# Does $\int_e^\infty{\frac{\ln(x)^\alpha}{x}dx}$ exists?

My task is to determine wetherthe integral $$\int_e^\infty{\frac{\ln(x)^\alpha}{x}\,dx}$$ does exist or not in depencence of $$\alpha\in \mathbb{R}$$.

For that I wrote $$b$$ instead of $$\infty$$ and then calculated the integral using substitution, which should be $$\frac{1}{\alpha+1}((\ln(b))^{\alpha+1}-1)$$. Now for the improper integral I get $$\lim\limits_{b->\infty}(\int_e^b(\frac{\ln(x)^\alpha}{x}dx))=\frac{1}{\alpha+1}\cdot(\lim\limits_{b->\infty}\ln(b)^{\alpha+1}-1)$$

And because $$\lim\limits_{c->\infty}\ln(c)=\infty$$ and $$\lim\limits_{c->\infty} c^{\alpha+1} = \left\{ \begin{array}{ll} \infty & \alpha\geq -1 \\ 1 & \alpha=-1 \\ 0 & \alpha<-1 \end{array} \right.$$

So the limit only exists for $$\alpha<-1$$, is this right?

• This is the right result – Dr. Sonnhard Graubner Jan 12 at 13:20
• Why do you consider in the end the lim of a power of $c$ instead that of $\ln c$? – user Jan 12 at 13:25
• because ln(c) is infinity anyway – Yefexem Jan 12 at 13:28
• Does: $$\int{\frac{(\ln x)^{\alpha}}{x}dx}=\frac{(\ln x)^{\alpha +1}}{\alpha+1}+C$$ help? (FYI: I used IBP) – Rhys Hughes Jan 12 at 13:38

$$I=\int_e^\infty\frac{\ln^a(x)}{x}dx$$ $$u=\ln(x),\,dx=xdu$$ $$I=\int_1^\infty u^adu=\left[\frac{u^{a+1}}{a+1}\right]_1^\infty$$ and this is clearly divergent
• Except if $\alpha < -1$, I think. – Claude Leibovici Jan 12 at 15:39