calculate $\lim_{n\to\infty}\int_{[0,\infty)} \exp(-x)\sin(nx)\,\mathrm{d}\mathcal{L}^1(x)$

Question

We've had the following Lebesgue-integral given:

$$\int_{[0,\infty)} \exp(-x)\sin(nx)\,\mathrm{d}\mathcal{L}^1(x)$$

How can you show the convergence for $n\rightarrow\infty$?

We've tried to use dominated convergence but $\lim_{n\rightarrow\infty} \sin(nx)$ doesn't exist. Then we've considered the Riemann-integral and tried to show that $$\int_0^\infty |\exp(-x)\sin(nx)| \,\mathrm dx $$ exists but had no clue how to calculate it. So how can you show the existence of the Lebesgue-integral and calculate it?

Answer 1

$ |\exp(-x)\sin(nx)| \leq \exp(-x) $

Moreover, you can easily compute the integral for arbitrary $n$ by integrating by parts twice:

$$ \int_{[0,\infty)} \exp(-x)\sin(nx) = -\exp(-x)\sin(nx) |_{0}^{\infty} +n\int_{[0,\infty)} \exp(-x)\cos(nx) $$

$$ \int_{[0,\infty)} \exp(-x)\sin(nx) = n\int_{[0,\infty)} \exp(-x)\cos(nx) $$

$$ \int_{[0,\infty)} \exp(-x)\sin(nx) = n\exp(-x)\cos(nx) |_{0}^{\infty}-n^2\int_{[0,\infty)} \exp(-x)\sin(nx) $$

$$ (n^2 + 1) \int_{[0,\infty)} \exp(-x)\sin(nx) = n $$

So the integral equals $ \frac{n}{n^2 +1} $

Answer 2

This can be done without computing by Riemann-Lebesgue Lemma

$$\int_{\Bbb R}f(x)\sin(nx)dx\to0$$

for $f$ continuous almost every where