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Question

Derive equations

∫0Tsin⁡nωTtsin⁡mωTtdt={0m≠nT/2m=n\int_{0}^{T} \sin n \omega_{T} t \sin m \omega_{T} t d t=\left\{\begin{array}{ll}{0} & {m \neq n} \\ {T / 2} & {m=n}\end{array}\right.

,

∫0Tcos⁡nωTtcos⁡mωTtdt={0m≠nT/2m=n,\int_{0}^{T} \cos n \omega_{T} t \cos m \omega_{T} t d t=\left\{\begin{array}{ll}{0} & {m \neq n} \\ {T / 2} & {m=n}\end{array}\right.,

and ∫0Tcos⁡nωTtsin⁡mωTtdt=0\int_{0}^{T} \cos n \omega_{T} t \sin m \omega_{T} t d t=0 and hence verify the equations for the Fourier coefficient given by equations a0=2T∫0TF(t)dt,a_{0}=\frac{2}{T} \int_{0}^{T} F(t) d t, an=2T∫0TF(t)cos⁡nωTtdtn=1,2,…,a_{n}=\frac{2}{T} \int_{0}^{T} F(t) \cos n \omega_{T} t d t \quad n=1,2, \ldots, and bn=2T∫0TF(t)sin⁡nωTtdtn=1,2,….b_{n}=\frac{2}{T} \int_{0}^{T} F(t) \sin n \omega_{T} t d t \quad n=1,2, \ldots.

Solution

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We begin with the first given integral for the product of sines. We'll divide it into two cases, for m≠nm \neq n and for m=nm = n. For all of the integrals the frequency is defined as ωT=2πT\omega_T = \dfrac{2 \pi}{T}.

For m≠nm \neq n we have:

∫0Tsin⁡(nωTt)sin⁡(mωTt)dt=[sin⁡((n−m)ωTt)2(n−m)ωT−sin⁡((n+m)ωTt)2(n+m)ωT]0T=[sin⁡((n−m)ωTT)2(n−m)ωT−sin⁡((n+m)ωTT)2(n+m)ωT−sin⁡((n−m)ωT⋅0)2(n−m)ωT+sin⁡((n+m)ωT⋅0)2(n+m)ωT]=sin⁡((n−m)ωTT)2(n−m)ωT−sin⁡((n+m)ωTT)2(n+m)ωT−0=sin⁡((n−m)2πTT)2(n−m)ωT−sin⁡((n+m)2πTT)2(n+m)ωT=sin⁡((n−m)2π)2(n−m)ωT−sin⁡((n+m)2π)2(n+m)ωT=0−0∫0Tsin⁡(nωTt)sin⁡(mωTt)dt=0m≠n\begin{align*} \int_0^T \sin(n \omega_T t) \sin(m \omega_T t) dt &= \left[ \dfrac{\sin((n - m)\omega_T t)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m)\omega_T t)}{2 (n + m) \omega_T} \right]_0^T \\ &= \left[ \begin{array} {ll} \dfrac{\sin((n - m)\omega_T T)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m)\omega_T T)}{2 (n + m) \omega_T} \\ - \dfrac{\sin((n - m)\omega_T \cdot 0)}{2 (n - m) \omega_T} + \dfrac{\sin((n + m)\omega_T \cdot 0)}{2 (n + m) \omega_T} \end{array} \right] \\ &= \dfrac{\sin((n - m)\omega_T T)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m)\omega_T T)}{2 (n + m) \omega_T} - 0 \\ &= \dfrac{\sin((n - m)\dfrac{2 \pi}{T} T)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m)\dfrac{2 \pi}{T} T)}{2 (n + m) \omega_T} \\ &= \dfrac{\sin((n - m) 2 \pi)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m) 2 \pi)}{2 (n + m) \omega_T} \\ &= 0 - 0 \\ \int_0^T \sin(n \omega_T t) \sin(m \omega_T t) dt &= 0 \qquad m \neq n \end{align*}

mm and nn are both integers so their sum or difference will be an integer and any sine of a product of 2π2 \pi with an integer will be 0.

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