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This is my first post here.

I have some troubles with this property of the Fourier coefficients. Indeed, let $f(x)$ be a continuous real function, with compact support $[a,b] \subset (0,2\pi)$, and $ b_k := \frac{1}{\pi}\int_0^{2\pi}f(x)\sin kx \; \mathrm{d}x $

My question is: does the series $ \sum_{k=1}^\infty \frac{b_k}{k}$ converge? If the answer is yes, is its sum equal to $\int_0^{2\pi}\frac{\pi-x}{2}f(x) \; \mathrm{d}x$?

How can I prove/disprove this? What do you suggest? Thanks a lot.

  • 0
    @Chris True : )2012-04-11

1 Answers 1

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For arbitrary continuous function $\varphi\in C([0,2\pi])$ denote $ a_0(\varphi)=\frac{1}{\pi}\int\limits_{0}^{2\pi}\varphi(x)dx $ $ a_k(\varphi)=\int\limits_{0}^{2\pi}\varphi(x)\cos(kx)dx\qquad b_k(\varphi)=\int\limits_{0}^{2\pi}\varphi(x)\sin(kx)dx $ Consider function $ F(x)=\int\limits_{0}^x f(t)dt-\frac{a_0(f)x}{2} $ Since $f\in C([0,2\pi])$ then $F\in C^1([0,2\pi])$. Note that $ F(x+2\pi)-F(x)=\int\limits_{0}^{2\pi}f(t)dt-a_0(f)\pi=0. $ So $F$ is $2\pi$ periodic. Since $F$ is $2\pi$ periodic continuously differentiable function its Fourier series uniformly converges to $F$. Thus we have for all $x\in[0,2\pi]$ the following equality $ F(x)=\frac{a_0(F)}{2}+\sum\limits_{k=1}^\infty\left(a_k(F)\cos(kx)+b_k(F)\sin(kx)\right) $ Integration by parts gives us $ a_0(F)=\frac{1}{\pi}\int\limits_{0}^{2\pi}F(x)dx= \frac{1}{\pi}\left(xF(x)|_0^{2\pi}-\int\limits_{0}^{2\pi}x\left(f(x)-\frac{a_0(f)}{2}\right)dx\right)= \frac{1}{\pi}\left(-\int\limits_{0}^{2\pi}xf(x)dx+\int\limits_{0}^{2\pi}\pi f(x)dx\right)= \int\limits_{0}^{2\pi}\frac{\pi-x}{\pi}f(x)dx $ $ a_k(F)=\int\limits_{0}^{2\pi}F(x)\cos(kx)dx= \left(F(x)\frac{\sin(kx)}{k}\right)_0^{2\pi}- \int\limits_{0}^{2\pi}\left(f(x)-\frac{a_0(f)}{2}\right)\frac{\sin(kx)}{k}dx= $ $ -\frac{1}{k}\int\limits_{0}^{2\pi}f(x)\sin(kx)dx=-\frac{1}{k}b_k(f) $ $ b_k(F)=\int\limits_{0}^{2\pi}F(x)\sin(kx)dx= \left(F(x)\frac{-\cos(kx)}{k}\right)_0^{2\pi}- \int\limits_{0}^{2\pi}\left(f(x)-\frac{a_0(f)}{2}\right)\frac{-\cos(kx)}{k}dx= $ $ \frac{1}{k}\int\limits_{0}^{2\pi}f(x)\cos(kx)dx=\frac{1}{k}a_k(f) $ Hence $ F(x)=\frac{a_0(F)}{2}+ \sum\limits_{k=1}^\infty\left(-\frac{b_k(f)}{k}\cos(kx)+\frac{a_k(f)}{k}\sin(kx)\right) $ Substitute $x=0$, then $ 0=\frac{a_0(F)}{2}-\sum\limits_{k=1}^\infty\frac{b_k(f)}{k} $ i.e. $ \sum\limits_{k=1}^\infty\frac{b_k(f)}{k}= \frac{a_0(F)}{2}= \int\limits_{0}^{2\pi}\frac{\pi-x}{2\pi}f(x)dx $

  • 0
    Not at all, Romeo :)2012-04-10