When is this function Lebesgue integrable (based on variables)?

Question

I didn't know how to start with this basic question :

Let $a,b > 0$ and $f$ a function :

$$f(x)={xe^{-ax}\over 1-e^{-bx}}.$$

Find $a$ and $b$ so that $f$ is Lebesgue integrable in $[0,+∞[$ .

Answer 1

I would say that it is integrable for all values of $a,b>0$.

For every $a,b,x>0$ $f(x)$ is positive, so it remains to show that its integral is finite. Let us denote ${\rm d}x={\rm d}\lambda(x)$ where $\lambda$ is the Lebesgue measure.

We may decompose the integral as $$ \int_{[0,+\infty)} \hspace{-20pt}f(x)\,{\rm d}x = \int_{[0,1]}\hspace{-12pt}f(x)\,{\rm d}x + \int_{(1,+\infty)}\hspace{-20pt}f(x)\,{\rm d}x $$ Let us first focus on the first one: as $\lim_{x\to 0^+}f(x)=\frac{1}{b}$, it follows that $f$ (which is defined for $x\neq 0$) can be extended continuously in $[0,\infty)$. In particular, it is continuous on the compact interval $[0,1]$ and therefore $$ \int_{[0,1]}\hspace{-12pt}f(x)\,{\rm d}x < \infty $$ Now, let us notice that for all $a,b>0$ $$ f(x) = \frac{xe^{-ax}}{1-e^{-bx}} \leq \frac{xe^{-ax}}{1-e^{-b}} \quad\forall x>1 $$ Therefore, $$ \int_{(1,+\infty)}\hspace{-20pt}f(x)\,{\rm d}x \leq \frac{1}{1-e^{-b}}\int_{(1,+\infty)}\hspace{-20pt}xe^{-ax}\,{\rm d}x = \frac{1}{1-e^{-b}}\frac{a+1}{a^2}e^{-a} <\infty $$ which completes the proof.

The last computation uses the Riemann integration; remember that a positive Riemann-integrable function is also Lebesgue-integrable.

Answer 2

The function $f$ being continuous on $]0,\infty[$, it is sufficient to study its behaviour at $0$ and $+\infty$.

We have $$ f(x) \sim xe^{-ax}\qquad (x\to \infty) $$ wich is integrable at the neighbourhood of $+\infty$.

And $$ f(x) \rightarrow \frac{1}{b}\qquad(x\to 0). $$ wich is integrable at the neighboorhood of $0$ (actually, $f$ can be continuously continuated in $0$).

From this, we conclude that $f$ is integrable on $]0,\infty[$.