I have to find out for which $b$, the improper integral exist.
$$\int_e^\infty \! \frac{(\ln (x))^b}{x}\ \, \mathrm{d}x$$
I know that the improper integral exists if the limit exist.
Is there a general methodic to find these $b$ or just trying?
Thx in advance
This is called Bertrand's Integral.
It's general form is given by, $$\int_e^{+\infty} \frac{1}{x^\alpha\log(x)^\beta} dx $$ and it converges if $(\alpha>1)$ or $(\alpha=1,\beta>1)$.
Hence, $\displaystyle\int_e^{+\infty}\frac{\log(x)^b}{x}dx $ converges for $b<-1$.
One can first solve the integral with the substitution $u=\ln(x)$, which gives us
$$\int u^b\ du=\frac{u^{b+1}}{b+1}\color{#888888}{+c}$$
Placing in bounds, we then get
$$\left.\int_e^a\frac{(\ln(x))^b}x\ dx=\frac{(\ln(x))^{b+1}}{b+1}\right|_{x=e}^{x=a}$$
as $a\to\infty$, this integral exists if $(\ln(x))^{b+1}$ goes to $0$, i.e. for $b<-1$.
For $b=-1$, the integral turns out to be $\ln(\ln(x))$, which diverges.
Hint For $b=-1$ a primitive of the function is $\ln(\ln x)$ and so we see the integral is divergent. By comparison we get also the divergence for $b>-1$. Now let $b<-1$, a primitive is $\frac1{b+1}(\ln x)^{b+1}$ and in this case the integral is convergent.
