# [SOLVED]understanding theorem in Fourier series

#### dwsmith

##### Well-known member
Suppose $f$ is continuous and periodic with period $2\pi$ on $(-\infty,\infty)$, and $f'$ exist and is in $\mathcal{P}\mathcal{C}[-\pi,\pi]$.
Then $\sum\limits_{k = -\infty}^{\infty}\lvert A_k\rvert < \infty$.

$f'$ has a Fourier series so let's call the coefficients $A_n'$. Then $f' = \sum\limits_{n = -\infty}^{\infty}A_n'e_n$ and $A_n' = \frac{1}{2\pi}\int_{-\pi}^{\pi}f'(\theta)e^{-in\theta}d\theta$. Let's integrate by parts letting $u = e^{-in\theta}$ and $dv = f'(\theta)$. Then $du = -ine^{-in\theta}d\theta$ and $v = f(\theta)$.
\begin{alignat*}{3}
A_n' & = & \frac{1}{2\pi}\left[\left. f(\theta)e^{-in\theta}\right|_{-\pi}^{\pi} + in\int_{-\pi}^{\pi}f(\theta)e^{-in\theta}d\theta\right]\\
& = & \frac{1}{2\pi}\left[\underbrace{f(\pi)e^{-in\theta} - f(-\pi)e^{in\theta}}_{\text{this part}} + in\int_{-\pi}^{\pi}f(\theta)e^{-in\theta}d\theta\right]
\end{alignat*}

I know that $f(\pi) = f(-\pi)$ but why is that piece zero? We have $e^{in\theta}$ and $e^{-in\theta}$.

#### Opalg

##### MHB Oldtimer
Staff member
Suppose $f$ is continuous and periodic with period $2\pi$ on $(-\infty,\infty)$, and $f'$ exist and is in $\mathcal{P}\mathcal{C}[-\pi,\pi]$.
Then $\sum\limits_{k = -\infty}^{\infty}\lvert A_k\rvert < \infty$.

$f'$ has a Fourier series so let's call the coefficients $A_n'$. Then $f' = \sum\limits_{n = -\infty}^{\infty}A_n'e_n$ and $A_n' = \frac{1}{2\pi}\int_{-\pi}^{\pi}f'(\theta)e^{-in\theta}d\theta$. Let's integrate by parts letting $u = e^{-in\theta}$ and $dv = f'(\theta)$. Then $du = -ine^{-in\theta}d\theta$ and $v = f(\theta)$.
\begin{alignat*}{3}
A_n' & = & \frac{1}{2\pi}\left[\left. f(\theta)e^{-in\theta}\right|_{-\pi}^{\pi} + in\int_{-\pi}^{\pi}f(\theta)e^{-in\theta}d\theta\right]\\
& = & \frac{1}{2\pi}\left[\underbrace{f(\pi)e^{-in\theta} - f(-\pi)e^{in\theta}}_{\text{this part}} + in\int_{-\pi}^{\pi}f(\theta)e^{-in\theta}d\theta\right]
\end{alignat*}

I know that $f(\pi) = f(-\pi)$ but why is that piece zero? We have $e^{in\theta}$ and $e^{-in\theta}$.
You actually have $e^{in\pi}$ and $e^{-in\pi}$, and these are the same because they are both equal to $(-1)^n.$