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

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?

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

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

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

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