Question

Here's a fun integral question I made for a competition a few years ago. I was inspired by this much easier question:

Warmup: Let f(x) be a continuous real-valued function on [0,1]. Calculate: \[\lim_{n\to\infty}\int_0^1x^nf(x)dx\]

Now for the real question: Let f(x) be a continuous real-valued function on [0,1]. Calculate \[\lim_{n\to\infty} n\int_0^1x^nf(x)dx\]

Answer 1

The first one is clearly zero.

The second one .... one sec

Answer 2

f(1)

Answer 3

Didn't take too long. I meant to say "calculate with proof" but I suppose that is trivial once you know the answers.

Answer 4

Yes, I know. But I can write the proofs for these.

These integrals are actually a kind of 'primitive' form of the Laplace transform.

Answer 5

we usually have e^(-st) as the function we're integrating against. If we write x = e^-s, then that's how we get back to these integrals.

Answer 6

Makes sense, never real thought about that connection too much. Very clever!

Want a harder question (that should take longer than 10 seconds for you to solve)?

Answer 7

Talking of writing proofs though, writing a good proof for that first integral isn't to obvious, I don't think.

Answer 8

It seems to me you need to write it as
\[ \int_0^{1-\delta} + \int_{1-\delta}^1 \]
Showing the first piece goes to zero is straightforward as the function \( g(x) = x^n f(x) \) converges uniformly to the zero function on \( [0,1-\delta] \). Then you just need to deal with the the other bit.

Answer 9

Anyway. What's the other question?

Answer 10

I don't think all of that machinery is too necessary. Since the function is continuous on a closed interval, it must be bounded. Then some manipulations followed by a squeeze theorem will give you the result.

Anyhoo, I'll type up the next question (an inequality) real quick

Answer 11

Yes, you don't need the uniform convergence. But I'd still break it up into the two pieces.

Answer 12

...which isn't to say there isn't another way to skin that cat. Just the way I'd do it.