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Prove the orthogonality relations for sine and cosine using complex exponentials

  1. Prove the integral formula,

        \[ \int_0^{2 \pi} e^{inx} e^{-imx} \, dx =  \begin{cases} 0 & \text{if } m \neq n, \\ 2 \pi & \text{if } m = n. \end{cases} \]

    for integers m and n.

  2. Prove the following orthogonality relations of sine and cosine using the relation in part (a), where m and n are integers with m^2 \neq n^2.

        \[ \int_0^{2 \pi} \sin (nx) \cos (mx) \, dx = \int_0^{2 \pi} \sin (nx) \sin (mx) \, dx = \int_0^{2 \pi} \cos (nx) \cos (mx) \, dx = 0, \]

        \[ \int_0^{2 \pi} \sin^2 (nx) \, dx = \int_0^{2 \pi} \cos^2 (nx) \, dx  = \pi \qquad \text{if } n \neq 0. \]


  1. Proof. First, if n = m then we have

        \[ \int_0^{2 \pi} e^{inx} e^{-imx} \, dx = \int_0^{2 \pi} e^{ix(n-m)} \, dx = \int_0^{2 \pi} dx = 2 \pi. \]

    If n \neq m then we have

        \begin{align*}  \int_0^{2 \pi} e^{inx} e^{-imx} \, dx &= \int_0^{2 \pi} e^{ix(n-m)} \, dx \\[9pt]  &= \int_0^{2\pi} e^{ikx} \, dx &(\text{for some } k \neq 0) \\[9pt]  &= \frac{1}{k} \left( e^{ikx} \Bigr \rvert_0^{2\pi} \right) \\[9pt]  &= \frac{1}{k} \left( e^{2 \pi i k} - e^0 \right) \\[9pt]  &= \frac{1}{k} (1-1) = 0. \qquad \blacksquare \end{align*}

  2. Proof. These are all direct computations using part (a). Here they are,

        \begin{align*}  \int_0^{2 \pi} \sin (nx) \cos (mx) \, dx &= \int_0^{2 \pi} \left( \frac{e^{inx} - e^{-inx}}{2i} \right) \left( \frac{e^{imx} + e^{-imx}}{2} \right) \, dx \\[9pt]  &= \frac{1}{4i} \int_0^{2 \pi} \big( e^{ix(n+m)} - e^{ix(m-n)} + e^{ix(n-m)} - e^{ix(-n-m)} \big) \, dx \\[9pt]  &= 0. \end{align*}

    (The final line follows by part (a) and since m^2 \neq n^2 by hypothesis which implies n \neq m, -n \neq m and -n \neq -m.) Next,

        \begin{align*}  \int_0^{2\pi} \sin (nx) \sin (mx) \, dx &= \int_0^{2 \pi} \left( \frac{e^{inx} - e^{-inx}}{2i} \right) \left( \frac{e^{imx} - e^{-imx}}{2i} \right) \, dx \\[9pt]  &= -\frac{1}{4} \int_0^{2 \pi} \big( e^{ix(n+m)} - e^{ix(n-m)} - e^{ix(m-n)} + e^{ix(-n-m)} \big) \, dx \\[9pt]  &= 0. \end{align*}

    The third formula,

        \begin{align*}  \int_0^{2 \pi} \cos (nx) \cos(mx) \, dx &= \int_0^{2 \pi} \left( \frac{e^{inx} + e^{-inx}}{2} \right) \left( \frac{e^{imx} + e^{-imx}}{2} \right) \, dx \\[9pt]  &= \frac{1}{4} \int_0^{2 \pi} \big( e^{ix(n+m)} + e^{ix(n-m)} + e^{ix(m-n)} + e^{ix(-n-m)} \big) \, dx \\[9pt]  &= 0. \end{align*}

    For the next one we use the identities for \cos^2 \theta and \sin^2 \theta derived in this exercise (Section 9.10, Exercise #4(b)).

        \begin{align*}  \int_0^{2 \pi} \sin^2 (nx) \, dx &= \int_0^{2 \pi} \frac{1}{2}(1 - \cos(2nx)) \, dx \\[9pt]  &= \pi - \int_0^{2 \pi} \cos (2nx) \, dx \\  &= \pi. \end{align*}

    Finally,

        \begin{align*}  \int_0^{2 \pi} \cos^2 (nx) \, dx &= \frac{1}{2} \int_0^{2 \pi} (1 - \sin^2 nx)) \, dx \\[9pt]  &= 2 \pi - \pi = \pi. \qquad \blacksquare \end{align*}

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