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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real-analysis
integration
definite-integrals
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0the function that I mean is integrable or not integrable? – 2012-12-13
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0It 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. – 2012-12-13
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0how I can calculate it? – 2012-12-13
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1It equals $\ln 2$: http://www.wolframalpha.com/input/?i=integrate+%28e%5E%28-x%29-e%5E%28-2*x%29%29%2Fx+from+0+to+infinity – 2012-12-13
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0I haven't understand how I can find Ln2 – 2012-12-14