1}{t}\right)\,dt$$.

The problem is to then show that this integral is equivalent to the summation representation of $$\gamma$$:
$$\gamma:= \lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{k} - \log n\right]$$

Assuming a lot of stuff, you can prove the first definition from the Laplace transform of log and the properties of the gamma function. However, this problem is much more difficult because they want you to show the equivalence between this integral and difference limit with what I think is very little background information.

I'm sure that this can be solved with elementary techniques, because it's in a textbook, but I am still stumped on how to do this. I've tried using elementary calculus, like just taking the integral and various series expansions, and even tried fitting the Euler-MacClaurin summation formula. I think there is just some simple trick, but I'm not sure.

Any insight would be much appreciated!

 

[adriank]

Do you know about the product expression for the Gamma function?
$$\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left[ \left(1 + \frac{z}{k}\right)^{-1} e^{z/k} \right]$$
Taking the log of this:
$$\log \Gamma(z) = -\gamma z - \log z + \sum_{k=1}^\infty \left[ - \log\left(1 + \frac{z}{k}\right) + \frac{z}{k} \right]$$
For [itex]z = 1[/tex], this becomes
$$0 = \log \Gamma(1) = -\gamma + \sum_{k=1}^\infty \left[ \frac{1}{k} - \log(k + 1) + \log k \right]$$
so that
$$\gamma = \sum_{k=1}^\infty \left[ \frac{1}{k} - \log(k + 1) + \log k \right] = \lim_{n \to \infty} \left[ \sum_{k=1}^n \frac{1}{k} - \log(n+1) \right] = \lim_{n \to \infty} \left[ \sum_{k=1}^n \frac{1}{k} - \log n \right]$$
(the last equality because $\lim_{n \to \infty} \log(n+1) - \log n = 0$).

The problem is reduced to showing that the infinite product definition and integral definition of the Gamma function are equivalent. Then $\gamma = -\Gamma'(1)$; the first definition of $\gamma$ you give is easily derived from the integral definition of the Gamma function.