Integrating Fouries series problem

[ognik]

As the 2nd part of a question, we start with the Fourier sin series expansion of dirac delta function $\delta(x-a)$ in the half-interval (0,L), (0 < a < L):

$ \delta(x-a) = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} sin \frac{n \pi x}{L} $

The questions goes on "By integrating both sides of this eqtn from 0 to x, show that the cos expansion of the square wave $ f(x) =\begin{cases}0,\; 0\le x \lt a \\ 1,\; a \lt 0 \lt L\end{cases}$

is $ f(x) = \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) sin(\frac{n\pi x}{L}) $

But when integrating both sides I get:

$ \int_{0}^{x}\delta(x-a).1 \,dx = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} \int_{0}^{x} sin \frac{n \pi x}{L} \,dx$

$ = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} (\frac{L}{n\pi}) [-cos \frac{n \pi x}{L}]^x_0 $

$ = -\frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) cos(\frac{n\pi x}{L}) $

It looks to me that I am correct, could someone please check this?

 

[mathmari]

But when integrating both sides I get:

$ \int_{0}^{x}\delta(x-a).1 \,dx = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} \int_{0}^{x} sin \frac{n \pi x}{L} \,dx$

$ = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} (\frac{L}{n\pi}) [-cos \frac{n \pi x}{L}]^x_0 $

$=$ -$\frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) cos(\frac{n\pi x}{L}) $

It looks to me that I am correct, could someone please check this?

The first sign of you result is wrong. It should be $$\frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} \sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} \sin(\frac{n\pi a}{L}) \cos(\frac{n\pi x}{L}) $$
The rest is correct! (Smile)

The questions goes on "By integrating both sides of this eqtn from 0 to x, show that the cos expansion of the square wave $ f(x) =\begin{cases}0,\; 0\le x \lt a \\ 1,\; a \lt 0 \lt L\end{cases}$

is $ f(x) = \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) sin(\frac{n\pi x}{L}) $

So there must be a typo...