Continuity of $f(x)=\int_0^{\infty} \cos (tx)\big(\frac{\sin (t) }{t}\big)^n dt$ Is the following function
 $$f(x)=\int_0^{\infty} \cos(tx)\Bigg(\frac{\sin (t)}{t}\Bigg)^n dt$$ continuous in $x$?
how do we prove this?
Thanks
 A: Explaining use of dominated convergence theorem ... since OP asks.
Assume $n > 1$ (and $n$ need not be an integer).  So
$$
\int_0^\infty \left|\left(\frac{\sin t}{t}\right)^n\right|\;dt < +\infty
$$
by comparison with $1$ near $0$ and with $t^{-n}$ near $\infty$.  We will use this for the dominating function when we do dominated convergence.
To show $f$ is continuous, it suffices to prove: for any sequence $x_k$ with $x_k \to x$, we have $f(x_k) \to f(x)$.  Now
$$
f(x_k) = \int_0^\infty \cos(t x_k)\left(\frac{\sin t}{t}\right)^n\;dt
$$
Note since $x_k \to x$ we have $tx_k \to tx$ and $\cos$ is continuous, so $\cos(tx_k) \to \cos(tx)$
for all $t$.
So then by the dominated convergence theorem $f(x_k) \to f(x)$, as required.
A: Hint:
For $\,n=1\,$ it's not continuous for all $\,x\in\mathbb{R}$ , we have to exclude $\{-1,+1\}$ .
$\displaystyle\int_0^{\infty} \cos(tx)\big(\frac{\sin t}{t}\big)^1 dt = \frac{\pi}{4}(\text{signum}(1-x)+\text{signum}(1+x)) $
$\displaystyle\int_0^{\infty} \cos(tx)\big(\frac{\sin t}{t}\big)^2 dt = \frac{\pi}{8}(|x-2| - 2|x| + |x+2|) $ , it's continuous.
What one has to proof is:
The function $f$ is continuous in $a$ if for every $\epsilon$ exists a $\delta$ such that for all $x$ with $|x-a|<\delta$ 
is $|f(x)-f(a)|<\epsilon$ .
A: Hint: Use the following $|\cos a - \cos b| =2|\sin\dfrac{a+b}{2}\sin\dfrac{a-b}{2}|\leq 2\cdot 1\cdot\dfrac{|a-b|}{2} = |a-b|. $
Using this, you can prove the function is not just continuous, but Lipschitz continuous in $x$, when $n>2.$
A: The dominated convergence theorem shows $f$ is continuous for $n>1.$
