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