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My textbook analysis wants to show with an example that demanding uniform continuity to swap limits and integrals is nog always needed. (as introduction/motivation to Lebesgue integration)

In the example they show that: $$\lim_{n\to \infty} \int_0^1 \frac{1}{\sqrt{x}e^{nx}} = \int_0^1 \lim_{n\to \infty}\frac{1}{\sqrt{x}e^{nx}} \quad\quad (n \in \mathbb{N})$$

I agree that the RHS equals 0 (the limit heads of to $0$ as $n \to \infty$). But why would the LHS equal zero? The book states:

"... It also appears that $\lim_{n\to \infty} \int_0^1 f_n =0$, meaning that it is allowed to swap integral en limit. However there is no uniform convergence ... "

I've checked the LHS with Wolfram Alpha, is there a way to calculate the integral of the LHS manually though (without swapping limit and integral)?

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  • $\begingroup$ Are you asking how to evaluate $\Large{\int_0^1 \frac{dx}{\sqrt{x} e^{n x}}}$? $\endgroup$ – Antonio Vargas Apr 8 '14 at 15:20
  • $\begingroup$ Indeed, that would be the problem. $\endgroup$ – dietervdf Apr 8 '14 at 16:07
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You could use $e^{nx} \ge 1 + nx$. Then for any $a > 0$ you have $$ \int_0^1 \frac{1}{\sqrt x e^{nx}} \, dx \le \int_0^a \frac{1}{\sqrt{x}} \, dx + \frac{1}{n} \int_a^1 \frac{1}{x \sqrt{x}} \, dx.$$ The first integral equals $2 \sqrt a$ and the second integral equals $\displaystyle \frac 2n \left[ \frac {1}{\sqrt a} - 1 \right]$. Thus $$ \int_0^1 \frac{1}{\sqrt x e^{nx}} \, dx \le 2 \sqrt a + \dfrac{2}{n\sqrt{a}}.$$ In particular, with $a = \dfrac 1n$ you have $$0 \le \int_0^1 \frac{1}{\sqrt x e^{nx}} \, dx \le \frac{4}{\sqrt{n}}.$$

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  • $\begingroup$ Okay, that's a pretty cool method :). But why is $e^{nx} \geq 1+nx$? $\endgroup$ – dietervdf Apr 8 '14 at 16:06
  • $\begingroup$ Look at the Taylor series of $e^{nx}$... $\endgroup$ – Umberto P. Apr 8 '14 at 16:07
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After playing around a bit I found the following, would this be valid?

Looking at $\int_0^1 x^{\frac{-1}{2}} e^{-nx}dx$ and substituting $nx = t$ would give: $$= \int^n_0 t^{\frac{-1}{2}}\sqrt{n} e^{-t} \frac{dt}{n}$$

With the consequence: $$\lim_{n\to \infty} \frac{1}{\sqrt{n}} \int_0^n t^{\frac{-1}{2}}e^{-t}dt = 0\cdot \Gamma\left(\frac{1}{2}\right) = 0.$$

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  • $\begingroup$ Yup, looks good to me! $\endgroup$ – Antonio Vargas Apr 8 '14 at 16:17

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