$ \int_{1}^{\infty} x^a \sin{\left(\frac{1}{x^b}\right)} dx $ Convergent and Divergent?

Question

For a given $a, b \in \mathbb{R}$ and $b \geq 0$, I am interested in determining the convergence or divergence of the integral

$$ \int_{1}^{\infty} x^a \sin{\left(\frac{1}{x^b}\right)} dx $$

Specifically, I would like to find the values of $a$ and $b$ for which the integral converges and diverges.

Additionally, I would appreciate any insights regarding the convergence or divergence of the integral when $b < 0$.

Thank you for your assistance!

Answer 1

$$ I=\int_{1}^{\infty} x^a \sin{\left(\frac{1}{x^b}\right)} dx~~~~~~~~~b\ge0 $$

If $ b=0$,

$$ I=\sin(1)\int_{1}^{\infty} x^a dx$$

This integral converges when $a<-1$. This is the easy case.

If $ b>0$, let $\displaystyle\theta=\frac{1}{x^b}$, we get

$$I=\frac{1}b\int_{0}^{1} \frac{\sin\theta}{\theta^{p+1}} ~d\theta,~~~~~p=\frac{a+1}{b}$$

Note that: $\displaystyle \theta-\frac{\theta^3}{6}\le\sin\theta\le\theta$

if $p<1$

$$I\le \frac{1}b\int_{0}^{1} \frac{\theta}{\theta^{p+1}} ~d\theta=\frac{1}b\int_{0}^{1} \frac{1}{\theta^{p}} ~d\theta=\frac{1}{b(1-p)}\longrightarrow\text{converge}$$

if $p\ge 1$

$$I\ge \frac{1}b\int_{0}^{1} \frac{\theta-\frac{\theta^3}{6}}{\theta^{p+1}} ~d\theta=\frac{1}b\int_{0}^{1} \left(\frac{1}{\theta^{p}}-\frac{1}6\theta^{2-p}\right)~d\theta\longrightarrow\text{diverge}$$

Summary:

Given that $b\ge0$, this integral converges if and only if $\boxed{a+1<b}$.

Answer 2

If $b=0$, the integral $\int_1^{+\infty} x^a\,dx $ converges if and only if $a < -1$. When $b > 0$ , for large $x$, you have that $\sin(1/x^b) \sim \frac{1}{x^b}$, hence, the integral has the same nature as $\int_1^{+\infty} x^{a-b}\, dx$, which is convergent if and only if $a-b < -1$, i.e. $a < b-1$.

Answer 3

In general $$\boxed{\textrm{the integral converges iff $a + 1 < b$}}.$$ We establish this fact separately for each possible sign of $b$:

• For $b > 0$ and large $x$, $$x^a \sin\left(\frac{1}{x^b}\right) - x^{a - b} \in o\left(x^{a - b}\right) .$$ In particular, the integral converges iff $a + 1 < b$.

• For $b = 0$, the integral is $\sin \int_1^\infty x^a \,dx$ and so converges iff $a < - 1$.

• For $b < 0$, the substitution $u = x^{-b}$ transforms the integral to $$-\frac{1}{b} \int_1^\infty u^{-\left(\frac{a + 1}{b} + 1\right)} \sin u \,du,$$ so it converges iff the exponent is $< -1$, i.e., iff $a + 1 < b$.

Remark In fact we can give explicit formulae for the integral, that apply when $a + 1 < b$:

• $b > 0$: $\displaystyle\frac{{}_1F_2 \left(\frac{b - a - 1}{2 b}; \frac{3}{2}, \frac{3 b - a - 1}{2 b}; -\frac{1}{4}\right)}{b - a - 1} $, where ${}_1 F_2$ denotes the hypergeometric function.
• $b = 0$: $\displaystyle -\frac{\sin 1}{1 + a} .$
• $b < 0$: $\displaystyle \frac{{}_1F_2 \left(\frac{b - a - 1}{2 b}; \frac{3}{2}, \frac{3 b - a - 1}{2 b}; -\frac{1}{4}\right)}{b - a - 1} - \frac{\pi \sec \frac{\pi (a + 1)}{2 b}}{2 b \Gamma\left(\frac{a + b + 1}{b}\right)}$