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How to evaluate this integral: $$\int_{0}^{\infty} \biggl\lfloor{\frac{n}{e^{x}}\biggr\rfloor} \ dx, $$where $n \in \mathbb{N}$.

The same integral when asked to evaluate for $n=2$ (say) i can do it by splitting the limits from $x = 0$ to $x = \log{2}$ where $e^{x}$ takes the value 1, and then from $x= \log{2}$ to $x = \infty$. But how to do this for the general case $n$. I thought of two ways:

  • Using induction on $n$. This will not work!

  • Splitting the limits from $x =0$, to $x = \log{n}$ also seems to cause some problems for me.

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up vote 12 down vote accepted

The integrand will only take on integral values m, and only for a finite measure of x values. The interval of x values in which the integrand equals exactly m is $ (- \ln \frac{m+1}{n} , - \ln \frac{m}{n} ] $. Split the integral over these intervals, evaluate (upper bound of interval minus lower bound times m) and turn them into a summation as $$ \sum_{m=1}^n m \left( \ln \frac{m+1}{m} \right) = \ln \prod_{m=1}^n \left( 1 + \frac{1}{m} \right)^m $$

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+1: I have deleted my answer which said something similar. Welcome to the site chroma :-) – Aryabhata Nov 4 '10 at 20:04
Very nice!. Would you mind to provide a resource to look at that kind of procedure? – Robert Smith Nov 5 '10 at 1:24
I'm not aware of a good resource for this procedure, but I've seen it done before with certain integrals involving the zeta and (Riemann) prime counting functions. Basically look at the integrand, see if its behavior can be distinguished on certain subintervals of the domain to be integrated over, and then split the integral up so each case can be done separately. It's very often used when the integrand involves a piecewise continuous function. – chroma Nov 5 '10 at 15:57
@chroma The giant product actually simplifies nicely (by telescopic cancellation) to $$ \frac{(n+1)^n}{n!}. $$ – Srivatsan Sep 10 '11 at 15:35

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