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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.

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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 $\alpha > 1$.

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