I do not know how to proof that integral $\displaystyle \int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx $ exists. I have used Taylor or convergence tests but I do not see the next steps.
I do not know how to proof that integral $\displaystyle \int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx $ exists. I have used Taylor or convergence tests but I do not see the next steps.
I guess now I should do: $\displaystyle \int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx =\underbrace{ \int^{A}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{1} +\underbrace{\int^{\infty}_{A} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{2} $ for $\displaystyle A>1$. With Your observation I conclude the first integral converges but how about the second one?
Thank you for your last post.
Using $\displaystyle \lim_{x \rightarrow \infty} \frac{\ln ^{\pi}x}{x^3}=0 $ I can prove the integer satisfied necessary condition but it is not sufficient. On the other hand in convergence tests limes to zero say nothing (I mean limit comparison test). Maybe I am wrong?
What I intended to say is that for x 'large enough' is...
$\displaystyle \frac{\ln^{\pi} x}{x^{3}}< \frac{1}{x^{3-\varepsilon}}$ (1)
... no matter 'how small' is $\displaystyle \varepsilon>0$. Now the integral...
$\displaystyle \int_{a}^{\infty}\frac{dx}{x^{\alpha}}\ ,\ a>0$ (2)
.... converges for all $\displaystyle \alpha>1$, so that also Your integral converges...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$