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

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

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

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 $\frac{n}{n^2 +1}$