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- May 21st 2008, 07:44 PMszpengchaohelp with this integral problem
- May 21st 2008, 09:56 PMkalagota
for the first one, it just follows directly from $\displaystyle m \leq f(x) \leq M$ by taking the definite integrals in all parts (since integral preserves the direction of the inequality). Also note that constants can go out of the integral sign.

for the second part, since $\displaystyle g(x) \geq 0$, the inequality does not change i.e. $\displaystyle mg(x) \leq f(x)g(x) \leq Mg(x)$ and you can take the integrals.

the next two are applications.. you can do it. - May 21st 2008, 11:09 PMIsomorphism
First apply parts to get,

$\displaystyle \int_{0}^{\frac1{\sqrt{n}}} nf(x) e^{-nx} \, dx = -f(x) e^{-nx} \bigg{|}_{0}^{\frac1{\sqrt{n}}} + \int_{0}^{\frac1{\sqrt{n}}} f'(x) e^{-nx} \, dx$

$\displaystyle \lim_{n \to \infty} -f(x) e^{-nx} \bigg{|}_{0}^{\frac1{\sqrt{n}}}$

$\displaystyle = \lim_{n \to \infty} -f(\frac1{\sqrt{n}}) e^{-\sqrt{n} x} + f(0) = f(0)$

[That is because $\displaystyle \lim_{n \to \infty} -f(\frac1{\sqrt n})e^{-\sqrt{n}x} = 0$ since $\displaystyle e^{-\sqrt{n}x} \to 0$ as $\displaystyle n \to \infty$]

We claim $\displaystyle \int_{0}^{\frac1{\sqrt{n}}} f'(x) e^{-nx} \, dx = 0$

First observe that $\displaystyle 0 < e^{-nx} < 1$

So $\displaystyle 0 < \int_{0}^{\frac1{\sqrt{n}}} e^{-nx}f'(x)\, dx < \int_{0}^{\frac1{\sqrt{n}}} f'(x) \, dx$

$\displaystyle 0 < \lim_{n \to \infty} \int_{0}^{\frac1{\sqrt{n}}} e^{-nx}f'(x)\, dx < \lim_{n \to \infty} f\left(\frac1{\sqrt{n}}\right) - f(0) = f(0)\text{(Why?) }- f(0) = 0$

Thus claim proved by sandwich theorem...

So

$\displaystyle \lim_{n \to \infty} \int_{0}^{\frac1{\sqrt{n}}} nf(x) e^{-nx} \, dx = f(0)$

EDIT: I realised this halfway through...(Worried) My proof is incomplete... I assumed (1) f is differentiable (2) f'(x) > 0.... Let me think about it a little more.... - May 22nd 2008, 05:29 AMkalagota
as i have said, the questions are just applications of the previous.

note that, $\displaystyle f$ is*continuous*in a**closed**and**bounded**interval, then you can use Maximum-Minimum theorem, that is $\displaystyle f$ is also bounded. Also, $\displaystyle f$ is integrable.

Let $\displaystyle m, M$ be real numbers such that $\displaystyle m \leq f(x) \leq M$ for $\displaystyle x \in [0,1]$. Applying the first and second facts, we have $\displaystyle m \int_0^{1/\sqrt{n}} n e^{-nx}dx \leq \int_0^{1/\sqrt{n}} nf(x)e^{-nx}dx \leq M \int_0^{1/\sqrt{n}} n e^{-nx}dx$ (since $\displaystyle 1/\sqrt{n} \in [0,1]$)

i just wonder if it equal $\displaystyle f(0)$ or $\displaystyle 0$ unless $\displaystyle f(0) = 0$ - May 22nd 2008, 12:44 PMIsomorphism