91Ó°ÊÓ

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

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

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

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

Multiply the given equation with sin(nx).

$\frac{1}{2\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}f\left(x\right)\sin \left(nx\right)dx=\frac{1}{2\mathrm{π}}\underset{k=1}{\overset{∞}{∑}}{∫}_{-\mathrm{π}}^{\mathrm{π}}\left[b_k\sin \left(nx\right)\sin \left(kx\right)+a_k\sin \left(nx\right)\cos \left(kx\right)\right]dx$

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

Apply integral on sines.

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

Calculate further as shown below.

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

For $k≠n$, where $k∈N$and $n∈N$, the above result is equal to zero.

For k = n, second term = 0.

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

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

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

$$\begin{gathered} \underset{k-n→0}{\lim }\frac{1}{2\mathrm{π}}\left[\frac{\sin \left\{\mathrm{π}\right(k-n\left)\right\}}{k-n}\right]=\frac{\mathrm{π}}{2\mathrm{π}} \\ =\frac{1}{2} \\ {∫}_{-\mathrm{π}}^{\mathrm{π}}\left(\sin kx\right)\sin \left(nx\right)dx=\left\{\begin{array}{l}0,\text{for}k≠n\\ \frac{1}{2}\text{for}k=n\end{array}\right. \end{gathered}$$

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