# Lebesgue dominated convergence integral

How would one compute this integral using Lebesgue dominated convergence theorem? $$\lim_{n\rightarrow \infty}\Big[\int_0^n\Big( 1+\frac{2x}{5n} \Big)e^{-x/2}dx\Big]$$ My understanding is we can't just set upper limit of the integral to be infinity and replace the integrand with just $e^{-x/2}$. But the upper limit being $n$ confuses me a lot.

• Hint: $\int_A f(x) = \int_{\mathbb R} 1_A(x) f(x)$ ... – user251257 Nov 21 '16 at 19:51

## 2 Answers

Let $f_n(x)=\left(1+\frac{2x}{5n}\right)e^{-x/2}\cdot\mathbb1_{(0,n)}(x)$.

Clearly we have for all $n$ and $x>0$ that $0 \le f_n(x)\le f(x)=\left(1+\frac{2}{5}x\right)e^{-x/2}$

And we can prove by integrating by parts that $\int_0^\infty f(x) dx=\frac{18}{5}<\infty$

So $|f_n|\le f$ which is integrable.

Also for $n>x$ we have $f_n(x)=\left(1+\frac{2x}{5n}\right)e^{-x/2}\rightarrow e^{-x/2}$ when $n\rightarrow\infty$

Then $\lim_{n\rightarrow\infty}\int f_n(x)dx=\int\lim_{n\rightarrow\infty}f_n(x)dx=\int_0^\infty e^{-x/2}dx=2$

• Thank you very much. Just one question to clarify, can always say that $\lim_{n \rightarrow \infty} \chi_{(0,n)}(x)=\chi_{(0,\infty)}(x)$? I'm assuming this is one of the identities you are using in the solution to find the limit function. – BoBoB Nov 21 '16 at 20:10
• You might consider replacing $n>x$ with $x\le n$ in the second to last sentence. Everything else looks good. +1 ... -Mark – Mark Viola Nov 21 '16 at 20:12
• Yes. Observe that for $x>0$ you may have only $\chi_{(0,n)}(x)$ only for $n\le x$ (which is a finite number). As soon as we have $n>x$ all the sequence terms are $1$, so the limit goes to 1. – Momo Nov 21 '16 at 20:15
• @Dr. MV I'm not sure why do you say I should use $\le$ instead of $<$? – Momo Nov 21 '16 at 20:20
• Because the integral extends to $x=n$. – Mark Viola Nov 21 '16 at 20:21

Hint: Write the integral as $\int_0^\infty \left(1 + \frac{2x}{5n}\right) e^{-x/2} I\{x\le n\}\,dx$ and apply the dominated convergence theorem.