Prove convergence by bounding with another function

Question

• $\displaystyle \int_{10}^\infty \frac{\ln (\frac{x-5}{x+5})}{x} \, dx$

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So I know that it's negative on the whole segment, converges and -> 0, but i can't find converging function to bound this one in order to prove convergence. Seems like it less than $ -\frac{1}{x} $, but this integral don't converge.

Answer 1

Show that $\frac{|\ln (\frac{x-5}{x+5})|}{x} x^2$ has a positiv limit $L$ as $x \to \infty$. Hence there is $c >10$ such that $\frac{|\ln (\frac{x-5}{x+5})|}{x} \le \frac{2L}{x^2}$ for all $x \ge c$.

Therefor $\int_{10}^\infty \frac{\ln (\frac{x-5}{x+5})}{x} \, dx$ converges abolutely.

Answer 2

$$\ln (1+ x) \leq \sqrt{2x}$$ Thus $$\left|\ln\left( \frac{x-5}{x+5}\right)\right|\leq\sqrt{\frac{20}{x+5}}$$

Answer 3

Note that for $x\to \infty$

$$\ln \left(\frac{x-5}{x+5}\right)=\ln \left(1-\frac{10}{x+5}\right)\sim -\frac{10}{x+5}$$

thus

$$\frac{ \ln (\frac{x-5}{x+5})}{x}\sim -\frac{10}{x(x+5)}$$

then the given integral coverges by limit comparison test with $\frac1{x^2}$.