1

I have to show that for $f \in C_{2\pi}$ and $b_{n}$ Fourier coefficient the following holds:

$$\sum_{n=1}^{\infty} \dfrac bn=\dfrac 1{2\pi}\int_{0}^{2\pi} (\pi-x)f(x)dx - (1)$$

I have used for $g(x)=(\pi-x)=2\sum_{n=1}^{\infty} \dfrac {\sin(nx)}n$ for every $x \in (0,2\pi)$.

After substituting I have:

$$\dfrac 1\pi \int_{0}^{2\pi} \sum_{n=1}^{\infty} \dfrac{\sin(nx)}n f(x)dx$$

My question is can I interchange integral with sum now by using Fubini Theorem to obtain: $$\sum_{n=1}^{\infty} \dfrac 1n \dfrac 1{\pi} \int_{0}^{2\pi} \sin(nx) f(x)dx$$

Marco Cantarini
  • 33,062
  • 2
  • 47
  • 93
Melina
  • 937
  • 2
    Yes, you can. (You have two occurrences of $\frac{1}{n}$ in the last line, one should be eliminated.) The question is how to justify the interchange. The most convenient argument depends on whether you use the Lebesgue or the Riemann integral. In either case, you need to know a bit about how (how good) the Fourier series of $(\pi - x)$ converges. – Daniel Fischer Oct 08 '15 at 09:34
  • @DanielFischer Thank you. Yes, I have to remove $\dfrac 1n$ in the last line. – Melina Oct 08 '15 at 10:22

0 Answers0