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The following question from $\textit{Statistical Inference }$by Casella & Berger was given to me on an assignment last semester. On the surface, it appeared to be a straight-forward question about transformations, but I stumbled across an integral I could not understand. Here is the question:

If $X \sim \text{exponential(1)}$, show that $Y=\mu-\beta \log X$ has the $\text{Gumbel$(\mu,\beta)$}$ distribution, where $-\infty<\mu<\infty$ and $\beta>0$.

We have that $f_X(x)=e^{-x}$ for $x \geq 0$. Now, let $Y=\mu-\beta \log X$. Then, $g(x)=\mu-\beta \log x$, which is monotone on $0<x<\infty$. We will use the standard theorem for transformations, with $g^{-1}(y)=e^{\frac{\mu-y}{\beta}}$ on $-\infty <y<\infty$. Then, the transformed pdf is given by: $$f_Y(y)=e^{-e^{\frac{\mu-y}{\beta}}}\left|\frac{d}{dy}e^{\frac{\mu-y}{\beta}}\right|=\frac{1}{\beta}e^{-e^{\frac{\mu-y}{\beta}}}e^{\frac{\mu-y}{\beta}} \text{ on $-\infty<y<\infty$} $$

I then attempted to find the mean of the distribution as follows. \begin{eqnarray*} \text{E}(Y)&=& \frac{1}{\beta}\int_{-\infty}^\infty ye^{-e^{\frac{\mu-y}{\beta}}}e^{\frac{\mu-y}{\beta}}dy \hspace{0.5in} \text{Let $u=e^{\frac{\mu-y}{\beta}} \implies du=-\frac{1}{\beta}e^{\frac{\mu-y}{\beta}}dy$} \\ &=& \int_0^\infty \left(\mu e^{-u}-\beta e^{-u}\log u \right)du=\mu \int_0^\infty e^{-u}du-\beta\int_0^\infty \left(e^{-u}\log u\right) du \\ &=& \mu-\beta\int_0^\infty \left(e^{-u}\log u\right) du \end{eqnarray*} According to Wikipedia, the mean is given by $\mu+\beta\gamma$, where $\gamma\approx 0.5772$ is the Euler-Mascheroni constant. Therefore, the last integral above is equal to $-\gamma$, and indeed, that is the answer Wolfram|Alpha provides. I am wondering if there is any way to solve the integral $I=\int_0^\infty (e^{-u}\log u) du$ using standard methods. I tried making every substitution I could think of and kept getting stuck. I even showed this to my instructor; he was not able to show that it equals $-\gamma$ either. I should also mention that prior to doing this question, I had never heard of the Euler-Mascheroni constant. Any help would be appreciated. Thank you!

  • I don't have proof, but the integral is also on Wikipedia – Simply Beautiful Art Jan 02 '17 at 22:34
  • This is not a scientific argument, but I find it hard to believe that it can be computed by only elementary means. Do you rule out more advanced methods, such as complex-analytic ones (not that I would know any useful one for this problem)? – Alex M. Jan 02 '17 at 22:39
  • No, not necessarily. When I took it to my instructor, he only attempted elementary methods, and that might have been the reason we did not get anywhere. I'm mostly just curious to see how we arrive at that constant as the final result. –  Jan 02 '17 at 22:41
  • @DouglasFir: By all means, also check the answers given here and the links therein. As I suspected, clearly not a trivial integral. – Alex M. Jan 02 '17 at 22:50
  • @AlexM. Thank you! I think both of those links pretty much cover my question--was not able to find them before I posted mine. –  Jan 02 '17 at 22:51
  • Even more links: http://math.stackexchange.com/questions/697383/integral-from-0-to-infty-of-lnx-ex and http://math.stackexchange.com/questions/980593/some-integral-representations-of-the-euler-mascheroni-constant and http://math.stackexchange.com/questions/419026/computing-int-0-infty-frac-log-x-exp-x-dx. They should contain almost everything important about the subject. – Alex M. Jan 02 '17 at 22:56

1 Answers1

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We start with \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-x} x^t \mathrm{d} x = \Gamma(t+1) \end{equation} Differentiate with respect to $t$ \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-x} x^t \mathrm{ln}(x) \mathrm{d} x = \frac{d\Gamma(t+1)}{dt} = \Gamma(t+1) \psi^{(0)}(t+1) \end{equation}

Taking the limit $t \to 0$ yields \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = \Gamma(1) \psi^{(0)}(1) = -\gamma \end{equation}

Note that this solution appeared in a book written by @Zaid Alyafeai

poweierstrass
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  • So the next question would be to explain why $\psi ^{(0)} (1) = - \gamma$... Essentially, your answer turns a mysterious equality into another equally mysterious equality. – Alex M. Jan 02 '17 at 22:44
  • There are a large number of definitions of the Euler-Mascheroni constant. So one has to select a particular definition as given. Think of how one would define $e$. – poweierstrass Jan 02 '17 at 22:50