91Ó°ÊÓ

The given function is $h\left(x\right)=\underset{n=-∞}{\overset{∞}{∑}}f\left(x+2n\mathrm{π}\right)$

For the given function$\mathit{u}\left(x\right)$define its Fourier transformation as

$\mathit{F}\left\{u\left(x\right)\right\}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{π}}}\underset{\mathbf{-}\mathbf{∞}}{\overset{\mathbf{∞}}{\mathbf{∫}}}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{k}\mathbf{x}}\mathit{u}\left(x\right)\mathit{d}\mathit{x}$

A function is said to be periodic functions if it repeats value after specific interval of time.

$$\begin{gathered} h\left(x\right)=\underset{n=-∞}{\overset{∞}{∑}}f\left(x+2n\mathrm{π}\right) \\ h\left(x+2\mathrm{π}\right)=\underset{n=-∞}{\overset{∞}{∑}}f\left(x+2\left(n+1\right)\mathrm{π}\right) \\ =\underset{n=-∞}{\overset{∞}{∑}}f\left(x+2m\mathrm{π}\right)m=n+1 \end{gathered}$$

Thus the function is indeed periodic.

$$\begin{gathered} h\left(x\right)=\underset{n=-∞}{\overset{∞}{∑}}{c}_n{e}^{inx} \\ {c}_n=\frac{1}{2\mathrm{π}}\underset{0}{\overset{2\mathrm{π}}{∫}}\left(\underset{k=-∞}{\overset{∞}{∑}}f\left(x+2k\mathrm{π}\right)\right)e^{-inx}dx \\ =\frac{1}{2\mathrm{π}}\underset{k=-∞}{\overset{∞}{∑}}\underset{0}{\overset{2\mathrm{π}}{∫}}f\left(x+2k\mathrm{π}\right)e^{-inx}dx \end{gathered}$$

As$u=x+2\mathrm{Ï€}kdu=dx$

$$\begin{gathered} c_n=\frac{1}{2\mathrm{π}}\underset{k=-∞}{\overset{∞}{∑}}\underset{2\mathrm{π}k}{\overset{2\mathrm{π}\left(k+1\right)}{∫}}f\left(u\right)e^{-in\left(u-2\mathrm{π}k\right)}du \\ =\frac{1}{2\mathrm{π}}\underset{k=-∞}{\overset{∞}{∑}}{e}^{ink2\mathrm{π}}\underset{2\mathrm{π}k}{\overset{2\mathrm{π}\left(k+1\right)}{∫}}f\left(u\right)e^{-inu}du \\ =\frac{1}{2\mathrm{π}}\underset{k=-∞}{\overset{∞}{∑}}\underset{2\mathrm{π}k}{\overset{2\mathrm{π}\left(k+1\right)}{∫}}f\left(u\right)e^{-inu}du \\ =\frac{1}{2\mathrm{π}}\underset{-∞}{\overset{∞}{∫}}f\left(u\right)e^{-inu}du=g\left(n\right) \end{gathered}$$

Where we used $e^{ink2\mathrm{Ï€}}=1$since is an integer

Simply put x=0

$$\begin{gathered} h\left(x\right)=\underset{n=-∞}{\overset{∞}{∑}}{c}_n{e}^{inx} \\ =\underset{k=-∞}{\overset{∞}{∑}}f\left(x+2\mathrm{π}k\right) \end{gathered}$$

This implies that.$\underset{-∞}{\overset{∞}{∑}}g\left(n\right)=\underset{-∞}{\overset{∞}{∑}}f\left(2\mathrm{π}k\right)$

Therefore,for the function, h(x)

Part (a)$c_n=g\left(n\right)$where,$g\left(\mathrm{Î\pm }\right)$is the Fourier transform of f(x)

Part (b)$\underset{-∞}{\overset{∞}{∑}}g\left(n\right)=\underset{-∞}{\overset{∞}{∑}}f\left(2\mathrm{π}k\right)$