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} $
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