# Prove that $\int_0^{\infty} \frac{\sin x}{x^p}\, dx$ converges for $0<p<2$

I would really appreciate feedback/guidance. This is from a past year exam and my professor didn't cover much on improper integrals. I'm trying to prove that $\int_0^{\infty} \frac{\sin x}{x^p}\, dx$ converges for $0<p<2$ using the comparison test for improper integrals. I know that from the test, if $|f(x)| \leq g(x) \forall x\geq a$ then $\int_a^{\infty} g(x)\, dx$ converges.

EDIT: Here's my second attempt at the proof after much discussion in the comments below.

We know that $\frac{\sin x}{x^p}$ will be continuous on 1 because $\lim_{x\rightarrow 1} \frac{\sin x}{x^p} = lim_{x\rightarrow 1} x^{-p} \sin(x) = 1^{-p}\sin(1) = \sin(1)$.

However, for 0 we have that:

$\lim_{x \rightarrow 0} \frac{\sin x}{x^p} = \frac{\sin 0} {0^{p}} = \frac{0}{0^{-p}} = \frac{1}{0^{p-1}}$. Since we can't have 0 in the denominator, we only have continuity on 0 if p-1 < 1, which implies p<2.

And so, by a theorem, we know that $\frac{\sin(x)}{x^p}$ is Riemann integrable on [0,1] for p<2. Furthermore, by the improper limit comparison test, if we prove that $\int_1^{\infty} f(x)\, dx$, then we can show that $\int_0^{\infty} f(x)\, dx$ converges. Hence we have that

$\lim_{b \rightarrow \infty} \int_1^{b} \frac{\sin(x)}{x^p}\,dx = \lim_{b \rightarrow \infty} \int_1^{b} \frac{\sin(x)}{x^p}\,dx =\lim_{b\rightarrow \infty} -\frac{cos x}{{p}} |_p^{b} - \int_1^{b} \frac{\cos x}{(p-1)x^{p+1}}\,dx$

And so I know that since $\int \frac{cos x}{x^{p+1}} < \int \frac{1}{x^{p+1}}$, and since the RHS will converge if and only if p+1 > 1, we have that p>0.

And so, we conclude that the integral converges for $0<p<2$.

• Your integration by parts is wrong. $$\int_1^b \frac{\sin x}{x^p}\,dx = \frac{-\cos x}{x^p}\biggr\rvert_1^b - p\int_1^b \frac{\cos x}{x^{p+1}}\,dx$$ – Daniel Fischer Dec 18 '16 at 23:32
• I don't see how you can ignore $x=0$ like this. In fact, about half of the difficulty in showing convergence is due to the blowup of $\sin(x)/x^p$ at $x=0$ for $p>1$. Certainly there are many functions such that $\int_1^\infty f <\infty$ but $\int_0^\infty f = \infty$; say, $f(x) = 1/x^2$. – Gyu Eun Lee Dec 18 '16 at 23:35
• @Gyu Eun Lee Hi there! I apologize, I didn't intend to ignore x=0. It slipped my mind that we would have an issue at x=0. Thanks ofr pointing it out! – Nikitau Dec 18 '16 at 23:40
• @DanielFischer Ah, that was careless of me. Thank you! – Nikitau Dec 18 '16 at 23:42

## 1 Answer

Integrate by parts with $u=x^{-p}$ and $v=-\cos(x)$. Then, we can write

$$\int_1^L \frac{\sin(x)}{x^p}\,dx=\left.\left(-\frac{\cos(x)}{x^p}\right)\right|_{1}^{L}-p\int_1^L \frac{\cos(x)}{x^{p+1}}\,dx$$

• Thanks! It seems like I got my integration by parts completely mixed up. Judging from what we get from integration by parts, we can write that $\int_1^{L} \frac{cos(x)}{x^{p+1}} < \int_1^{L} \frac{1}{x^{p+1}}$, and we know that the RHS converges only if p+1 > 1, which implies p>0. But how do we show that we also need p<2? – Nikitau Dec 18 '16 at 23:44
• @Nikitau That constraint comes from the behaviour at $0$. – Daniel Fischer Dec 18 '16 at 23:44
• @DanielFischer This is probably a dumb question, but is it the behaviour of x at zero? Oh then I guess what I wrote earlier in my original post didn't make sense, where I said if $\int_1^{\infty} f(x) dx$ converges then $\int_0^{\infty} f(x) dx$ converges? – Nikitau Dec 18 '16 at 23:48
• @Nikitau It makes sense, but it's only true if the behaviour on $[0,1]$ is good enough. Look at $\frac{\sin x}{x^p}$ as $x \to 0^+$. For what $p$ is that bounded? For what $p$ does it tend to $+\infty$? How fast? How fast may it grow and still remain (improperly) integrable? – Daniel Fischer Dec 18 '16 at 23:52
• @DanielFischer Hmm. I'm guessing we would get $0^{1-p}$ as x goes to $0^+$? Since we can't have 0 in a denominator, we would need 1-p>0? and so 1>p? And $\frac{sin x}{x}$ goes to infinity for all values of p>0? – Nikitau Dec 18 '16 at 23:57