For outsiders, mathematics is mostly the 'art' of manipulating random strings of weird symbols, and two symbols that seem to summarize the quintessential esoterica of mathematics are and . When you get into actually doing some real math, you discover that the aura of mysticism surrounding mostly stems from the fact that mathematicians are horrible at making intuitive symbols. For example, the sigma could've easily replaced by something like But the symbol, often dubbed the 'long-S', actually has some meaning of its own. It doesn't just rephrase some well known idea in a strange way; it indicates a very powerful and novel idea that was never encountered before. In this post, we're going to see some proofs where the idea of integrals makes a stellar appearance.


Have you ever had a friend (or worse, a teacher) claiming that equals ? Quite an irrational statement, isn't it? I know how desperate and annoying it is to try to convince them of what's really going on. So here's a novel way to silence your friend once and for all.

Consider the definite integral . It can be evaluated as follows (mostly copied from Wikipedia):

And here's the surprise. Since the integrand is positive on , the integral must be positive, implying .

The question of evaluating the integral was asked on Putnam 1968. It is surprisingly simple, no doubt, but the inference drawn is simply surprising.



Let's say we wish to obtain a good lower bound on the expression . Why we need that is irrelevant, but it can of use in, say, some analytical number theory proof. Specifically, we wish to prove that Note how it would give us a bound of roughly for the original expression.

But how do we prove this? There are many ways, but the following proof is particularly nifty:

Call the expression simply , and consider the integral When we expand out the expression , the largest degree of would be , so the integrals of the terms would be of the form for integers and . That means would be an integer; but it's clearly positive, so . Now note that AM-GM gives , so This proves our claim.

I got to know of this proof at the IMOTC this year; don't you think it's too clever not to share?



The calculation of the antiderivative of is one of the first exercises you'd encounter after studying integration by parts in a calculus class. It's easy to work out, no doubt; but what follows is another simple proof which I find particularly elegant.

But first, I must mention something called Fubini's theorem in analysis (if you already know what it is, feel free to skip this paragraph). In normal calculus done on 2 dimensional -plane, integrals are used to compute areas under curves. When you move on to 3 dimensions, you might be curious about the volume under the graph of some function . There are two, morally dual ways of computing this: you might divvy up the volume in thin sheets parallel to the -plane , compute each of them by normal 2-d integrals, and sum over them in the direction of -plane. Or you do this other way round; first dividing parallel to -plane and then integrating parallel to -plane. It's intuitively obvious that both processes should yield the same answer; and that's what Fubini's theorem says. Informally speaking, this allows us to 'switch' integrals as we often do with sums (under certain conditions, but we won't go into the details). It's the continuous version of double-counting, if you will.

Now we move to the proof. Note that , so we have Now what does the last integral mean? It's the volume under the surface , but computed with rather weird limits. For each , the inside integral goes from to , while itself moves from to . For an intermediate value of , say , goes only up to . It's not hard to graph the whole thing and see that this means travels inside the triangle with vertices at . So when we switch the integral, they must be done in a peculiar way:That's the same thing we'd get from Integration by Parts.

This beautiful proof is stolen from Math.SE.