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Show why the Fourier integral formula fails to represent the function $f(x)=1$ ($-\infty<x<\infty)$. Which condition in the theorem is not satisfied by the function?

What i tried

The theorem states that,

Let $f$ denote a function which is piecewise continuous on every bounded interval of the $x$ axis, and suppose that it is absolutely integrable over that axis; that is, the improper integral

$$\int_{-\infty}^\infty |f(x)| dx$$

convergers, then the fourier integral

$$\frac{1}{\pi}\int_0^{\infty}\int_{-\infty}^{\infty} f(s)\cos(a(s-x) \,\mathrm ds\,\mathrm da$$

convergers to the mean value

$$\frac{f(x+)+f(x-)}{2}$$

of the one-sided limits of f at each point ($-\infty<x<\infty)$ where both of the one-sided derivatives $f'_{R}(x)$and $f'_{L}(x)$ exist.

I mentioned that $f(x)=1$ is not bounded and is not piecewise to begin with thus it cannot satisfy the conditions.also the integral $$\int_{-\infty}^\infty |1| dx$$ does not converge. Am i correct, and is there a better explanation?

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    $\begingroup$ It Is bounded and piecewise continuous (the fact that there's just one piece is not relevant). The last thing about the integral not being finite is the issue. $\endgroup$ Sep 22, 2016 at 13:17
  • $\begingroup$ Okay thanks. I got the rough idea. The integral is not finite thus it tends to infinity and hence not convergent which dosent satisfy the conditions. But how do i phrase my explanation properly to make it clearer and more precise? $\endgroup$
    – ys wong
    Sep 22, 2016 at 13:30
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    $\begingroup$ Just say "this function does not satisfy the conditions of the theorem because it does not have a finite integral." $\endgroup$ Sep 22, 2016 at 13:31

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