Show the existence and compute $\lim_{n \rightarrow \infty} \int_{(0,\infty)}(\frac{ne^x+1}{ne^{2x}+4x^2})dλ(x)$

Question

I have to determine whether the limit for $n \rightarrow \infty$ exists and, if so, I have to compute it

The Integral is:

$$K_n= \int_{(0,\infty)}\left(\frac{ne^x+1}{ne^{2x}+4x^2}\right)dλ(x)$$

I want to check if I have done everything right. So I can use the theorem of dominant convergence. Since $f_n=\frac{ne^x+1}{ne^{2x}+4x^2}$ are measurable $f_n$ converges to $\frac{1}{e^x}$ pointwise and $f_n \leq g(x):=2$ we can use the theorem:

we have that the limit exists and it is

\begin{align}\lim_{n \rightarrow \infty } \int_{(0,\infty)}\left(\frac{ne^x+1}{ne^{2x}+4x^2}\right)dλ(x)&=\int_{(0,\infty)}\lim_{n \rightarrow \infty }\left(\frac{ne^x+1}{ne^{2x}+4x^2}\right)dλ(x)\\&= \int_{(0,\infty)} \frac{1}{e^x}dλ(x)\end{align}

Is everything right or have I done some errors?

Answer 1

No you cannot say, that as $|f_{n}|\leq 2$ , you can apply DCT. That is not true. You need $g$ to be integrable. A constant function is not integrable unless the measure space is finite.

What you can to do is that $|f_{n}|\leq e^{-x}+\frac{1}{n}e^{-2x}\leq e^{-x}+e^{-2x}$ which is integrable and hence by the Dominated Convergence Theorem , you have the required conclusion such that $\lim\int_{(0,\infty)}f_{n}\,d\lambda = \int_{(0,\infty)}e^{-x}\,d\lambda=1$

Answer 2

Your uniform upper bound $g$ has to be integrable over $(0,\infty)$, which is not the case for $2.$

Replace it by $$\frac{e^x+1}{e^{2x}}.$$