Question

Show that in (5.2 ) the average values of$\mathit{s}\mathit{i}\mathit{n}\mathit{m}\mathit{x}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{n}\mathit{x}$ and of $\mathit{c}\mathit{o}\mathit{s}\mathit{m}\mathit{x}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{n}\mathit{x}\mathbf{,}\mathbf{}\mathit{m}\mathbf{\ne }\mathit{n}$are zero (over a period), by using the complex exponential forms for the sines and cosines as in (5.2).

Short answer

The function is ${\int }_{-\pi }^{\pi }\left(\sin mx\right)\sin \left(nx\right)dx={\int }_{-\pi }^{\pi }\left(\cos mx\right)\cos nxdx=\left\{\begin{array}{l}0\text{for}m\ne n\\ \frac{1}{2}\text{for}m=n\end{array}\right.$.

Step-by-step solution

Given

The given expressions are $\sin \left(mx\right)\sin \left(nx\right)$ and $\cos \left(mx\right)\cos \left(nx\right),m\ne n$.

Definition of Integral

The definite integral of a real-valued function f(x)with respect to a real variable xon an interval [a, b] is expressed as ${\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}\mathbf{=}\mathit{F}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{-}\mathit{F}\mathbf{\left(}\mathit{a}\mathbf{\right)}$.

Here, $\int =$ Integration symbol, a= Lower limit, b= Upper limit, f(x) Integral and dx Integrating agent

Thus, ${\int }_a^{b}f\left(x\right)dx$is read as the definite integral of $f\left(x\right]$with respect to dx from a to b.

Integrate the Cosines

Apply integral on cosines.

$$\begin{gathered} \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(\cos mx\right)\cos nxdx=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(\frac{e^{imx}+e^{-imx}}{2}\right)\left(\frac{e^{inx}+e^{-inx}}{2}\right)dx \\ =\frac{1}{8\pi }{\int }_{-\pi }^{\pi }\left(e^{imx}+e^{-imx}\right)\left(e^{inx}+e^{-inx}\right)dx \\ =\frac{1}{8\pi }{\int }_{-\pi }^{\pi }\left(e^{ix(m+n)}+e^{ix(m-n)}+e^{ix(n-m)}+e^{-ix(m+n)}\right)dx \\ =\frac{1}{4\pi }{\left[\frac{e^{ix(m+n)}}{i(m+n)}+\frac{e^{ix(m-n)}}{i(m-n)}+\frac{e^{ix(n-m)}}{i(n-m)}-\frac{e^{-ix(m+n)}}{i(m+n)}\right]}_{-\pi }^{\pi } \end{gathered}$$

Calculate further as shown below.

$$\begin{gathered} \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(\cos mx\right)\cos nxdx=\frac{1}{8\pi }{\left[\left\{\frac{e^{ix(m+n)}}{i(m+n)}-\frac{e^{-ix(m+n)}}{i(m+n)}\right\}+\left\{\frac{e^{ix(m-n)}}{i(m-n)}+\frac{e^{ix(n-m)}}{i(n-m)}\right\}\right]}_{-\pi }^{\pi } \\ =\frac{1}{4\pi }{\left[\frac{\sin \left\{x\right(m+n\left)\right\}}{m+n}+\frac{\sin \left\{x\right(m-n\left)\right\}}{m-n}\right]}_{-\pi }^{\pi } \\ =\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \right(m+n\left)\right\}}{m+n}+\frac{\sin \left\{\pi \right(m-n\left)\right\}}{m-n}\right] \end{gathered}$$

For $m\ne n$, where $m\in N$and $n\in N$, the above result is equal to zero.

For $m=n$, first term =0.

$\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(\cos mx\right)\cos nxdx=\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \right(m-n\left)\right\}}{m-n}\right]$

Take the limit $\underset{m\to n\to 0}{\lim }\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \left(m-n\right)\right\}}{m-n}\right]$.

We know that $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$.

$\underset{m\to n\to 0}{\lim }\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \right(m-n\left)\right\}}{m-n}\right]=\frac{\pi }{2\pi }=\frac{1}{2}$

Integrate the sines

Apply integral on sines.

$$\begin{gathered} \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(\sin mx\right)\sin \left(nx\right)dx=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(\frac{e^{imx}-e^{-imx}}{2}\right)\left(\frac{e^{inx}-e^{-inx}}{2}\right)dx \\ =\frac{1}{8\pi }{\int }_{-\pi }^{\pi }\left(e^{imx}-e^{-imx}\right)\left(e^{inx}-e^{-inx}\right)dx \\ =\frac{1}{8\pi }{\int }_{-\pi }^{\pi }\left(e^{ix(m+n)}-e^{ix(m-n)}-e^{ix(n-m)}+e^{-ix(m+n)}\right)dx \\ =\frac{1}{4\pi }{\left[\frac{e^{ix(m+n)}}{i(m+n)}-\frac{e^{ix(m-n)}}{i(m-n)}-\frac{e^{ix(n-m)}}{i(n-m)}-\frac{e^{-ix(m+n)}}{i(m+n)}\right]}_{-\pi }^{\pi } \end{gathered}$$

Calculate further as follows:

$$\begin{gathered} \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(\sin mx\right)\sin \left(nx\right)dx=\frac{1}{8\pi }{\left[\left\{\frac{e^{ix(m+n)}}{i(m+n)}-\frac{e^{-ix(m+n)}}{i(m+n)}\right\}-\left\{\frac{e^{ix(m-n)}}{i(m-n)}-\frac{e^{ix(n-m)}}{i(m-n)}\right\}\right]}_{-\pi }^{\pi } \\ =\frac{1}{4\pi }{\left[\frac{\sin \left\{x\right(m+n\left)\right\}}{m+n}-\frac{\sin \left\{x\right(m-n\left)\right\}}{m-n}\right]}_{-\pi }^{\pi } \\ =\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \right(m-n\left)\right\}}{m-n}-\frac{\sin \left\{\pi \right(m+n\left)\right\}}{m+n}\right] \end{gathered}$$

For $m\ne n$, where $m\in N$ and $n\in N$, the above result is equal to zero.

For $m=n$, second term =0.

$\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(\sin mx\right)\sin \left(nx\right)dx=\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \right(m-n\left)\right\}}{m-n}\right]$

Take the limit $\underset{m-n\to 0}{\lim }\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \right(m-n\left)\right\}}{m-n}\right]$.

We know that$\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$.

$$\begin{gathered} \underset{m-n\to 0}{\lim }\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \right(m-n\left)\right\}}{m-n}\right]=\frac{\pi }{2\pi } \\ =\frac{1}{2} \end{gathered}$$

Hence, ${\int }_{-\pi }^{\pi }\left(\sin mx\right)\sin \left(nx\right)dx={\int }_{-\pi }^{\pi }\left(\cos mx\right)\cos nxdx=\left\{\begin{array}{l}0,\text{for}m\ne n\\ \frac{1}{2}\text{for}m=n\end{array}\right.$.