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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?

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$ |\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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  • $\begingroup$ Next time make sure to write the absolute value in the first place $\endgroup$ – Jytug Nov 27 '14 at 20:42
  • $\begingroup$ sry, my mistake now. you've totally answered my question ;) $\endgroup$ – per Nov 27 '14 at 20:46
  • $\begingroup$ I believe it is okay: one minus comes from integrating by parts, and another one comes from the antiderivative of $ \exp(-x) $ $\endgroup$ – Jytug Nov 27 '14 at 21:16
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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

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