# Using the comparison test for integrals with a numerator of x

As part of my calc II course I was taught a rule:

If $$a<0$$ then $$\int^\infty_a\frac1{x^p}\ dx$$is convergent if $$p>1$$ and divergent if $$p\leq1$$

Essentially this rule is telling us that if the integrand goes to zero fast enough, then the integral will converge. But, how do I know if the integral will go fast enough?
If I have the problem: $$\int_{{\,1}}^{{\,\infty }}{{\frac{{{{\bf{e}}^{ - x}}}}{x}\,dx}}$$and I want to do a compartison test to see if it converges, how do I get that first guess in order to decide if I need to make it bigger or smaller?

• Possible comparisons are $e^{-x}/x < 1/x$ and $e^{-x}/x < e^{-x}$. Which one do you think is helpful to determine convergence? – RRL Sep 26 '19 at 22:05
• $e^{-x}$ seems more helpful to me. – Burt Sep 26 '19 at 22:06
• Yes -- as you undoubtedly know that $\int_1^\infty e^{-x} \, dx$ converges. – RRL Sep 26 '19 at 22:08

We have that

$$\frac{{{{\bf{e}}^{ - x}}}}{x}=\frac1{xe^x}\le\frac1{x^2}$$

and therefore

$$\int_{{\,1}}^{{\,\infty }}{{\frac{{{{\bf{e}}^{ - x}}}}{x}\,dx}}\le \int_{{\,1}}^{{\,\infty }}{\frac1{xe^x}\le\int_{{\,1}}^{{\,\infty }}\frac1{x^2}}dx$$

As an alternative, we can use limit comparison test that is

$$\frac{\frac{{{{\bf{e}}^{ - x}}}}{x}}{\frac1{x^2}}=\frac{x}{e^x}\to 0$$

Hint. Note that $$e^x\ge x$$ for sufficiently large $$x.$$ Thus, we have that $$e^{-x}\le \frac1x$$ for $$x\to+\infty,$$ and consequently, that $$\frac{e^{-x}}{x}\le \frac{1}{x^2},$$ when $$x$$ is large enough.