Show that for f(x) satisfying Dirichlet's conditions
Show that for $f(x)$ satisfying Dirichlet's conditions on $(-\pi,\pi)$ with continuous third derivative (and periodic boundary matching of $f,f',f''$), $$ a_n=\frac1{\pi n^3}\int_{-\pi}^\pi f'''(x)\sin(nx)\,dx, \qquad b_n=-\frac1{\pi n^3}\int_{-\pi}^\pi f'''(x)\cos(nx)\,dx. $$
Explanation of Method
Three successive integrations by parts on $a_n=\frac1\pi\int f\cos(nx)\,dx$ (and similarly for $b_n$) each strip off one derivative from $f$ and one power of $n$ from the denominator; the boundary terms vanish at every stage because $\sin(n\pi)=\sin(-n\pi)=0$ and, when a $\cos$ boundary term appears, because the corresponding derivative of $f$ agrees at $\pm\pi$ by periodicity.
- Integrate $a_n$ by parts once, using $\sin(n\pi)=0$ to kill the boundary term.
- Integrate by parts a second time, using periodicity of $f'$ to kill that boundary term.
- Integrate a third time, again using $\sin(n\pi)=0$, to arrive at the $f'''$ formula; repeat the parallel three steps for $b_n$.
Step 1 · First integration by parts (for $a_n$)
With $u=f(x)$, $dv=\cos(nx)\,dx$, and $\sin(n\pi)=\sin(-n\pi)=0$: $$ a_n = \frac1\pi\int_{-\pi}^\pi f(x)\cos(nx)\,dx = -\frac1{\pi n}\int_{-\pi}^\pi f'(x)\sin(nx)\,dx. $$
Step 2 · Second and third integrations
With $u=f'(x)$, $dv=\sin(nx)\,dx$, and using $\cos(n\pi)=\cos(-n\pi)=(-1)^n$ together with periodicity $f'(\pi)=f'(-\pi)$ (so the boundary term cancels): $$ a_n = -\frac1{\pi n^2}\int_{-\pi}^\pi f''(x)\cos(nx)\,dx. $$ One more integration by parts (using $\sin(n\pi)=0$ again): $$ \boxed{a_n = \frac1{\pi n^3}\int_{-\pi}^\pi f'''(x)\sin(nx)\,dx} $$
Step 3 · The parallel derivation for $b_n$
The identical three-step process, starting from $b_n=\frac1\pi\int f\sin(nx)\,dx$ (using $\cos(n\pi)=\cos(-n\pi)$ and periodicity of $f$ to kill the first boundary term, $\sin(n\pi)=0$ for the second, and periodicity of $f''$ for the third), gives $$ \boxed{b_n = -\frac1{\pi n^3}\int_{-\pi}^\pi f'''(x)\cos(nx)\,dx} $$
Summary Table
| Property | Value |
|---|---|
| $a_n$ | $\dfrac1{\pi n^3}\displaystyle\int_{-\pi}^\pi f'''(x)\sin(nx)\,dx$ |
| $b_n$ | $-\dfrac1{\pi n^3}\displaystyle\int_{-\pi}^\pi f'''(x)\cos(nx)\,dx$ |
| Consequence | coefficients decay at least as fast as $1/n^3$ (since $f'''$ is bounded) |
Additional Notes
- This is the general theoretical mechanism behind every rapid-decay coefficient seen elsewhere in this collection: Problem 341's $1/n^4$ decay for $(\pi^2-x^2)^2$ (whose extension is $C^1$-periodic) and Problem 329's $1/n^3$ decay both trace directly to this integration-by-parts principle.
- The requirement that $f,f',f''$ match at $\pm\pi$ (periodic boundary conditions) is essential — Problem 324's $f(x)=x$ has $f(\pi)\ne f(-\pi)$, which is exactly why its coefficients decay only like $1/n$ rather than faster, despite $x$ itself being perfectly smooth on the open interval.
- By induction, a function with $k$ continuous, periodically-matching derivatives has Fourier coefficients decaying at least as $1/n^{k+1}$ — the general "smoothness implies fast Fourier decay" principle that underlies the entire progression of decay rates seen across Problems 324–357.