I'm trying to show that $$\int_0^{\infty}\left(1-\frac{x}{n}\right)^n\log(x)\,dx=\int_0^{\infty}e^{-x}\log(x)\,dx$$

My idea was to apply dominated convergence theorem, since we notice that $\lim_{n\to \infty}\left(1-\frac{x}{n}\right)^n =e^{-x}.$ The problem is that I'm not able to show that the integral $$\int_0^{\infty}e^{-x}\log(x)\,dx \lt \infty$$

Can anyone give me a Hint?


For $x>1$ use $\log(x) \leq x-1\leq x$. Then $\left|\int_1^\infty e^{-x} \log(x) dx \right| \leq \int_1^\infty e^{-x} x dx$. Now integrate by parts. For $0<x<1$ use $e^{-x}\leq 1$ and that $-\int_0^1 \log(x)dx<\infty$ (integration by parts) to get $|\int_0^1 \log(x) e^{-x} dx| \leq \int_0^1 |\log(x)| dx <\infty$.

As for a dominating function, it is very important that you note that $(1-x/n)^n$ is increasing. It does not follow from $(1-x/n)^n \to e^{-x}$ that $(1-x/n)^n \leq e^{-x}$ by itself.

  • $\begingroup$ Yes you are absolutely right, that was not a problem to show, but stupidly I didn't notice that $log(x) \lt x$. Thank you very much :-) $\endgroup$ – Bman72 Jan 20 '14 at 16:56
  • $\begingroup$ $log(x)\leq x$ is okay, $\int_0^\infty e^{-x}logxdx\leq\int_0^\infty e^{-x}xdx$ is also okay, but if you take absolute value this may not be true. I didn't get it, as on the left we have negative integral ($-5<2$ but if you take absolute value, $"<"$ is not true any more), we should say something that integral on the left is not $-\infty$, or some other argument to support your inequality, I am not sure. $\endgroup$ – seriously divergent Jan 23 '14 at 2:44
  • $\begingroup$ @seriouslydivergent, whoops, of course. Thank you. $\endgroup$ – abnry Jan 23 '14 at 2:52
  • $\begingroup$ Now it sounds perfect :) $\endgroup$ – seriously divergent Jan 23 '14 at 8:00

That integral is equal to $-\gamma$, where $\gamma$ is the Euler-Mascheroni constant. See: http://en.wikipedia.org/wiki/Euler-Mascheroni_constant#Integrals


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