One possibility is the integration by substitution formula. But this doesn't explain why $\int\_1^{\infty}x^{-3}dx$ is easy to evaluate (and rational) whilst $\sum\_1^{\infty}n^{-3}$ is unknown (and seemingly not a simple combination of well known numbers). It is true that $x^{-3}$ has a simple inverse derivative whilst $n^{-3}$ does not have a known simple inverse difference, but is there more to it than that?



Is there a reason why integrals are so much easier to evaluate than sums?
Well, some things are easier in one place, and other things are easier in the other. The many assertions in answers that integrals of $x^n$ are easier than the corresponding sums is misleading. In the continuous world, $x^n$ is a very natural object. In the discrete world, however, the natural object is the falling factorial: $x^{\underline{n}}$ is defined as $x(x-1)\dots(x-n+1)$. Compute the derivative of $x^{\underline{n}}$ and you get a mess. Compute the forward difference and you get the beautifully simple $f(x+1)-f(x) = (x+1)^{\underline{n}} - x^{\underline{n}} = n x^{\underline{n-1}}$.


That is, 
 $$\sum\_{i=1}^n i^9= \frac{1}{20} n^2 (n+1)^2 \left(n^2+n-1\right) \left(2 n^4+4 n^3-n^2-3 n+3\right)$$ 
is a little messy. Certainly messier than $\int x^9 dx$. On the other hand,  

 $$\sum\_{i=1}^n i^{\underline{9}} = \frac1{10} (n+1)^{\underline{10}}.$$
while the integral $\int x^{\underline{9}}dx$ is bad enough that I dare you to work it out by hand.


The truth here is that the discrete world is often easier (and more fundamental, too), but history has prejudiced us to be more familiar with the continuous world. [Zeilberger](http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/real.html) (who else?) has written passionately about this, and if you know of other sources for discrete calculus (online or off) please add them as comments to this answer!


