Limit of an integral

**problem** · January 4th 2013, 05:56 AM

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

**topsquark** · January 4th 2013, 07:21 AM

Originally Posted by problem

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

I can't solve it anyway, but to clarify: Is $e^{x^n}$
$= e^{(x^n)}$

or
$= \left ( e^x \right ) ^n$

-Dan

**problem** · January 4th 2013, 07:38 AM

It is $e^{(x^n)}$ .

**GJA** · January 4th 2013, 08:34 AM

Hi problem,

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

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

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

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