Proving a Lebesgue integral exists: $\int_0^{\pi/2}\left(\frac{1}{2(e^x -1)}-\frac{1}{\tan(x)\sin(x)}\right)d\lambda(x)$

Question

Can you tell me how I would go about proving that the Lebesgue integral $$\int_{0}^{\pi/2}\left(\frac{1}{2(e^x -1)} - \frac{1}{\tan(x)\sin(x)}\right)d\lambda(x)$$ exists? I've already shown that $\dfrac{1}{2(e^x -1)} - \dfrac{1}{\tan(x)\sin(x)}$ is continuous over the interval $(0,\pi/2)$ but I don't know how to proceed from here.

Thank you for the help.

Answer 1

As written, the given integral is divergent since the integrand, as $x \to 0^+$, admits the Laurent series expansion $$ \frac{1}{2(e^x -1)} - \frac{1}{\tan(x)\sin(x)}=-\frac1{x^2}+\frac1{2x}-\frac1{12}+O\left(x\right) $$ which is not Lebesgue integrable over $(0,\pi/2)$.