I want to solve the following integral by using the DCT (Dominated Convergence Theorem). We introduce the sequence: $f_n(x) = 1_{[0,1]}\frac{nx^nsin(nx)-n}{\sqrt{x+2n^2}}$. And we want to calculate $\lim_{n\rightarrow\infty}\int_\mathbb{R}f_nd\lambda$.
If we can show that there is a function $g$ that is integrable such that $|f_n|\leq g$ for all $n \in \mathbb{N}$, then we can use the DCT. So I claim that the function $g = 1_{[0,1]}\cdot \frac{1}{\sqrt{2}}$ is such a function.
$ |f_n| = |1_{[0,1]}\frac{n(x^nsin(nx)-1)}{\sqrt{x+2n^2}}| \leq |1_{[0,1]}\frac{n}{\sqrt{2n^2}}| = |1_{[0,1]}\sqrt{\frac{n^2}{{2n^2}}}| =|1_{[0,1]}{\frac{1}{\sqrt{2}}}| = 1_{[0,1]}{\frac{1}{\sqrt{2}}}= g$
We use that $|x^nsin(nx)-1|$ can not be bigger than $1$ (because of the domain of x). We also use that $\sqrt{x+2n^2} \leq \sqrt{2n^2}$, this once again follows from the domain of x.
So now that we've controlled the conditions for the DCT and we know that we can use the DCT, it's time to use it.
We need to find the limit $\lim_{n\rightarrow\infty} \frac{n(x^nsin(nx)-1)}{\sqrt{x+2n^2}}$. I claim that this is equal to $-\frac{1}{2}\sqrt{2}$. This follows from the following:
$\frac{n(x^nsin(nx)-1)}{\sqrt{x+2n^2}} = \frac{\frac{x^nsin(nx)}{n}-1}{\sqrt{\frac{x}{n^2}+2}} \rightarrow^{n\rightarrow\infty} \frac{-1}{\sqrt{2}} = -\frac{1}{2}\sqrt{2}$.
Thus now, the integral we want to calculate is: $\int_{[0,1]}-\frac{1}{2}\sqrt{2} \ \text{d}\lambda = (1-0)\cdot-\frac{1}{2}\sqrt{2} = -\frac{1}{2}\sqrt{2}$
My questions are:
Is this the correct way to approach such a question?
How exactly can I calculate the Lebesgue-integral without making use of the Riemann integral?
Thanks for your time,
K. Kamal.