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?