In three books, the Fourier series coefficients are in different forms. I don't understand why they are equivalent.
From book 1: \begin{align} a_n&=\frac{1}{L}\int_ {-L}^{L}f(t)\cos\Big(\frac{\pi n}{L}t\Big)\, dt\tag 1 \\ b_n&=\frac{1}{L}\int_ {-L}^{L}f(t)\sin\Big(\frac{\pi n}{L}t\Big)\, dt\tag 2 \end{align}
From book 2:
\begin{align} a_n&=\frac{2}{L}\int_ {-L/2}^{L/2}f(t)\cos\Big(\frac{2\pi n}{L}t\Big)\, dt\tag 3 \\ b_n&=\frac{2}{L}\int_ {-L/2}^{L/2}f(t)\sin\Big(\frac{2\pi n}{L}t\Big)\, dt\tag 4 \end{align}
From book 3: \begin{align} a_n&=\frac{2}{L}\int_ {0}^{L}f(t)\cos\Big(\frac{2\pi n}{L}t\Big)\, dt\tag 5 \\ b_n&=\frac{2}{L}\int_ {0}^{L}f(t)\sin\Big(\frac{2\pi n}{L}t\Big)\, dt\tag 6 \end{align}
Eq. $(1)-(2)$ don't have the extra factor of 2, why?
Can we find $(5)-(6)$ by changing the limits $[-L/2,L/2]=[0,L]$ in $(3)-(4)$?
Context is all. In the case of Fourier series there are several variations depending on the context. We assume we have a function $f(t)$ with period $T$. The Fourier series coefficients depend on the values of $f(t)$ on an interval of length $T$. In the case of book 1, $T=2L$ and the interval is $[-L,L]$. For the other two books, $T=L$. For book 2 the interval is $[-L/2,L/2]$ and for book 3 $[0,L]$. The common factor for all the integrals is $2/T.$ Expressing that in terms of $L$ yields some variations.