2
$\begingroup$

I'm supposed to test for convergence the following integral $$\int_1^{\infty}\frac{\ln x}{x\sqrt{x^2-1}}dx$$ I have tried using the comparison test with two different integrals but I've failed. I also tried using the Dirichlet test, however it doesn't work for this integral. I have thought about using the limit comparison test however I don't have any idea with what would I compare the expression I have.

Any hints?

$\endgroup$
  • 1
    $\begingroup$ Solve it, and you will find it does converge. With a simple calculus of residue you can find it's numerical solution to be $$\frac{\pi\ln(2)}{2}$$ $\endgroup$ – Von Neumann Jul 8 '17 at 11:01
  • $\begingroup$ The series converges by the comparison test. $\endgroup$ – codetalker Jul 8 '17 at 11:02
  • $\begingroup$ @Siddhant What is the integrand of the integral you've used to compare the original to? $\endgroup$ – ahra Jul 8 '17 at 11:04
  • 1
    $\begingroup$ @HenryTuring Calculus of residue is out of scope of the course I'm taking, and even the calculator at integral-calculator.com can't find the primitive function of this integral. $\endgroup$ – ahra Jul 8 '17 at 11:21
  • 1
    $\begingroup$ @HenryTuring I always have in mind that idea: Why two steps ( convergence and evaluation ) ?. They can merge in one step unless it's a cumbersome integral. $\endgroup$ – Felix Marin Jul 8 '17 at 22:06
6
$\begingroup$

Testing for convergence isn't so bad, simply note that for $x>\sqrt2$:

$$0<\frac1{x\sqrt{x^2-1}}<\frac1{x\sqrt{x^2-\frac12x^2}}=\frac{\sqrt2}{x^2}$$

Thus,

$$0<\int_{\sqrt2}^\infty\frac{\ln(x)}{x\sqrt{x^2-1}}~\mathrm dx<\sqrt 2\int_{\sqrt2}^\infty\frac{\ln(x)}{x^2}~\mathrm dx$$

Integration by parts,

$$\int_{\sqrt2}^\infty\frac{\ln(x)}{x^2}~\mathrm dx=\frac{\ln(2)}{2\sqrt2}+\int_{\sqrt2}^\infty\frac1{x^2}~\mathrm dx=\frac{\ln(2)}{2\sqrt2}+\frac1{\sqrt2}$$

For $1\le x\le\sqrt2$:

$$0\le\frac{\ln(x)}{x\sqrt{x^2-1}}\le1$$

$$0<\int_1^{\sqrt2}\frac{\ln(x)}{x\sqrt{x^2-1}}~\mathrm dx<\sqrt2-1$$

Thus, the integral converges and is bounded by $\displaystyle0<I<\frac{\ln(2)}2+\sqrt2$.

$\endgroup$
  • $\begingroup$ Thank you for your help! $\endgroup$ – ahra Jul 9 '17 at 11:40
  • $\begingroup$ No problem! :-) $\endgroup$ – Simply Beautiful Art Jul 9 '17 at 11:42
8
$\begingroup$

After substitution $x=\frac{1}{\sin{t}}$ use the following Euler: $\int\limits_0^{\frac{\pi}{2}}\ln{\sin{t}}\,\mathrm d t$.

The answer is $\frac{\pi}{2}\ln2$.

The Euler's work: $$\int\limits_0^{\frac{\pi}{2}}\ln{\sin{t}}\,\mathrm d{t}=2\int\limits_0^{\frac{\pi}{4}}\ln{\sin{2t}}\,\mathrm d{t}=$$ $$=\frac{\pi}{2}\ln2+2\int\limits_0^{\frac{\pi}{4}}\ln{\sin{t}}\,\mathrm d{t}+2\int\limits_0^{\frac{\pi}{4}}\ln{\cos{t}}\,\mathrm d{t}=$$ $$=\frac{\pi}{2}\ln2+2\int\limits_0^{\frac{\pi}{4}}\ln{\sin{t}}\,\mathrm d{t}+2\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln{\sin{t}}\,\mathrm d{t}=$$ $$=\frac{\pi}{2}\ln2+2\int\limits_0^{\frac{\pi}{2}}\ln{\sin{t}}\,\mathrm d{t}.$$ Thus, $$\int\limits_0^{\frac{\pi}{4}}\ln{\sin{t}}\,\mathrm d{t}=-\frac{\pi}{2}\ln2,$$ which says that your integral converges.

$\endgroup$
  • $\begingroup$ Hm... yes, this should work out nicely. $\endgroup$ – Simply Beautiful Art Jul 8 '17 at 11:43
1
$\begingroup$

For $x$ close to $1:$ Since $\ln x \sim x-1$ as $x\to 1^+,$ the integrand has limit $0$ from the right at $1.$ Hence there is no problem at $1.$

For large $x,$ the integrand looks like

$$\tag 1 \frac{\ln x }{x^2}.$$

Now $\ln x \to \infty$ as $x\to \infty,$ but it does so at a laughably small rate. So $\ln x < x^{1/2}$ for large $x.$ Thus $(1)$ is bounded above by $1/x^{3/2}$ for large $x.$ Since $\int_1^\infty (1/x^{3/2})\, dx <\infty,$ we have convergence of the original integral by the comparsion test.

$\endgroup$
0
$\begingroup$

I separated the integral: $$(1)\int_1^{3}\frac{\ln x}{x\sqrt{x^2-1}}dx, (2)\int_3^{\infty}\frac{\ln x}{x\sqrt{x^2-1}}dx$$

$$(1)\backsim \int_1^{3}\dfrac{1}{\sqrt{x^2-1}}dx$$ which converges because it is the derivative of $\sin^{-1}(x)$

$$(2)\backsim \int_3^{\infty}\frac{1}{x^2\log^{-1}(x)}dx $$ which converges because $2$ (the exponent)$>1$

So the entire integral converges.

$\endgroup$

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.