Show 1/(exp(x) -a) sin x is integrable over [0, infinity)

**Amanda1990** · February 22nd 2009, 01:35 AM

I can show that for n > 0, $exp(-nx) \sin x$ is Lebesgue integrable over $[0,\infty)$ and that $\int_{0}^\infty exp(-nx) \sin dx = (1+n^2)^{-1}$ .

Now I need to show that for $0 \leq a \leq 1, (exp(x) -a )^{-1} sin x$ is integrable over $[0, \infty)$ and that

$\int_{0}^\infty (exp(x) -a)^{-1} \sin x dx = \sum a^{n-1}/(1+n^2)$ , where the sum is from n=1 to infinity. I'm struggling over both the last bits - what's the idea here?

**Media_Man** · May 30th 2009, 05:41 AM

$\int_0^\infty \frac1{e^x-a}\sin x dx$ = $\int_0^\infty e^{-x}\frac{1}{1-ae^{-x}}\sin x dx$ = $\int_0^\infty e^{-x}(\sum_{n=1}^\infty a^{n-1}e^{-(n-1)x})\sin x dx$ = $\sum_{n=1}^\infty a^{n-1}\int_0^\infty e^{-nx}\sin x dx$ = $\sum_{n=1}^\infty a^{n-1}\frac1{1+n^2}$ QED

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Show 1/(exp(x) -a) sin x is integrable over [0, infinity)

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