0
$\begingroup$

I have check if any of these improper integrals converge:

$$\int_{1}^{3} \frac{1}{x\sqrt{(x-1)\ln{x}}}$$ and $$\int_{3}^{\infty} \frac{1}{x\sqrt{(x-1)\ln{x}}}$$

To be honest, I got stuck with the first one. What I did:

  1. try to calculate its value - no result, thus I go proceeded to comparison test
  2. $$\int_{1}^{3} \frac{1}{x\sqrt{(x-1)\ln{x}}}\geq\int_{1}^{3} \frac{1}{x\sqrt{x \ln{x}}}=\int_{1}^{3} \frac{1}{x^{3/2}\sqrt{\ln{x}}}$$ it also didn't give me any result so, I tried to compare it with $\int \frac{1}{x\ln(x)}$
  3. $$\int_{1}^{3} \frac{1}{x\sqrt{(x-1)\ln{x}}}\leq \int_{1}^{3} \frac{1}{x\ln{x}}=\ln(\ln(3))-\ln(\ln(1))=\ln(\ln(3))<1$$

Thus I get that it converges in $[1,3]$ interval and diverges in $[3,\infty]$. In book I get different answer: diverges in $[1,3]$ interval and converges in $[3,\infty]$.

What did I do wrong? Thank you for your help.

$\endgroup$
5
  • $\begingroup$ for the second one, you can show that the area is finite until such time as the function (with a substitution) becomes < x^(-3/2) - and after that it has to converge at infinity $\endgroup$ – Cato Sep 6 '16 at 15:19
  • $\begingroup$ Your inequality in 3. is wrong -- how do you justify $\sqrt{(x-1)\ln x} \geq \ln x$? Instead, you may want to use $\ln(1+h) \leq h$, so that $\ln x=\ln (1+(x-1)) \leq x-1$ here, and then $\sqrt{(x-1)\ln x} \leq \sqrt{(x-1)^2}=x-1$. $\endgroup$ – Clement C. Sep 6 '16 at 15:20
  • $\begingroup$ As for the first one, observe that the expansion of the logarithm in the neighbourhood of 1 is $ \ln (x)=x-1+o(x-1)$. So you have a non integrable singularity: $((x-1)\ln(x)))^{-1/2}\simeq ((x-1)^2)^{-1/2}=(x-1)^{-1}$ $\endgroup$ – guestDiego Sep 6 '16 at 15:21
  • $\begingroup$ in point 3 - what happens when x gets very close to 1? e,g 1.00001 - you can see that (x - 1) is very small and ln x is very small, so will it be less than the RHS? $\endgroup$ – Cato Sep 6 '16 at 15:23
  • $\begingroup$ in point 3, if you integrate from 1 to 2, then 2 to 3 is finite is easy to see, then for the 1 to 2 part, the integral > 1 / (2sqrt(x - 1)) which is unbounded $\endgroup$ – Cato Sep 6 '16 at 15:26
1
$\begingroup$

The book in this case is right. For limits at infinity the condition is $$\int^{\infty}\frac{dx}{x^k}$$ converges iff $k>1$, just like harmonic series. You can make a comparison with $$x^{-\frac{3}{2}}$$ so $1< \frac{3}{2}$ it converges.

For a singular point the condition is $$\int_{0}\frac{dx}{x^k}$$ converges for $k <1$ here note that $\frac{\ln x}{x-1}\to 1$ as $x\to 1$ so your integral can be compared with $$\int_{1}\frac{dx}{x-1}$$ which diverges.

$\endgroup$
0
$\begingroup$

$$I_1=\int_{1}^{3}\frac{dx}{x\sqrt{(x-1)\log x}}=\int_{0}^{2}\frac{dx}{(x+1)\sqrt{x\log(x+1)}}$$ is divergent since in a right neighbourhood of the origin $\log(x+1)$ behaves like $x$, hence both $\frac{1}{\sqrt{x\log(x+1)}}$ and $\frac{1}{(x+1)\sqrt{x\log(x+1)}}$ are not integrable over $(0,\varepsilon>0)$. On the other hand, the second integral is convergent since $$ 0\leq \int_{3}^{+\infty}\frac{dx}{x\sqrt{(x-1)\log x}}\leq\int_{3}^{+\infty}\frac{dx}{x\sqrt{(x-1)}}=2\arctan\frac{1}{\sqrt{2}}\leq\sqrt{2}. $$

$\endgroup$

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.