Integral of the sequence of functions $f_n(x)=ae^{-nax}-be^{-nbx}$

Question

Let be the sequence of functions $f_n:(0, +\infty) \rightarrow \mathbb{R} \ \ (n\geq1)$ $$f_n(x)=ae^{-nax}-be^{nbx}, \ \ \ x>0, \ \ \ \ \ 0<a<b$$ Prove that: $$a) \ \ \forall n\geq 1 \ \int_0^{+\infty}|f_n|<\infty \ and \ \int_0^{+\infty}f_n=0$$ $$b) \ \ \sum_{n\geq1}\int_0^{+\infty}|f_n|=+\infty$$ Prove that the series $\sum_{n\geq1}f_n(x)$ is convergent for $x >0$, and that the series defines a function $f:(0, +\infty)\rightarrow \mathbb{R}$ as: $$f(x)=\sum_{n\geq1}f_n(x)$$ Prove that $\int_0^{+\infty}|f|<+\infty$ and that $\int_0^{+\infty}f=\log(b/a)$.

So, I managed to prove the point $(a)$ using the triangle inequality, and explicitly calculating the integral. I also proved that $(b)$ by proving that $\int_0^{+\infty}|f_n| \geq \frac1n$ by using the reverse triangle inequality and explicitly calculating the integral. Now I'm stuck on the last two points, and I don't know how to continue, because if try to explicitly calcualte the sum and then the integral, and to use the triangle inequality I found and upper bound that is divergent. Any help?

Answer 1

By summing the resulting geometric series, we can find that $f(x)=\frac{a}{e^{ax}-1} - \frac{b}{e^{bx}-1}$. Write $\int_0^\infty |f|=\int_0^1 |f|+\int_1^\infty |f|$. For the first term, the only point that is potentially problematic is around zero. But notice that in a right-neighborhood of zero, we have that $f(x)=\frac{1}{2}(b-a)+\mathcal{O}(x)$ by a Taylor expansion. So the first term is finite. For the second term, use the fact that for $x\ge 1$, $|f|\le \frac{e^{-ax}}{1-e^{-a}} + \frac{e^{-bx}}{1-e^{-b}} \le C e^{-ax} $ for some $C(a,b)>0$. Hence the second term is finite, and the left hand side as well. I don't have an answer right now for the second question.

Answer 2

notice that: $$\sum_{n\ge1}f_n=a\sum_{n\ge1}\left(e^{-ax}\right)^n-b\sum_{n\ge1}\left(e^{-bx}\right)^n$$ now since $a,b>0$ you see that for $x>0\Rightarrow 0<e^{-ax},e^{-bx}<1$ and so you have a geometric series (whose convergence can be easily justified) so use the formula for sum of a geometric series and you will get your result. Now it should be easy to get an explicit formula for $f$ and you will see that $\int_{\mathbb R^+}|f|$ follows easily.

As for the final part, consider a function: $$I(a,b)=\int_0^\infty f\,dx$$ this reminds me somewhat of the Frullani integral but not quite