As the title of this post, how could I prove $\Big(\dfrac{\sin t}{t}\Big)^{n}$ is Lebesgue integrable over $\mathbb{R}$ for $n\geq 2$?
It has been clear that $\dfrac{\sin t}{t}$ is not Lebesgue integral over $\mathbb{R}$, from this post: Showing $\frac{\sin x}{x}$ is NOT Lebesgue Integrable on $\mathbb{R}_{\ge 0}$, but I don't think there is a post about the case of $n\geq 2$.
My attempt:
So for $n\geq 2$, $$\int_{\mathbb{R}}\Big|\dfrac{\sin t}{t}\Big|^{n}dt\leq \int_{\mathbb{R}}\dfrac{1}{|t^{n}|}dt<\infty,$$ but then this post $1/|x|^n$ is not integrable seems suggest that the above last inequality is wrong, since this post argued over $\mathbb{R}^{d}$ and concluded that $\dfrac{1}{|t|^{n}}$ is not integrable for all $n\in (0, d],$ and of course includes the case of $n\geq 2$.
What should I do?
Thank you!
Edit 1:
Oh, by the way, since $\Big(\dfrac{\sin(t)}{t}\Big)^{n}$ is an even function, I think it is sufficient to show that it is integrable over $\mathbb{R}_{\geq 0}$.
Edit 2:
Okay I think I figured it out.
Since $\Big(\dfrac{\sin t}{t}\Big)^{n}$ is an even function, we only need to show $$\int_{0}^{\infty}\Big|\dfrac{\sin t}{t}\Big|^{n}dt<\infty.$$
Firstly, for any $a>0$, $$\int_{a}^{\infty}\Big|\dfrac{\sin t}{t}\Big|^{n}dt\leq\int_{a}^{\infty}\Big|\dfrac{1}{t}\Big|^{n}dt<\infty, $$ since $n\geq 2$ and $a>0$.
On the other hand, at $t=0$, $\Big(\dfrac{\sin t}{t}\Big)^{n}$ is defined and bounded since $\dfrac{\sin t}{t}$ is defined and bounded. (This is because $\lim_{t\rightarrow 0^{-}}\dfrac{\sin t}{t}=\lim_{t\rightarrow 0^{+}}\dfrac{\sin t}{t}=1$)
Therefore, $$\int_{0}^{\infty}\Big|\dfrac{\sin t}{t}\Big|^{n}dt<\infty,$$ and thus $$\int_{-\infty}^{\infty}\Big|\dfrac{\sin t}{t}\Big|^{n}dt<\infty.$$
The only problem here is that I did not formally show that "$\Big(\dfrac{\sin t}{t}\Big)^{n}$ is integrable at $0$", since you cannot integrate at a point, so perhaps I need to integrate on $[0,\epsilon]$?
I would be really appreciate if anyone can give me a formal proof of $\Big(\dfrac{\sin t}{t}\Big)^{n}$ being integrable at $0$. Otherwise I will be stick to my current proof and answer my own post.
Since there is no further discussion, I am gonna answer my own post.
Since $\Big(\dfrac{\sin t}{t}\Big)^{n}$ is an even function, we only need to show $$\int_{0}^{\infty}\Big|\dfrac{\sin t}{t}\Big|^{n}dt<\infty.$$
Firstly, for any $a>0$, $$\int_{a}^{\infty}\Big|\dfrac{\sin t}{t}\Big|^{n}dt\leq\int_{a}^{\infty}\Big|\dfrac{1}{t}\Big|^{n}dt<\infty, $$ since $n\geq 2$ and $a>0$.
On the other hand, at $t=0$, $\Big(\dfrac{\sin t}{t}\Big)^{n}$ is defined and bounded since $\dfrac{\sin t}{t}$ is defined and bounded. (This is because $\lim_{t\rightarrow 0^{-}}\dfrac{\sin t}{t}=\lim_{t\rightarrow 0^{+}}\dfrac{\sin t}{t}=1$)
Therefore, $$\int_{0}^{\infty}\Big|\dfrac{\sin t}{t}\Big|^{n}dt<\infty,$$ and thus $$\int_{-\infty}^{\infty}\Big|\dfrac{\sin t}{t}\Big|^{n}dt<\infty.$$