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 determine the convergence/divergence of two improper integrals, both with the same attribute:

$$\int_2^\infty f(x)\,dx=\int_2^\infty \frac{x-\lfloor x\rfloor-\frac{1}{2}}{\ln{x}}dx$$

$$\int_1^\infty g(x)\,dx=\int_1^\infty \frac{(-1)^{\lfloor x \rfloor}}{x\ln{x}}dx$$

The common attribute these two have is that their numerators are alternating between two constant values as the functions progresses twoards infinity.

Note the numerators are bounded by some constant values, and the denominators are monotonically decreasing. Thus, I thought I could use Dirichlet's test for convergence of improper integrals to determine convergence.

In $f(x)$ I can do that, since both the numerator and denominator are continuous as that's what the test requires for convergence. However, in $g(x)$ I can't, since the numerator isn't continuous.

Any advice? :)

Thanks in advance. This is my first question as a Mathematics user.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

For the first integral $$\int_2^{\infty} \frac{\{x\}- 1/2}{\ln(x)} dx$$ Define $\displaystyle B_1(x) = \frac{\{x\}^2}{2} - \frac{\{x\}}{2} + \frac1{12}$. Then we get that $$\int_2^{\infty} f(x) dx = \int_2^{\infty} \frac{d(B_1(x))}{\ln(x)} = \left. \frac{B_1(x)}{\ln(x)} \right \rvert_2^{\infty} + \int_2^{\infty} \frac{B_1(x)}{x \ln^2(x)} dx = \frac1{12 \ln(2)} + \int_2^{\infty} \frac{B_1(x)}{x \ln^2(x)} dx$$ which converges clearly since $$\left \lvert \int_2^{\infty} \frac{B_1(x)}{x \ln^2(x)} dx \right \rvert \leq \frac1{12} \left \lvert \int_2^{\infty} \frac{1}{x \ln^2(x)} dx \right \rvert < \infty$$

share|improve this answer
    
Thanks, but as I've stated in the question - My problem is with the second integral. :) –  Ory Band May 5 '12 at 11:28
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.