0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Possible comparisons are $e^{-x}/x < 1/x$ and $e^{-x}/x < e^{-x}$. Which one do you think is helpful to determine convergence? $\endgroup$ – RRL Sep 26 '19 at 22:05
  • $\begingroup$ $e^{-x}$ seems more helpful to me. $\endgroup$ – Burt Sep 26 '19 at 22:06
  • 2
    $\begingroup$ Yes -- as you undoubtedly know that $\int_1^\infty e^{-x} \, dx$ converges. $\endgroup$ – RRL Sep 26 '19 at 22:08
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.