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I'm trying to show that the integral $$ \int_1^\infty \frac{ (\log(y))^n }{y^2} \ dy $$ is convergent for every real number $ n \geq 1$. If $ n < 2$, I can bound $ |\log(y)|$ by $y$ and hence show that the integral does converge, but I'm not sure how to construct a tighter bound for the case $ n \geq 2$.

Any help is appreciated!

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You may substitute $u=\log(y)$ and obtain $\int_0^\infty u^n e^{-u}\,du$. You may find this integral easier to bound (but not necessarily.) Actually, you could use repeated integration by parts to evaluate it exactly if $n$ is an integer. – alex.jordan Dec 3 '12 at 1:23
Do you know that for large enough $y$, that no matter how small of a positive number $r$ is, that $\log(y)<y^r$? Apply this with $r=1/(2n)$. – alex.jordan Dec 3 '12 at 1:28
up vote 4 down vote accepted

Let $y=e^t$. We then get that $$I(n) = \int_0^{\infty} \dfrac{t^n}{e^{2t}} e^t dt = \int_0^{\infty} \dfrac{t^n}{e^t} dt = \Gamma(n+1)$$ There are many ways to prove this. One simple way is integration by parts to get that $$I(n) = n I(n-1)$$ Using this we get that $I(n) = n(n-1)(n-2) \cdots (k+1)I(k)$ where $k \in [0,1)$. Hence, we just need to bound $$\int_0^{\infty} t^k e^{-t} dt$$ where $k \in [0,1)$. Now note that \begin{align} \int_0^{\infty} t^k e^{-t} dt & = \int_0^{1}t^k e^{-t} dt + \int_1^{\infty}t^k e^{-t} dt\\ & \leq \int_0^1 t^k dt + \int_1^{\infty} t e^{-t} dt\\ & = \dfrac1{k+1} + \dfrac2e \end{align} Hence, we have that $I(n)$ is bounded by $$n(n-1)(n-2) \cdots (k+1) \left( \dfrac1{k+1} + \dfrac2e\right)$$

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We can easily prove $\log[x]/x^{2}\rightarrow 0$. For $\log[x]^{n}$ compared with $x^{2}$, taking logarithms on both sides give you $n\log\log[x]$ compared with $2\log[x]$. And the ratio must vanish.

However, this is not strong enough to prove the above integral exists: $\frac{1}{x}$ also vanish. The original function $\frac{\log[x]^{n}}{x^{2}}$ is monotonely decreasing for $n\ge 1$. So we may apply the integral test. The ratio test now gives the ratio to be $$\frac{(y+1)^{2}}{y^{2}}*\frac{1}{\log[y]}$$ As $y\rightarrow \infty$ this must be $<0$. So the original integral must converge.

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