Question

Write out the details of the derivation of equation 5.10.

Short answer

The equation becomes $b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\sin \left(nx\right)dx$.

Step-by-step solution

Given

The given equation is $b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\sin \left(nx\right)dx$.

Definition of Integral

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

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(x) f(x) with respect to dx from a to b.

Multiply the given equation

Multiply the given equation with sin(nx).

$\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\sin \left(nx\right)dx=\frac{1}{2\pi }\sum _{k=1}^{\infty }{\int }_{-\pi }^{\pi }\left[b_k\sin \left(nx\right)\sin \left(kx\right)+a_k\sin \left(nx\right)\cos \left(kx\right)\right]dx$

In the above relation, the second part is always zero and first part is non zero (if k = n ).

Integrate the sines and simplify

Apply integral on sines.

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

Calculate further as shown below.

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

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

For k = n, second term = 0.

$=\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \right(k-n\left)\right\}}{k-n}\right]$

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

It is known that $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$.

$$\begin{gathered} \underset{k-n\to 0}{\lim }\frac{1}{2\pi }\left[\frac{\sin \left\{\pi \right(k-n\left)\right\}}{k-n}\right]=\frac{\pi }{2\pi } \\ =\frac{1}{2} \\ {\int }_{-\pi }^{\pi }\left(\sin kx\right)\sin \left(nx\right)dx=\left\{\begin{array}{l}0,\text{for}k\ne n\\ \frac{1}{2}\text{for}k=n\end{array}\right. \end{gathered}$$

Hence, $b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\sin \left(nx\right)dx$.