Calculate improper Lebesgue integral

Question

Trying to learn Lebesgue integrals, and want to calculate this integral:
$$\int_0^\infty \cos(2x)e^{-x}\,d\lambda(x)$$
where $\lambda$ denotes the Lebesgue measure.
My guess is that Riemann integrable if for every interval $[0,N],\; N\in \mathbb{N}$, we have that $\lim_{N\to \infty} \int_0^N |f(x)|\,dx<\infty$, then the Riemann integral is equal to the Lebesgue. But then how do I go on from there?

Answer 1

$\int_0^{\infty } \cos (2x) e^{-2x} \, d\lambda (x) =\lim _{N\to \infty} \int_0^{N} \cos (2x) e^{-2x} \, d \lambda (x)$ in the Lebesgue sense. And $\int_0^{N} \cos (2x) e^{-2x} \, d \lambda (x)$ can be evaluated as a Riemann integral because Riemann integrability implies Lebesgue intergability (and integrals in the two senses are equal). Use two integration by parts or use a standard table of integrals for this Riemann integral.

Answer 2

We have that: $$ \forall x\in\mathbb{R}_+, |\cos(2x)e^{-x}|\leq e^{-x} $$ And $x\mapsto e^{-x}$ is Lebesgue measurable, thus $x\mapsto\cos(2x)e^{-x}$ is Lebesgue measurable. Then: $$ \begin{align} \int\limits_{\mathbb{R}_+}\cos(2x)e^{-x}dx&=\left[\frac{1}{2}\sin(2x)e^{-x}\right]_0^\infty-\frac{1}{2}\int\limits_{\mathbb{R}_+}\sin(2x)(-e^{-x})dx \\ &= 0 + \frac{1}{2}\left(\left[\frac{-1}{2}\cos(2x)e^{-x}\right]_0^\infty-\frac{1}{2}\int\limits_{\mathbb{R}_+}\cos(2x)e^{-x}dx\right) \\ &= \frac{1}{4}-\frac{1}{4}\int\limits_{\mathbb{R}_+}\cos(2x)e^{-x}dx \end{align} $$ In conlusion: $$ \int\limits_{\mathbb{R}_+}\cos(2x)e^{-x}dx=\frac{1}{5} $$ .