I'm going through the answer to a Fourier series question and have come across some notation which I haven't seen before.
The question is to represent the periodic function
$$f(x) = \left\{\begin{array}{ll} -1 - \sin(x) & : - \pi < x \le 0\\ 1 - \sin(x) & : 0 < x \le \pi \end{array} \right.$$
(with the period $2\pi$) as a Fourier series.
I'm finding only the $b_{n}$ terms since it is clearly an odd function.
It quickly reduces to
$$ b_{n} = \frac{2}{\pi} \int_0^{\pi}{\sin(nx)}\,dx - \frac{1}{\pi} \int_{-\pi}^{\pi}{\sin(x) \cdot \sin(nx)}\,dx $$
and the next line in the answer is:
$$ = \bigg[-\frac{2}{n\pi} \cos(nx)\bigg]^{\pi}_{0} - \delta_{m1}$$
I can see how the first integral reduces to the first part — it's just the delta that I've not seen before.
I tried to solve the second integral myself, using trig rules, and I got
$$ - \frac{1}{\pi} \int_{-\pi}^{\pi}{\sin(x) \cdot \sin(nx)}\,dx =- \frac{2\sin(n \pi)}{\pi(-1 + n^2)}$$
(ie. this should be the second 'half' of my $b_n$).
I'm not sure how to reduce this to the (correct) answer of
$$ f(x) = \sum_{0}^{\infty} \frac{2}{n\pi} (1 − (−1)^n) \sin(nx) - \sin(x) $$
The part I'm struggling with is how to get from the second 'half' of my $b_n$ to the $-\sin(x)$ term. Perhaps knowing what the $\delta$ is will help?
The $\delta_{ml}$ is shorthand notation for the following: if $m=l$, then $\delta_{ml} = 1$; if $m\neq l$, then $\delta_{ml} = 0$. Note that if $n\neq 1$, then $\int_{-\pi}^{\pi}\sin(x)\sin(nx)\,dx = 0$ which is most easily verified by using complex exponentials. If however $n=1$, then you're integrating $\int_{-\pi}^{\pi} \sin(x)^2\,dx$ (which is not zero). Does this help?