# Dominated convergence theorem for $\lim\int_{0}^{n}(1-x/n)^{n}x^sdx$, $s<-1$.

When I was studying the dominated convergence theorem, I was glad to come across this problem:

Use the dominated convergence theorem (DCT) to show that $$\lim \int_{0}^{n} \left(1-\frac{x}{n} \right)^{n}x^sdx=\int_{0}^{\infty}e^{-x}x^sdx$$ where $s<-1$.

I let $\displaystyle f_n(x)= \left(1-\frac{x}{n} \right)^{n}$, clearly $\lim f_n(x)=e^{-x}$. And $e^{-x}$ is integrable.

The book I am using stated the dominated convergence theorem as follows:

Theorem: Let $f_n$ be a sequence of measurable functions such that $f_n \to f$ almost everywhere. If there exist a real valued function $g$ define on a measure space such that for each $n$, $|f_n| \le g$ almost everywhere then $f$ is integrable and $$\int f=\lim\int f_n$$

To be able to use the DCT it necessary to find any real valued function $g$ that dominate $f_n$. That is where I get stuck, can somebody help me out?

EDIT: The $\lim$ here means limit as n tend to infinity.

-
What's the limit? $n \to \infty?$ –  Argon May 21 '12 at 15:35
Your statement of the Dominated Convergence Theorem isn't quite correct. You need $g$ to be absolutely integrable; it cannot just be any real-valued function. –  Henry T. Horton May 21 '12 at 15:44
Note that $(1-x/n) \leq \exp(-x/n)$ for all $x \in [0,1)$ and hence $(1-x/n)^n \leq \exp(-x)$. –  user17762 May 21 '12 at 15:45
$\Huge \text{Huge font}$ :D Could you make it a bit $\small \text{smaller}$? –  dtldarek May 21 '12 at 15:45
@HassanMuhammad Yes. To make the argument clear, $$\int_0^n (1-x/n)^n x^s dx = \int_0^{\infty} f(x) dx \leq \int_0^{\infty} \exp(-x)x^s dx$$ where $$f(x) = \begin{cases} (1-x/n)^n x^s & \text{ for } 0 \leq x \leq n \\ 0 & \text{ otherwise} \end{cases}$$ Also as Didier points out, you need $s>-1$. Otherwise $\displaystyle \int_0^{\infty} \exp(-x)x^s dx$ diverges. –  user17762 May 21 '12 at 16:00

Let me write out my comment so that this question gets an answer. The dominated convergence theorem says the following.

If a sequence of measurable functions $\displaystyle \{f_n(x)\}_{n=1}^{\infty}$, on the measure space $(\Omega, \mathcal{F}, \mu)$, converge point-wise to a function $f(x)$ and every $f_n(x)$ is dominated by some integrable function $g(x)$ i.e. $\lvert f_n(x) \rvert \leq g(x)$, $\forall n$, $\forall x \in \Omega$ and $\displaystyle \int_{\Omega} g d\mu < \infty$, then $\displaystyle \int_{\Omega} f d \mu$ exists and $$\lim_{n \rightarrow \infty} \int_{\Omega} f_n d \mu = \int_{\Omega} f d \mu$$

Note that it is important that the dominating function $g$ is integrable as Henry T. Horton points out in the comments. Also, the condition $\lvert f_n \rvert \leq g$ everywhere, can be relaxed to $\lvert f_n \rvert \leq g$ $\mu$-almost everywhere.

Also, we need $s > -1$. Otherwise the integrals given in the question diverge.

Now lets apply this to the problem at hand. We want to evaluate $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx$.

Define $$f_n(x) = \begin{cases} \left( 1 - \frac{x}n\right)^n x^s & \text{ if }x \in [0, n]\\ 0 & \text{ otherwise}\end{cases}$$

Hence, $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx$.

Now note that $f_n(x)$ converges point-wise to $e^{-x} x^s$ on $[0, \infty)$. This is so since $$\displaystyle \lim_{n \rightarrow \infty} \left(1 - \frac{x}{n} \right)^n = \exp(-x).$$

More importantly, $f_n(x)$ is dominated by $e^{-x} x^s$ on $[0,\infty)$ i.e. $\lvert f_n(x) \rvert \leq e^{-x} x^s$. This follows from the fact that $$1 - t \leq \exp(-t)$$ whenever $0 \leq t \leq 1$. Hence, we get that $$\left(1 - \frac{x}{n} \right) \leq \exp \left(-\frac{x}{n} \right)$$ which in-turn gives us $$\left(1 - \frac{x}{n} \right)^n \leq \exp \left(-x \right).$$ Hence, $\displaystyle f_n(x) \leq \exp(-x) x^s$. Also, $\displaystyle \int_0^{\infty} \exp(-x) x^s dx < \infty$ for all $s > -1$. Hence, in this case, the dominating function is the same as the limit function.

Putting all these things together, we get the desired result.

\begin{align} \displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx & = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx & \text{(From the definition of $f_n(x)$)}\\ & = \int_0^{\infty} \lim_{n \rightarrow \infty} f_n(x) dx & \text{(Since $f_n(x)$ is dominated by $f(x)$)}\\ & = \int_0^{\infty} f(x) dx & \text{(Since $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = f(x)$)}\\ & = \int_0^{\infty} \exp(-x) x^s dx \end{align}

-
Thanks for the answer, I really appreciated it. –  Hassan Muhammad May 21 '12 at 21:05
How can I know $\int_{0}^{\infty}e^{-x}x^sdx \lt \infty$ for $s>-1$? –  Hassan Muhammad May 22 '12 at 19:40
$e^{-x} < \frac1{P_{s+2}(x)}$ where $P_k(x)$ is a polynomial in $x$ of degree $k$. Now prove that the integral $\int_0^{\infty} \frac{x^s}{P_{s+2}(x)} dx$ is convergent. –  user17762 May 22 '12 at 19:51
Is $P_{k}(x)$ any polynomial of degree k? Take for example $P(x)=x^2$. –  Hassan Muhammad May 22 '12 at 20:18
For instance, if $s=0$, you can choose $P_2(x) = 1+x^2$. Make sure that there is no singularity at the origin when you choose $P_k(x)$. –  user17762 May 22 '12 at 20:26