# Convergence of 2 improper integrals

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.

• 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
– Cato
Sep 6, 2016 at 15:19
• 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 $\sqrt{(x-1)\ln x} \leq \sqrt{(x-1)^2}=x-1$. Sep 6, 2016 at 15:20
• 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: $((x-1)\ln(x)))^{-1/2}\simeq ((x-1)^2)^{-1/2}=(x-1)^{-1}$ Sep 6, 2016 at 15:21
• 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?
– Cato
Sep 6, 2016 at 15:23
• 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
– Cato
Sep 6, 2016 at 15:26

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.
$$I_1=\int_{1}^{3}\frac{dx}{x\sqrt{(x-1)\log x}}=\int_{0}^{2}\frac{dx}{(x+1)\sqrt{x\log(x+1)}}$$ 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 $$0\leq \int_{3}^{+\infty}\frac{dx}{x\sqrt{(x-1)\log x}}\leq\int_{3}^{+\infty}\frac{dx}{x\sqrt{(x-1)}}=2\arctan\frac{1}{\sqrt{2}}\leq\sqrt{2}.$$