emmostatwf

2022-09-06

Explain how ${\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-x}}{x}dx$ diverges? This is because the Laplace transform of $\frac{1}{t}$ can be reduced to this integral which has to diverge. But the limit comparison test with ${e}^{-x}$ shows that the integral converges.

The issue is at zero:
${\int }_{0}^{1}\frac{{e}^{-x}}{x}\phantom{\rule{thickmathspace}{0ex}}dx\ge \frac{1}{e}{\int }_{0}^{1}\frac{dx}{x}=\mathrm{\infty }$
Therefore ${\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-x}}{x}\phantom{\rule{thickmathspace}{0ex}}dx={\int }_{0}^{1}\frac{{e}^{-x}}{x}\phantom{\rule{thickmathspace}{0ex}}dx+{\int }_{1}^{\mathrm{\infty }}\frac{{e}^{-x}}{x}\phantom{\rule{thickmathspace}{0ex}}dx$ diverges

Drew Williamson

Regarding the second part of the question, to add to first's answer,
To use the limit comparison test with ${e}^{-x}$, we need
$\frac{{e}^{-x}}{x}\le {e}^{-x}$
throughout the interval over which integral takes place.
However, this is true only for x>1, and hence the test cannot be used.

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