Proving convergence of improper integral with Dirichlet's test

Question

Given the following integral:

$$ \int_1^\infty x^{-1}\sin\left(x^{\frac{3}{2}}\right)\sin(x)~dx $$

I need to prove it converge using Dirichlet's test. This is what I tried:

Let $g(x)=x^{-1}$, then it is diffrentiable, and also $\lim_{x\to\infty}g(x)=0$. For $g'(x)$ it is the case that $\int_a^\infty|g'(x)|dx$ is absolutley converge.

Now let $f(x)=\sin\left(x^{\frac{3}{2}}\right)\sin x$. Then $f$ is continuous. My only struggle is to show that $F(x)=\int_1^x \sin\left(t^{\frac{3}{2}}\right)\sin tdt$ is bounded, which I believe to be true bet not $100\%$ sure.

Answer 1

Here is another solution , perhaps more elegant:

Let $f(x)=x^{\frac{1}{2}}\sin\left(x^{\frac{3}{2}}\right)$ and $g(x)=x^{-\frac{3}{2}}\sin x$.

$f$ is an elementary function so it is continuous. For $F(x)$ we get:

$$ F(x)=\int_1^x t^{\frac{1}{2}}\sin\left(x^{\frac{3}{2}}\right)dt=-\frac{2}{3}\cos\left(x^{\frac{3}{2}}\right)+\frac{2}{3}\cos\left(1\right) $$

So $F$ is bounded. For $g(x)$, $g$ is diffrentiable and $\lim_{x\to\infty}g(x)=0$. All that left to do is to show that $g'(x)$ is absolutly convergent, which is the case here. Then $f,g$ satisfies all the conditions and $\int_1^{\infty}f(x)g(x)dx$ converge.

Answer 2

$$I=\int_1^\infty \frac{1}{t}\sin\left(t^{\frac{3}{2}}\right)\sin tdt$$

Let $t=u^2$

$$I=2\int_1^\infty \frac{1}{u}\sin\left(u^3\right)\sin u^2du$$

Let

$$F(a)=2\int_1^a \sin\left(u^3\right)\sin u^2du=\int_1^a \cos\left(u^3-u^2\right)-\cos\left(u^3+u^2\right)du$$

If $a$ is finite, $|F(a)|\le 2a$ is finite. So we only need to consider the infinite case, namely $F(\infty)$.

Let

$$I_1=\int_1^\infty \cos\left(u^3-u^2\right)du,~~~I_2=\int_1^\infty \cos\left(u^3+u^2\right)du$$ Integration by part $$I_1=\int_1^\infty \frac{1}{3u^2-2u}d\sin\left(u^3-u^2\right)=\frac{\sin\left(u^3-u^2\right)}{3u^2-2u}\big{|}_1^\infty+\int_1^\infty \frac{(3u-2)\sin\left(u^3-u^2\right)}{(3u^2-2u)^2}du$$

where

$$\left|\int_1^\infty \frac{(3u-2)\sin\left(u^3-u^2\right)}{(3u^2-2u)^2}du\right|\le \int_1^\infty \frac{(3u-2)}{(3u^2-2u)^2}du\longrightarrow \text{converge}$$

We can treat $I_2$ by the similar way.