I should determine whether this is a convergent or divergent integral. The problem is that I don't know how to start. i need to use the comparison test but i don't know where to start. $$ \int_{0}^{1} \frac{e^{-x}}{x}dx $$
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2Compare with ${1\over e}\int_0^1 {1\over x}dx$. – David Mitra May 11 '14 at 16:16
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@DavidMitra Nice. Why don't you submit that as an answer...? – BCLC May 11 '14 at 16:30
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@DavidMitra how can i continue from here? how to show that your example is not converge? – user2637293 May 11 '14 at 16:35
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$\int_0^1 1/x,dx=[ \ln x]0^1=\ln 1 -\lim\limits{x\rightarrow0^+}\ln x=\cdots$. – David Mitra May 11 '14 at 16:39
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Since $e^{-x}$ is decreasing on $[0,1]$ and positive there, for $0\le x\le1$ you have $e^{-x}/x\ge e^{-1}/x$. Consequently, you can compare with the integral $\int_0^1 {dx\over e\cdot x}$ to show your integral diverges.
David Mitra
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