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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

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  • $\begingroup$ is it $$\int_{0}^\infty \cos(tx)\left(\frac{\sin(t)}{t}\right)^ndt$$? $\endgroup$ Oct 10 '17 at 14:22
  • $\begingroup$ Hint: do you know some limit theorems for integrals? $\endgroup$
    – GEdgar
    Oct 10 '17 at 14:31
  • $\begingroup$ What is $n?\,\,$ $\endgroup$
    – zhw.
    Oct 10 '17 at 16:25
  • $\begingroup$ @Dr. Sonnhard Graubner Yes. $\endgroup$
    – perlman
    Oct 10 '17 at 19:21
  • $\begingroup$ @GEdgar I thought of dominant convergence theorem, but the formulation I know is quite different (we have a sequence of functions etc). $\endgroup$
    – perlman
    Oct 10 '17 at 19:21
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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.

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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 $|f(x)-f(a)|<\epsilon$ .

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  • $\begingroup$ Hello. I know the definition of continuity. But I can't see how this help us with $n>1$. $\endgroup$
    – perlman
    Oct 10 '17 at 19:23
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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 $n>2.$

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The dominated convergence theorem shows $f$ is continuous for $n>1.$

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