Fourier Series Integration
Theorem
Let’s say a periodic function $f$ with a period $2L$ is piecewise continuous in the interval $[-L,\ L)$. Then, the definite integral of $f$ can be expressed as follows.
$$ \int_{t_{1}}^{t_{2}} f(t) dt= c_{0}(t_{2}-t_{1}) +\sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n}\left( e^{i\frac{n\pi t_{2}}{L}}-e^{i\frac{n\pi t_{1}}{L}} \right) $$
Here, $c_{0},\ c_{n}$ is the complex Fourier coefficient.
Thus, the definite integral of $f(t)$ is the same as summing the definite integrals of each term of the Fourier series of $f(t)$. It is important to note that the series on the right side is not the Fourier series of the left side.
Proof
Let’s define a function $F$ as follows.
$$ F(t):= \int_{0}^t f(s)ds-c_{0}t $$
Then, it can be shown that $F$ is a periodic function with a period $2L$.
$$ \begin{align*} F(t+2L) &= \int_{0}^{t+2L} f(s)ds - c_{0}(t+2L) \\ &= \int_{0}^t f(s)ds+\int_{t}^{2L}f(s)ds - c_{0}t -2Lc_{0} \\ &= \int_{0} ^t f(s)ds-c_{0}t=F(t) \end{align*} $$
Since $f$ is piecewise continuous, $F$ is continuous. Therefore, $F$ satisfies the Dirichlet conditions and can be expanded into a Fourier series.
$$ F(t)=C_{0}+\sum \limits_{n \ne 0} C_{n}e^{i\frac{n\pi}{L}t} $$
The reason for separating the constant term will be known later. When calculating the coefficients,
$$ \begin{align*} C_{n}&=\dfrac{1}{2L}\int_{-L}^{L} F(t)e^{-i\frac{n\pi}{L}t} dt \\ &= \dfrac{-L}{in\pi} \dfrac{1}{2L} \left[ F(t)e^{i\frac{n\pi}{L}t} \right]_{-L}^{L} +\dfrac{L}{in\pi}\dfrac{1}{2L} \int_{-L}^{L} F^{\prime}(t)e^{-i\frac{n\pi }{L}t}dt \\ &= \dfrac{L}{in\pi}\dfrac{1}{2L} \int_{-L}^{L} \left( f(t)-c_{0} \right) e^{-i\frac{n\pi}{L}t}dt \\ &= \dfrac{L}{in\pi}\dfrac{1}{2L} \int_{-L}^{L} f(t)e^{-i\frac{n\pi}{L}t}dt -\dfrac{L}{in\pi}\dfrac{c_{0}}{2L} \int_{-L}^{L} e^{-i\frac{n\pi}{L}t} dt \\ &= \dfrac{L}{in\pi}\dfrac{1}{2L} \int_{-L}^{L} f(t)e^{-i\frac{n\pi}{L}t}dt \\ &= \dfrac{L}{in\pi}c_{n} \end{align*} $$
Therefore,
$$ F(t)=C_{0} +\sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} e^{i\frac{n\pi}{L}t} $$
Using the definition of $F(t)$ to calculate the definite integral of $f(t)$ gives
$$ \begin{align*} \int_{t_{1}}^{t_{2}} f(t) dt &= \int_{0}^{t_{2}} f(t) dt -\int_{0}^{t_{1}} f(t)dt \\ &= F(t_{2})- F(t_{1}) + c_{0}(t_{2}-t_{1}) \\ &= \left( C_{0}+\sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} e^{i\frac{n\pi}{L}t_{2}} \right) -\left( C_{0}+\sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} e^{i\frac{n\pi}{L}t_{1}} \right) +c_{0}(t_{2}-t_{1}) \\ &= c_{0}(t_{2}-t_{1}) + \sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} \left( e^{i\frac{n\pi}{L}t_{2}}-e^{i\frac{n\pi}{L}t_{1}} \right) \end{align*} $$
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As mentioned at the beginning of the document, note that the series on the right is not the Fourier series of $\int_{t_{1}}^{t_{2}}f(t)dt$. The Fourier series of $\displaystyle \int_{t_{1}}^{t_{2}} f(t) dt -c_{0}(t_{2}-t_{1})$ is $\sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} \left( e^{i\frac{n\pi}{L}t_{2}}-e^{i\frac{n\pi}{L}t_{1}} \right)$.