2

I'm working on Project Euler, problem 48:

The series, 11 + 22 + 33 + ... + 1010 = 10405071317.

Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.

This would be the equivalent of: $$ f(n) = \sum_{i = 1} ^ n i^i $$

The original question aside, it got me thinking. The rate of change from n to n+1 would be i^i, so if I could take the integral of ii, I could get a generalized formula for any given n. Upon plugging it into Wolfram Alpha, I got a monstrous expansion, but it said, "no result found in terms of standard mathematical functions". It's been years since I've taken calculus and I'm mostly a computer science guy to begin with, but is this premise sound? And if not, where did I go wrong?

Dan D
  • 213
  • 3
  • The integral approximates the sum, but doesn't equal it exactly. 2. The integral is a beast, with no nice value in terms of familiar functions.
    • – vadim123 Nov 04 '14 at 19:42
  • 1
    In terms of actually solving the problem this is a red herring – amcalde Nov 04 '14 at 19:45
  • I've used a completely unrelated method to solve the problem. This was just a curious thing. – Dan D Nov 04 '14 at 19:46
  • The function is difficult / impossible to integrate using pen and paper. And the integral is related to the sum, but it's not exactly the same. For instance, the difference between the sum $\sum_{i=1}^{n}\frac1i$ and the integral $\int_1^n\frac1xdx$ closes in on what is called the Euler-Mascheroni constant. Likewise, the difference between your sum and your integral is non-zero, although I summer in your car that difference diverges. – Arthur Nov 04 '14 at 19:46