Lebesgue dominated convergence integral

Question

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.

Answer 1

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

Answer 2

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.