Please help me find Fourier series mistake

[ognik]

Find the Fourier sin series expansion of dirac delta function $\delta(x-a)$ in the half-interval (0,L), (0 < a < L):

Now $b_n = \frac{1}{L} \int_0^L f(x)sin \frac{n \pi x}{L}dx $ - but L should be $\frac{L}{2}$ for this exercise...

So I would get $ \frac{2}{L} \int_0^L f(x)sin \frac{n \pi x}{\frac{L}{2}}dx $ - but the book shows $ \frac{2}{L} \int_0^L \delta(x-a)sin \frac{n \pi x}{L}dx $

I believed that the interval was $ \frac{2}{L}$ - what am I missing please?

 

[mathmari]

Find the Fourier sin series expansion of dirac delta function $\delta(x-a)$ in the half-interval (0,L), (0 < a < L):

Now $b_n = \frac{1}{L} \int_0^L f(x)sin \frac{n \pi x}{L}dx $

The formula for the coefficients $b_n$ is $$b_n=\frac{2}{L}\int_{0}^Lf(x)\sin \left (\frac{n\pi x}{L}\right )dx$$

 

[ognik]

My question is - why does the formula use $\frac{L}{2}$ as the interval before the integration $(\frac{2}{L} \int ...)$ but not also use $\frac{L}{2}$ in divisor of the sin term which I understand is also the interval?

(Given that I am writing the exam in a couple of hours, a quick reply would be appreciated :-) )

 

[mathmari]

It is related to the fact that for an even function $g(x)$ it holds that

$$ \int_{-a}^a g({x}) d x = 2 \int_0^a g ({x}) d x$$