# Limit of an integral

• Jan 4th 2013, 06:56 AM
problem
Limit of an integral
Show that $\lim_{n \to \infty} \int_0^1 \frac{x^2e^{x^n}}{1+x^3} dx=\frac{1}{3} \ln 2$
• Jan 4th 2013, 08:21 AM
topsquark
Re: Limit of an integral
Quote:

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
• Jan 4th 2013, 08:38 AM
problem
Re: Limit of an integral
It is $e^{(x^n)}$.
• Jan 4th 2013, 09:34 AM
GJA
Re: Limit of an integral
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!