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 am interested in finding the values of $a, b$ such that the integral

$$ \int_0^{\infty}\frac{{\left|\log x\right|}^b}{x^a} dx $$ converges.

My idea was to separate this integral: $$ \int_0^{1}\frac{\left|\log x\right|^b}{x^a} dx + \int_1^{\infty}\frac{\left|\log x\right|^b}{x^a} dx. $$ From the first part, we can see that $a$ needs to be less than $1$. Indeed, any powers of $x$ will dominate the $\log$. So $a<1$ is a necessary condition. To see that the integral converges for every $a<1$, let $\epsilon>0$ be such that $a+\epsilon <1$. Then $x^{\epsilon}{\left|\log x\right|}^{\beta}\to 0$ when $x\to \infty$. We can then rewrite the first integral as $$ \int_0^{1}\frac{x^{\epsilon}\left|\log x\right|^b}{x^{a+\epsilon}} dx, $$ which converges.

The problem comes from the second integral. It seems to me that when $a<1$, the second integral will never converge, since $\int_1^{\infty}\frac{1}{x^a} dx$ and $\int_1^{\infty}\left|\log x\right|^b dx$ both never converges.

Any suggestions would be much appreciated!

share|improve this question
1  
The only suggestion I can think of is to write $\log x$ instead of $log(x)$. Your reasoning is correct, and the integral is divergent for all values of $a$ and $b$. –  Julián Aguirre Jan 11 '13 at 21:51
add comment

1 Answer 1

up vote 4 down vote accepted

As you have done, split it into two as follows. $$\int_0^{\infty} \dfrac{\vert \log(x) \vert^b}{x^a} dx = \underbrace{\int_0^1 \dfrac{\vert \log(x) \vert^b}{x^a} dx}_I + \overbrace{\int_1^{\infty} \dfrac{\vert \log(x) \vert^b}{x^a} dx}^J$$ $$I = \underbrace{\int_0^1 \dfrac{\vert \log(x) \vert^b}{x^a} dx = \int_1^{\infty} \dfrac{\vert \log(x) \vert^b}{x^{2-a}} dx}_{x \to 1/x}$$ Now the integral $J$ converges only for $a>1$ whereas the integral $I$ converges only for $2-a>1$ i.e. $a<1$. Hence, you can never make this integral convergent. There is no hope to define even a principal value since both the integrals are always positive for any $a$ and $b$. Hence, your original integral will always diverge irrespective of $a$ and $b$.

share|improve this answer
    
Thanks that's what I thought! But I got confused by one of Folland's Real analysis exercises. In chapter 6, he asks, if $0<p_0<p_1\leq\infty$, find an example of functions $f$ on $(0,\infty)$ such that $f\in L^p$ iff $p_0<p<p_1$. As an hint, he says to consider functions of the form $f(x)= x^{-a}|log\,x|^b$. But these functions are never in $L^p$! –  Zoltan Jan 12 '13 at 4:40
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.