# Existence, finiteness of Lebesgue integral on $(0,\infty)$ of $f(x)=\frac{1-e^{-x}}{x}-\frac{1}{1+x}$

Ok, here is the function at hand:

$$f(x)=\frac{1-e^{-x}}{x}-\frac{1}{1+x}$$

My problem is to determine if the Lebesgue integral on the interval $(0,\infty)$ exists, and if it does, whether or not it is finite.

My first issue is some confusion on what it means for the integral to exist in the first place. I feel like finiteness implies existence and vice versa, but the question would not be posed in such a manner if it did not allow for infinite integrals.

In doing these types of problems, I am less interested in the tools to explicitly calculate the integral, but more in understanding the theory. So, if someone could help a self-studier like myself with breaking this problem down, I would be forever grateful.

Specifically, any detail as to the specific justification of why the integral exists (sequences of simple functions, continuous function on a finite interval, etc.) is especially helpful.

From here I am going to attempt a number of similar exercises, but it would go a long way to see how one such exercise can be done.

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As JBC says: for $f$ integrable, there are two parts: $f$ is measurable, and $\int|f| < \infty$. Since $f$ is continuous on $(0,\infty)$ it is measurable, so that part is clear. –  GEdgar Jun 23 '12 at 21:50

Your function is clearly measurable by continuity. \begin{align}\int_0^{+\infty}\left|\frac{1-e^{-x}}{x}-\frac{1}{1+x}\right|dx&=\int_0^{+\infty}\left|\frac{1-e^{-x}-xe^{-x}}{x(1+x)}\right|dx\\&\le\int_0^{+\infty}\left|\frac{1-e^{-x}}{x(1+x)}\right|dx+\int_0^{+\infty}\left|\frac{e^{-x}}{1+x}\right|dx\\&=\int_0^{+\infty}\frac{1-e^{-x}}{x(1+x)}dx+\int_0^{+\infty}\frac{e^{-x}}{1+x}dx\end{align}

Notice that $1-e^{-x}\sim_0x$, then $\frac{1-e^{-x}}{x(1+x)}\sim_0\frac{1}{1+x}$, so the first integral is convergent at $0$. And $\frac{1-e^{-x}}{x(1+x)}\sim_\infty\frac{1}{x^2}$, so the first integral is convergent at $+\infty$.
The second integral is clearly finite : $\frac{e^{-x}}{1+x}\le e^{-x}$ and $\int_0^{+\infty}e^{-x}dx<\infty$.

Conclusion : $$\int_0^{+\infty}\left|\frac{1-e^{-x}}{x}-\frac{1}{1+x}\right|dx<\infty$$ So the lebesgue integral exists and is finite.

Remarks :

1. We first define the Lebesgue integral for measurable functions valued in $[0,\infty]$ (which always exist, but may be $+\infty$).
2. For measurable functions with values in $\mathbb R$, we say that $f$ is Lebesgue integrable if $\int|f|<+\infty$ and then $\int f=\int f_+-\int f_-$ where $f_+=\max(0,f)$ and $f_-=\max(0,-f)$ (we show that if $f$ is measurable then $|f|$, $f_+$ and $f_-$ are too).
3. Let $I=[a,b[$ ($b$ can be $+\infty$) and $f:I\rightarrow\mathbb R$ such as $f$ is Riemann integrable on $[a,x]$ for all $a<x<b$, then $\int_I|f|<\infty$ if and only if $\lim_{x\rightarrow b^-}\int_a^x|f|<\infty$. In this case $\int_I|f|=\lim\int_a^x |f|$ and $\lim\int_a^xf$ exists and is equal to $\int_If$.

So for your exercise we have to check 2 using 3.

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apologies for the simple question, but what is the notation $\sim_0 x$ refer to? –  Justin Jun 23 '12 at 23:17
Have a look at en.wikipedia.org/wiki/Asymptotic_analysis for the explanation of this notation. Then we know that if $f>0$ on a neighborhood of $b$ and if $f\sim_b g$ then they have the same behavior concerning the convergence of the integral at the neighborhood of $b$. –  JBC Jun 23 '12 at 23:21
Thank you. I had never seen the notation before, but it makes sense now. –  Justin Jun 23 '12 at 23:25
From what I see on math.SE, this notation doesn't seem very used. In France, I feel that we often use this notation and properties that come with it (behaviors of series with non-negative terms, improper integrals of non-negative functions...). –  JBC Jun 23 '12 at 23:32
I can see the second integral and the finiteness/existence. What specifically is the argument for finiteness of the first integral? Am I right that we can apply dominated convergence theorem to the second integral? Or do we just need to show that it is finite and measurable? I am somewhat unclear on the first integral, however. I see the asymptotic behavior at 0, $\infty$ but these exercises are under a section about MCT/DCT, etc. so I'm just making sure I dont need these theorems specifically. –  Justin Jun 24 '12 at 17:53