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.

Im interested to know for which $\alpha \in \mathbb{R}$ the following integral converges: $$\int_0^\infty \frac{dx}{1+ (x^\alpha \sin x)^2}.$$

In the answers to this post it was shown that the integral diverges for $\alpha=1$ (and it can be similarly shown that the integral diverges for any $\alpha \leq 1$. Following this question, I started wondering for which $\alpha$ the integral converges. Note that the approaches used to answer the case $\alpha=1$ do not work as they bound the integral from below with a diverging series whereas to show convergence we need to bound it from above.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

Clearly it diverges for $\alpha \leq 0$. Now let $\alpha > 0$ and $n = 1, 2, \cdots$. For each $x$ satisfying $|x - n\pi| \leq \pi / 2$ we have $$ c_1 n^{\alpha} |x - n \pi| \leq x^{\alpha} |\sin x| \leq c_2 n^{\alpha} |x - n\pi|,$$ where $c_1$ and $c_2$ are a positive constants depending only on $\alpha$. This is because $$ \frac{2}{\pi}|x - n\pi| \leq |\sin x| \leq |x - n\pi|$$ and $$\frac{\pi}{2} n \leq \left( 1 - \frac{1}{2n}\right) \pi n \leq x \leq \left( 1 + \frac{1}{2n}\right) \pi n \leq \frac{3\pi}{2} n $$ for such $x$. Thus we have $$ \sum_{n=1}^{\infty} \int_{-\pi/2}^{\pi/2} \frac{dx}{1 + (c_2 n^{\alpha} x)^2} \leq \int_{\pi/2}^{\infty} \frac{dx}{1 + (x^{\alpha} \sin x)^2} \leq \sum_{n=1}^{\infty} \int_{-\pi/2}^{\pi/2} \frac{dx}{1 + (c_1 n^{\alpha} x)^2},$$ and by simple change of variable gives $$ \frac{1}{c_2} \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}} \int_{-c_2 \pi n^{\alpha} /2}^{c_2 \pi n^{\alpha}/2} \frac{dt}{1 + t^2} \leq \int_{\pi/2}^{\infty} \frac{dx}{1 + (x^{\alpha} \sin x)^2} \leq \frac{1}{c_1} \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}} \int_{-c_1 \pi n^{\alpha} /2}^{c_1 \pi n^{\alpha}/2} \frac{dt}{1 + t^2}.$$ Since $$ \int_{-c_2 \pi /2}^{c_2 \pi /2} \frac{dt}{1 + t^2} \leq \int_{-c_2 \pi n^{\alpha} /2}^{c_2 \pi n^{\alpha}/2} \frac{dt}{1 + t^2} \qquad \text{and} \qquad \int_{-c_1 \pi n^{\alpha} /2}^{c_1 \pi n^{\alpha}/2} \frac{dt}{1 + t^2} \leq \int_{-\infty}^{\infty} \frac{dt}{1 + t^2}, $$ this implies $$c_3 \zeta(\alpha) \leq \int_{\pi/2}^{\infty} \frac{dx}{1 + (x^{\alpha} \sin x)^2} \leq c_4 \zeta(\alpha) $$ for some positive constants $c_3$ and $c_4$ depending only on $\alpha$. Therefore the integral converges if and only if $\alpha > 1$.

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.