Laplace transform of $\frac{1}{t}$ and resulting improper integral

Question

Could someone please explain how $$\int_0^{\infty}\frac{e^{-x}}{x}dx$$ 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.

Please help.

Thanks in advance.

Answer 1

Regarding the second part of the question, to add to carmichael561'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 $x\gt1$, and hence the test cannot be used.

Answer 2

The issue is at zero: $$ \int_0^1\frac{e^{-x}}{x}\;dx\geq \frac{1}{e}\int_0^1\frac{dx}{x}=\infty $$ Therefore $\int_0^{\infty}\frac{e^{-x}}{x}\;dx=\int_0^{1}\frac{e^{-x}}{x}\;dx+\int_1^{\infty}\frac{e^{-x}}{x}\;dx$ diverges.