Show that $\displaystyle \lim_{n \to \infty} \int_0^1 \frac{x^2e^{x^n}}{1+x^3} dx=\frac{1}{3} \ln 2$

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- Jan 4th 2013, 05:56 AMproblemLimit of an integral
Show that $\displaystyle \lim_{n \to \infty} \int_0^1 \frac{x^2e^{x^n}}{1+x^3} dx=\frac{1}{3} \ln 2$

- Jan 4th 2013, 07:21 AMtopsquarkRe: Limit of an integral
- Jan 4th 2013, 07:38 AMproblemRe: Limit of an integral
It is $\displaystyle e^{(x^n)}$.

- Jan 4th 2013, 08:34 AMGJARe: Limit of an integral
Hi problem,

We can get at the exercise using the Lebesgue Dominated Convergence Theorem. Let $\displaystyle f_{n}=\frac{x^{2}e^{x^{n}}}{1+x^{3}},$ and note that $\displaystyle f_{n}\geq 0$ for all $\displaystyle n$ so that $\displaystyle |f_{n}|=f_{n}$ for all $\displaystyle n.$

1) Since we're working on $\displaystyle [0,1]$, $\displaystyle x^{n}\leq 1$ always. Hence, $\displaystyle |f_{n}|=f_{n}\leq \frac{x^{2}e}{1+x^{3}},$ which is integrable on $\displaystyle [0,1];$ i.e. $\displaystyle \frac{x^{2}e}{1+x^{3}}$ is our "dominating function."

2) Since we're working on $\displaystyle [0,1] $, $\displaystyle x^{n}\rightarrow 0$ as $\displaystyle n\rightarrow\infty$ for all $\displaystyle x\neq 1.$ Furthermore, $\displaystyle 1^{n}=1$ for all $\displaystyle n.$ Hence, the pointwise limit of the sequence $\displaystyle \{f_{n}\}$ is $\displaystyle f(x)=\frac{x^{2}}{1+x^{3}}$ for $\displaystyle x\neq 1$ and $\displaystyle f(1)=\frac{e}{2}$. Now the set $\displaystyle \{1\}$ is of measure 0, so $\displaystyle f(1)=\frac{e}{2}$ is actually irrelevant as far as the integral is concerned.

Now it is a matter of checking to see if we've satisfied the conditions of Lebesgue's DCT, applying it, then evaluating the integral of the limit function properly to get what we're after.

Does this get things going in the right direction? Let me know if anything is unclear. Good luck!