# For which $f$ does $\sum_{n=0}^{\infty} f(a_{1},…,a_{k};n) = \int_{0}^{\infty} f(a_{1},…,a_{k};x) dx$?

Do there exist real valued functions $f(a_{1},...,a_{k};x)$ with real parameters $\{a_{1},...,a_{k}\}$ such that"

$$\sum_{n=0}^{\infty} f(a_{1},...,a_{k};n) = \int_{0}^{\infty} f(a_{1},...,a_{k};x) dx$$

It would also be nice if there was more than one set of parameters that the above identity held. I've been looking through Gradshteyn and Ryzhik to see if I can find pairs of integrals and series for which it might hold, but haven't had much success so far.

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It seems a bit vague because I guess there are tons of functions that could work (piecewise constant functions for example). Now maybe you're asking for a function with a concise formula. –  Joel Cohen Oct 4 '11 at 19:25
Forget about $a_1$, $\ldots$, $a_k$. So you are asking for functions $f:{\mathbb R}_{\geq0}\to{\mathbb R}$ with $\sum_{n=0}^\infty f(n)=\int_0^\infty f(x)\ dx$. –  Christian Blatter Oct 4 '11 at 19:34
Quite related... –  Ｊ. Ｍ. Oct 4 '11 at 22:22
This doesn't solve the problem, but it's similar. IIRC, $\sum_{n=0}^{\infty} \frac{1}{n^n} = \int_0^1 x^x dx$. I think there's a similar result for $\int_0^1 x^{-x} dx$. It's proved by writing $x^x = exp(x ln x)$, expanding the $exp$ in its Taylor series, and evaluating the resulting integrals. –  marty cohen Oct 8 '11 at 23:42
@marty cohen: the sophomore's dream was the inspiration for this question, in fact. –  deoxygerbe Oct 9 '11 at 0:07