Question Number 922 by 123456 last updated on 25/Apr/15
$$\mathrm{find}\:\mathrm{the}\:\mathrm{fourier}\:\mathrm{serie}\:\mathrm{of} \\ $$$${f}\left({t}\right)=\mathrm{sinh}\left({t}\right) \\ $$$$\mathrm{into}\:\mathrm{the}\:\mathrm{interval}\:\left(−\mathrm{1},+\mathrm{1}\right) \\ $$
Answered by prakash jain last updated on 25/Apr/15
![F(w)=Σ_(k=−∞) ^∞ a_k e^(jkw_0 t) w_0 =((2π)/T)=π a_k =(1/2)∫_(−1) ^( +1) sinh (t)e^(−kjπt) dt, j=(√(−1)) sinh (t)=((e^t −e^(−t) )/2) a_k =(1/4)[∫_(−1) ^( 1) e^((1−jπ)kt) dt−∫_(−1) ^1 e^((1+jπ)kt) dt] =(1/4)[(e^((1−jπ)k) /((1−jπ)k))−(e^(−(1−jπ)k) /((1−jπ)k))−(e^((1+jπ)k) /((1+jπ)k))+(e^(−(1+jπ)k) /((1+jπ)k))] Further simplification can be done.](Q925.png)
$${F}\left({w}\right)=\underset{{k}=−\infty} {\overset{\infty} {\sum}}{a}_{{k}} {e}^{{jkw}_{\mathrm{0}} {t}} \\ $$$${w}_{\mathrm{0}} =\frac{\mathrm{2}\pi}{{T}}=\pi \\ $$$${a}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}} ^{\:+\mathrm{1}} \mathrm{sinh}\:\left(\mathrm{t}\right){e}^{−{kj}\pi{t}} {dt},\:{j}=\sqrt{−\mathrm{1}} \\ $$$$\mathrm{sinh}\:\left({t}\right)=\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}} \\ $$$${a}_{{k}} =\frac{\mathrm{1}}{\mathrm{4}}\left[\int_{−\mathrm{1}} ^{\:\mathrm{1}} {e}^{\left(\mathrm{1}−{j}\pi\right){kt}} {dt}−\int_{−\mathrm{1}} ^{\mathrm{1}} {e}^{\left(\mathrm{1}+{j}\pi\right){kt}} {dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{e}^{\left(\mathrm{1}−{j}\pi\right){k}} }{\left(\mathrm{1}−{j}\pi\right){k}}−\frac{{e}^{−\left(\mathrm{1}−{j}\pi\right){k}} }{\left(\mathrm{1}−{j}\pi\right){k}}−\frac{{e}^{\left(\mathrm{1}+{j}\pi\right){k}} }{\left(\mathrm{1}+{j}\pi\right){k}}+\frac{{e}^{−\left(\mathrm{1}+{j}\pi\right){k}} }{\left(\mathrm{1}+{j}\pi\right){k}}\right] \\ $$$$\mathrm{Further}\:\mathrm{simplification}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}. \\ $$