Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
add comment

2 Answers 2

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 $$n(n-1)(n-2) \cdots (k+1) \left( \dfrac1{k+1} + \dfrac2e\right)$$

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.