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?

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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



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

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 $1\le x\le\sqrt2$:


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

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

  • $\begingroup$ 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

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.

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

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.


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.


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