Functions with integral as an infinite sum of its own values

Question

Just curious, how can we figure out functions satisfying the following condition?

$\sum_{i=1}^{\infty}f(n)=\int_0^1f(x)dx$

I got one, f(x) = x^-x (I'll post my solution on request)

Are there more functions satisfying the same condition?

Answer 1

Why, there are many. $f(x)$ that equals $1$ for $x\in[0,1)$ and $0$ elsewhere will do. $f(x)=e^{-ax}$ with carefully chosen constant $a\approx0.930821$ will do. There are more such functions than you seem to think, and more than you will be thinking after reading these words, and even more than that.

Sophomore's dream (as your example is commonly called) has been famous for quite a while due to the nature of the function, which would not let us to come up with a closed form either for the sum or the integral, and yet the two are equal. Remove the function; what remains is trivial. It does not even make sense to look for other such functions.

Really, take an arbitrary function with finite integral on $[0,1]$. Extend it with arbitrary values at 3, 4, 5, and so on, just make sure that the sum is finite, too. Now assign $f(2)$ so as to make the sum equal to the integral. You're done. Oh, and you may define the function in an arbitrary way at the remaining non-integer points outside $[0,1]$, if you would like, or just leave it like that. Doesn't sound very exciting, does it?

So it goes.