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.
Is ${\int_{0}^{+\infty}\frac{e^{-x} - e^{-2x}}{x}dx}$ finite?
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0I haven't understand how I can find Ln2 – 2012-12-14
5 Answers
$ \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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0please how can I calculate it? – 2012-12-14
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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0It doesn't have a closed-form solution in terms of the usual elementary functions. See http://en.wikipedia.org/wiki/Exponential_integral – 2012-12-14
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
$(0-1)\log{\frac{1}{2}} = \log{2}$
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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0please how can I calculate it? – 2012-12-14
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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0You can see the last step, $\int_x^{2x} \frac{1-e^{-u}}{u} du = 0$, e.g. by Taylor expanding the numerator. – 2013-04-08