## anonymous 4 years ago 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 $\lim_{n\to\infty} n\int_0^1x^nf(x)dx$

1. JamesJ

The first one is clearly zero. The second one .... one sec

2. JamesJ

f(1)

3. anonymous

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

4. JamesJ

Yes, I know. But I can write the proofs for these. These integrals are actually a kind of 'primitive' form of the Laplace transform.

5. JamesJ

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.

6. anonymous

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)?

7. JamesJ

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

8. JamesJ

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.

9. JamesJ

Anyway. What's the other question?

10. anonymous

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

11. JamesJ

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

12. JamesJ

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