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I want to know if $\displaystyle{\int_{0}^{+\infty}\frac{e^{-x} - e^{-2x}}{x}dx}$ is finite, or in the other words, if the function $\displaystyle{\frac{e^{-x} - e^{-2x}}{x}}$ is integrable in the neighborhood of zero.

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the function that I mean is integrable or not integrable? – salma Dec 13 '12 at 20:33
It seems to me that it should be. Our primary concern is that the function might "blow up" around $x=0$. However, we can check the value of the integrand at zero. By L'Hopital's rule, $\lim_{x \to 0} \frac{e^{-x}-e^{-2x}}{x} = \lim_{x \to 0} \frac{-1+2}{1}$. So, the function seems to behave well around zero, so I suspect it will be integrable on a neighborhood about zero. Also, I checked on Wolfram Alpha and it says that it is integrable. – user35959 Dec 13 '12 at 20:39
how I can calculate it? – salma Dec 13 '12 at 20:42
It equals $\ln 2$:*x%29%29‌​%2Fx+from+0+to+infinity – Eric Angle Dec 13 '12 at 20:47
I haven't understand how I can find Ln2 – salma Dec 14 '12 at 6:25

5 Answers 5

$$ \frac1x\,(e^{-x}-e^{-2x})=\frac1x\,(1-x+O(x^2)-(1-2x+O(x^2)))=\frac1x\,(x+O(x^2))=1+O(x). $$ So the function can be extended to $x=0$ in a continuous way, and it thus integrable on any interval $[0,k]$.

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please how can I calculate it? – salma Dec 14 '12 at 10:24

Let $f(x) = \frac{e^{-x} - e^{-2x}}{x}$.

L'Hopital gives $\lim_{x \to 0} f(x)= 1$. Hence in some neighborhood $B(0,\epsilon)$ , $|f(x)| <2$. For $x\geq \epsilon$, we have $\frac{1}{x} \leq \frac{1}{\epsilon}$, and the function $x \mapsto e^{-x} - e^{-2x}$ is clearly integrable.

Hence $\int_0^\infty |f(x)| dx \leq 2 \epsilon + \frac{1}{\epsilon}\int_{\epsilon}^\infty |e^{-x} - e^{-2x}| dx $, and it follows that $f$ is integrable.

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please how can I calculate it? – salma Dec 14 '12 at 8:14
It doesn't have a closed-form solution in terms of the usual elementary functions. See – copper.hat Dec 14 '12 at 8:39

In general, when $f$ is "well-behaved" at zero and infinity:

$$\int_0^{\infty} dx \frac{f(a x) - f(b x)}{x} = (f(\infty)-f(0)) \log{\frac{a}{b}}$$

You can see this from this (rough) "proof":

$$\begin{align}\int_0^{\infty} dx \frac{f(a x) - f(b x)}{x} &= \int_0^{\infty} dx \: \int_b^a du \, \frac{d}{du} f(u x) \\ &= \int_b^a du \: \int_0^{\infty} dx \, \frac{d}{dx} f(u x)\\ &= \int_b^a \frac{du}{u} (f(\infty)-f(0)) \end{align}$$

The result follows. In this case, $f(x) = e^{-x}$, $a=1$, and $b=2$; the integral is then

$$(0-1)\log{\frac{1}{2}} = \log{2}$$

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It suffice to expand the function locally:$$e^{-x}-e^{-2x}=(1-x+x^{2}/2)+..-(1-2x+4x^{2}/2)-...=x-3x^{3}/2+...$$ where $...$ are terms of power at least cubic. It is not difficult to see the above expression divided by $x$ should be locally integrable around 0.

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please how can I calculate it? – salma Dec 14 '12 at 10:25

Claim: $$\int_0^{\infty} \frac{e^{-u}-e^{-2u}}{u} du = \ln(2).$$

Proof: Let \begin{align} C &\equiv \int_0^{\infty} \frac{e^{-u}-e^{-2u}}{u} du\\\ \\ &=\lim_{x=0}\left[ \operatorname{Ei}(1,x) - \operatorname{Ei}(1,2x)\right], \end{align} where $$ \operatorname{Ei}(1,x) \equiv \int_x^\infty \frac{e^{-u}}{u} du. $$ Now, let $$f(x) \equiv \int_1^x \frac{e^{-u}}{u} du.$$ Note:$$\frac{d\operatorname{Ei}(1,x)}{dx} = - \frac{df}{dx},$$ so $$f(x) = -\operatorname{Ei}(1,x) + c,$$ where $c\in \mathbb{R}$. Then $f(1) = -\operatorname{Ei}(1,1)+c$. However, $$ f(1) = \int_1^1 \frac{e^{-u}}{u} du = 0. $$ $\therefore c=\operatorname{Ei}(1,1)$, i.e. $$ \operatorname{Ei}(1,x) = \operatorname{Ei}(1,1) - \int_1^x \frac{e^{-u}}{u} du $$ Considering that $$ \ln(x) = \int_1^x \frac{1}{u} du, $$ we have $$ \operatorname{Ei}(1,x) = -\ln(x) + \operatorname{Ei}(1,1) + \int_1^x\frac{1-e^{-u}}{u} du \tag{$\star$}. $$ $(\star)$ applied to the definition of $C$ gives: \begin{align} \int_0^{\infty} \frac{e^{-u}-e^{-2u}}{u} du &=\lim_{x=0}\left[ \operatorname{Ei}(1,x) - \operatorname{Ei}(1,2x)\right]\\ &=\lim_{x=0}\left[ \ln(2)-\ln(1) - \int_x^{2x} \frac{1-e^{-u}}{u}du \right]\\ &=\ln(2). \end{align} Q.E.D.

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You can see the last step, $\int_x^{2x} \frac{1-e^{-u}}{u} du = 0$, e.g. by Taylor expanding the numerator. – Douglas B. Staple Apr 8 '13 at 18:07

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