Assuming knowledge of the cyclic behavior of $cos(x)$, integration by parts, and $\int_0^{\infty} f<\infty$ is enough here? Consider

\begin{align} & \int_0^\infty f(x)\cos(nx)dx = \left.\cos(nx)\int_0^\infty f(x)dx\right|_0^\infty+n\int_0^\infty f(x)\sin(nx)dx\\[2mm] & = \left.\cos(nx)\int_0^\infty f(x)dx\right|_0^\infty + \left.n\sin(nx)\int_0^\infty f(x)dx\right|_0^\infty-n^2\int_0^\infty f(x)\cos(nx)dx\\[2mm] & \text{So we have: }\\[2mm] & \left(1+n^2\right)\int_0^\infty f(x)\cos(nx)dx = \left.\cos(nx)\int_0^\infty f(x)dx\right|_0^\infty + \left.n\sin(nx)\int_0^\infty f(x)dx\right|_0^\infty\\[2mm] & \text{ and then }\\[2mm] & \int_0^\infty f(x)\cos(nx)dx =\frac{\left.\cos(nx)\int_0^\infty f(x)dx\right|_0^\infty + \left.n\sin(nx)\int_0^\infty f(x)dx\right|_0^\infty}{\left(1+n^2\right)} \end{align}

Taking the limit on both sides gives the result. Does anyone see a flaw in this argument?

  • $\begingroup$ If this argument works then I was wondering how much of a change in $f(x)$ would be required to make it false? $\endgroup$
    – JEM
    Apr 18 '14 at 21:37
  • $\begingroup$ After looking at this for a while. I think I've found a mistake. Choosing $u=\cos(nx),du = -n\sin(nx)$ and $dv = fdx, v=\int_0^\infty fdx$ does not produce $n\int_0^\infty f\sin(nx)dx$ but $n\int_0^\infty\left(\sin(nx)\int_0^\infty fdx\right)dx$ instead. The theorem is true but dont think parts will do it anymore. $\endgroup$
    – JEM
    Apr 24 '14 at 18:16
  • $\begingroup$ I think this proof might require some approximation theory. Back to the drawing board. $\endgroup$
    – JEM
    Apr 24 '14 at 21:52

The argument looks fine, although the notation is somewhat sloppy: for instance I would write the first step as

$$\int_0^{\infty} f(x)\cos(nx)\,dx = \cos(nx)\int_0^x f(y)\,dy\ \biggr\vert_{x=0}^{\infty} + n\int_0^{\infty} f(x)\sin(nx)\,dx.$$

The claim becomes false if $f(x)$ is locally integrable but not $L^1$, i.e. if $\lim_{x\to \infty} \int_0^x f(y)\,dy$ does not exist.

  • $\begingroup$ By 'integrable' you mean something like 'improperly integrable', right? Being integrable is synonymous with being in $L^1$ as far as I know. $\endgroup$
    – Potato
    Apr 18 '14 at 22:10
  • $\begingroup$ @Potato Yes, I mean locally integrable. Thanks. $\endgroup$
    – user7530
    Apr 18 '14 at 22:55

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