Convergence of 2 improper integrals

Question

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.

Answer 1

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.

Answer 2

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