# Showing that the equation $f(x)=1$ satisfy the condition of the fourier integral

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?

• 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. Sep 22, 2016 at 13:17
• 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? Sep 22, 2016 at 13:30
• Just say "this function does not satisfy the conditions of the theorem because it does not have a finite integral." Sep 22, 2016 at 13:31