How can I prove this integral diverges?
$$ \int_0^\infty \frac{e^{-x}}{x} dx = \infty $$
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$$ \int_{0}^{\infty}\frac{e^{-x}}{x}= \int_{0}^{1}\frac{e^{-x}}{x}+\int_{1}^{\infty}\frac{e^{-x}}{x} \\ > \int_{0}^{1}\frac{e^{-x}}{x} \\ > e^{-1}\int_{0}^{1}\frac{1}{x} $$ which diverges. |
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For $0\lt x\lt 1$, our function is $\gt e^{-1}\frac{1}{x}$. Thus $\int_\epsilon^1 \frac{e^{-x}}{x}\,dx \gt -e^{-1}\log(\epsilon)$. But $-\log(\epsilon)$ blows up as $\epsilon$ approaches $0$ from the right. |
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$s>0$ $$ \int_0^\infty e^{-sx} dx = \frac{1}{s} $$ $$ \int _1^\infty \int_0^\infty e^{-sx} dx ds = \int _1^\infty \frac{1}{s}ds $$ $$ \int_0^\infty \int _1^\infty e^{-sx} ds dx = \int _1^\infty \frac{1}{s}ds $$ $$ \int_0^\infty (\frac{e^{-sx}}{-x}) |_{s=1}^\infty dx = \int_1^\infty \frac{1}{s}ds $$ $$ \int_0^\infty \frac{e^{-x}}{x} dx = \ln\infty -\ln 1 = \infty$$ |
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