Convergence of $\displaystyle \int_0^\infty\frac{\sin x}{x^p + \sin x} dx$

Question

I'm trying to find bound on $p \gt 0$ where $\displaystyle \int_0^\infty\frac{\sin x}{x^p + \sin x} dx$ converges.

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Around zero we can move to equivalent(in terms of convergence) integral $\displaystyle \int_0^\varepsilon\frac{dx}{x^{p-1} + 1}$ using $x \gt \sin x \gt \frac x 2$. After checking out all choices of placement $p$, I decided that integral converges around zero. When going to infinity i ignored $\sin x$ in bottom part of fraction and looked for inequalities for $\displaystyle I_k \sim \int_{2\pi k}^{2\pi (k + 1)}\frac{\sin x}{x^p} dx$. For clarity I changed the variable to $t = 2\pi x$ and ignored the resulting constant in integral. Then I divided interval into $[k:k+1/2]$;$[k+1/2:k+3/2]$;$[k+3/2:k+4/2]$ On each interval I bounded $\frac{1}{x^p}$ with something like $\frac{1}{(k+i)^p}$ and integrated $\sin x$ to get another non-important constant. In result I got $$\frac{1}{(k+1/2)^p} - \frac{1}{(k+3/2)^p}\lt I_k \lt \frac{1}{(k)^p} - \frac{1}{(k+2)^p}\\$$ Then I calculated differences and threw out constant in bottom part of fractions because $k$ is arbitrarily big: $$\frac{(k + 3/2)^p - (k + 1/2)^p}{k^{2p}} < I_k < \frac{(k + 2)^p - k ^p}{k^{2p}}.$$ By using some Taylor series for $(1 + i/k)^p$ and ignoring constants: $$I_k \sim \frac{k^{p-1}}{k^{2p}} = \frac{1}{k^{p+1}}.$$ And series like this converges when $p > 0$. The problem is that in textbook the answer is $p \gt 1/2$. So I want someone to validate my proof.

Answer 1

I agree with your analysis when $x \approx 0$. Your analysis for $x \gg 1$ however has a flaw: you tried to use the approximation $$ \frac{\sin x}{x^p + \sin x} \approx \frac{\sin x}{x^p} $$ However, for $p \leq 1$ the integral $\int_1^\infty \frac{\sin x}{x^p} ~\mathrm{d}x $ is only conditionally convergent. Which means that the convergence can be unstable.

This is a problem because one observes that when $\sin x > 0$, you have $$ \left| \frac{\sin x}{x^p + \sin x} \right| < \left| \frac{\sin x}{x^p} \right| $$ while for when $\sin x < 0 $ you have $$ \left| \frac{\sin x}{x^p + \sin x} \right| > \left| \frac{\sin x}{x^p} \right| $$ and so there is a systematic bias (to by more negative) of $\frac{\sin x}{x^p + \sin x}$ compared to $\frac{\sin x}{x^p}$. If this bias is not integrable, then you have a problem.

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Here's one way to do the analysis:

For $x \gg 1$, observe that the integrand has the (absolutely convergent) expansion

$$ \frac{\sin x}{x^p + \sin x} = S - S^2 + S^3 - S^4 + \cdots $$

where $S := \dfrac{\sin x}{x^p}$.

You can split the sum into two parts, where up to $S^M$ for some large $M$ and the second from $M$ on. For any positive $p$, if you take $n \geq M > 1/p$ then each of the $S^n$ terms are absolutely integrable, and so you can interchange summations with integrals and there's no harm done. So it remains to consider the finite sum

$$ \int_{a \gg 1}^\infty \sum_{k = 1}^M S^k (-1)^{k-1} ~\mathrm{d}x $$

Since we have a finite sum we can interchange integration with summation (as long as we leave the taking of limit on the outside

$$ = \lim_{b \to +\infty} \sum_{k = 1}^M \int_a^b (-1)^{k-1} S^k ~\mathrm{d}x $$

Now, due to the oscillatory nature of $(\sin x)^k$ for $k$ is odd, you should be able to show that all those terms integrate to something that converges as $b \to \infty$ for any $p > 0$. (This is basically an integral version of the alternating series test.)

However, you have the problem with the terms where $k$ is even. Each of those terms contribute an integrand that is signed (negative). So as long as any of those terms diverge in integral, the original integral must diverge.

The term $S^2$ in particular concerns the integral

$$ \int \frac{\sin^2(x)}{x^{2p}} ~\mathrm{d}x $$

which you can bound below by the series $\sum j^{-2p}$, and hence diverges whenever $2p \leq 1$.

Answer 2

Due to $$\lim_{x\to0^+}\frac{\sin x}{x^p+\sin x}=\lim_{x\to0^+}\frac{\frac{\sin x}x}{x^{p-1}+\frac{\sin x}x}= \begin{cases} 0,& 0<p<1\\ \frac12,& p=1\\ 1,& p>1 \end{cases}, $$ the point $0$ is a continuous point for the function $\frac{\sin x}{x^p+\sin x}(p>0)$. So we consider the improper integral $$\int_{1}^{\infty}\frac{\sin x}{x^p+\sin x}dx.$$ Note that: $$\frac {\sin x}{x^p+\sin x} =\frac{\sin x}{x^p}-\frac{\sin^2x}{x^p(x^p+\sin x)}.$$ $$\int_1^{\infty}\frac {\sin x}{x^p} \, \mathrm{d}x\ \mbox{converges}\iff p>0,$$ and $$\int_1^{\infty}\frac{\sin^2x}{x^p(x^p+\sin x)}dx\ \mbox{converges}\iff p>\frac{1}{2}.$$ $\textbf{Proof}$: $$0\leq\frac{\sin^2x}{x^p(x^p+\sin x)}\leq\frac{1}{x^p(x^p+\sin x)}\sim\frac{1}{x^{2p}},$$ and $$\int_1^{\infty}\frac {1}{x^{2p}} \, \mathrm{d}x\ \mbox{converges}\iff p>\frac{1}{2}.$$

On the other hand, when $0<p\leq\frac12,$ $$\int_1^{\infty}\frac{\sin^2x}{x^p(x^p+\sin x)}dx\ \mbox{is not convergent}.$$ Proof as follows: $$\frac{\sin^2x}{x^p(x^p+\sin x)}\geq\frac{\sin^2x}{x^p(x^p+1)}\geq\frac{\sin^2x}{2x^{2p}}=\frac{1}{4x^{2p}}-\frac{\cos(2x)}{4x^{2p}},$$ $$\int_1^{\infty}\frac {1}{x^{2p}}dx\ \mbox{is not convergent for}\ 0<p\leq \frac12,$$ and Dirichlet's test implies $$\int_1^{\infty}\frac{\cos(2x)}{4x^{2p}}dx\ \mbox{is convergent for}\ 0<p\leq \frac12.$$

So, your integral converges if and only if $a>\frac12$.