As an excercise of Measure Theory, I have to calculate for all $a\in [-1,1]$ that: $$\int_{(0,+\infty)} \frac{\sin (t)}{e^t - a} d\lambda(t) = \sum_{n\geq 1} \frac{a^{n-1}}{n^2+1} $$
where $\lambda$ denotes the Lebesgue measure.
I think that I should use a Taylor Expansion, but I am not reaching any point. Using the Taylor Expansion of the exponential does not help because of the parameter $a$. And using the Taylor Expansion of $\sin$ and the Beppo-Levi Theorem I reach the point where it should be integrated $$\int_{(0,+\infty)} \frac{t^{2n+1}}{e^t-a} d\lambda(t)$$
Any posible help would be appreciated :)
I think you just can use the geometric series $$\sum_{k=1}^{\infty} (ae^{-x})^k = \frac{a e^{-x}}{1-a e^{-x}} = \frac{a}{e^x - a}$$