Re-arranging terms in Integral

Question

Is it ok if i write $\int_{0}^\infty \frac { \ln {(1+x)} }{ x^2 } $ to $\int_{0}^\infty \ { \ln {(1+x)} }*{\frac{1} x^2 } $ and then apply integration byparts rule?

Answer 1

It would be better to write it as $$ \int\frac{\ln(1+x)}{x^2}\,dx = \int\ln(1+x)\cdot \frac{1}{x^2}\,dx $$ and then apply integration by parts, with $1/x^2$ as differential factor and $\ln(1+x)$ as finite factor, so $$ \int\ln(1+x)\cdot \frac{1}{x^2}\,dx=\ln(1+x)\frac{-1}{x}- \int\frac{1}{1+x}\frac{-1}{x}\,dx $$ Now $$ \frac{1}{x(1+x)}=\frac{1}{x}-\frac{1}{1+x} $$ so the integral is $$ \int\frac{\ln(1+x)}{x^2}\,dx=-\frac{\ln(1+x)}{x}+\ln x-\ln(1+x)= -\frac{\ln(1+x)}{x}+\ln\frac{x}{1+x} $$ Since $$ \lim_{x\to0^+}\left(-\frac{\ln(1+x)}{x}+\ln\frac{x}{1+x}\right)=-\infty $$ the integral is not convergent at $0$.

It is convergent at $\infty$, though, because $$ \lim_{x\to\infty}\left(-\frac{\ln(1+x)}{x}+\ln\frac{x}{1+x}\right)=0 $$

Thus, for instance, $$ \int_1^\infty\frac{\ln(1+x)}{x^2}\,dx=2\ln2 $$ but the integral over $(0,\infty)$ diverges.

Answer 2

your integral is divergent since

when $x \to 0^+$

$\frac{\ln(1+x)}{x^2} $ is equivalent to $\frac{1}{x}$

and

$\int_0\frac{dx}{x}$ is divergent.