Question Number 174508 by Gbenga last updated on 03/Aug/22
$$ \\ $$$$\mathscr{L}\left(\boldsymbol{\mathrm{sin}}^{\boldsymbol{\mathrm{n}}} \left(\boldsymbol{\mathrm{x}}\right)\right)=? \\ $$
Answered by Mathspace last updated on 03/Aug/22
![sin^n x=(((e^(ix) −e^(−ix) )/(2i)))^n =(1/((2i)^n ))Σ_(k=0) ^n C_n ^k (e^(ix) )^k (−e^(−ix) )^(n−k) =(1/((2i)^n ))Σ_(k=0) ^n C_n ^k (−1)^(n−k) e^(ikx−i(n−k)x) =(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k e^(i(2k−n)x) =(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k {cos(2k−n)x +isin(2k−n)x} ⇒L(sin^n x) =(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k L(cos(2k−n)x +i(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k L(sin(2k−n)x L(cos(2k−n)x) =∫_0 ^∞ cos(2k−n)t e^(−xt) dt =Re(∫_0 ^∞ e^(−xt+i(2k−n)t) dt) and ∫_0 ^∞ e^((−x+i(2k−n))t) dt =[(1/(−x+i(2k−n)))e^((−x+i(2k−n))t) ]_0 ^∞ =−(1/(−x+i(2k−n))) =(1/(x−i(2k−n)))=((x+i(2k−n))/(x^2 +(2k−n)^2 )) ⇒Re(....)=(x/(x^2 +(2k−n)^2 )) ⇒ L(sin^n x) =(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k (x/(x^2 +(2k−n)^2 )) +i(((−1)^n )/((2i)^n ))Σ_(k=0) ^n (−1)^k C_n ^k ((2k−n)/(x^2 +(2k−n)^2 ))](Q174510.png)
$${sin}^{{n}} {x}=\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\right)^{{n}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \left({e}^{{ix}} \right)^{{k}} \left(−{e}^{−{ix}} \right)^{{n}−{k}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{n}−{k}} {e}^{{ikx}−{i}\left({n}−{k}\right){x}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} {e}^{{i}\left(\mathrm{2}{k}−{n}\right){x}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} \left\{{cos}\left(\mathrm{2}{k}−{n}\right){x}\:+{isin}\left(\mathrm{2}{k}−{n}\right){x}\right\} \\ $$$$\Rightarrow{L}\left({sin}^{{n}} {x}\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} {L}\left({cos}\left(\mathrm{2}{k}−{n}\right){x}\right. \\ $$$$+{i}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} {L}\left({sin}\left(\mathrm{2}{k}−{n}\right){x}\right. \\ $$$${L}\left({cos}\left(\mathrm{2}{k}−{n}\right){x}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{cos}\left(\mathrm{2}{k}−{n}\right){t}\:{e}^{−{xt}} {dt} \\ $$$$={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}+{i}\left(\mathrm{2}{k}−{n}\right){t}} {dt}\right) \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{x}+{i}\left(\mathrm{2}{k}−{n}\right)\right){t}} {dt} \\ $$$$=\left[\frac{\mathrm{1}}{−{x}+{i}\left(\mathrm{2}{k}−{n}\right)}{e}^{\left(−{x}+{i}\left(\mathrm{2}{k}−{n}\right)\right){t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=−\frac{\mathrm{1}}{−{x}+{i}\left(\mathrm{2}{k}−{n}\right)} \\ $$$$=\frac{\mathrm{1}}{{x}−{i}\left(\mathrm{2}{k}−{n}\right)}=\frac{{x}+{i}\left(\mathrm{2}{k}−{n}\right)}{{x}^{\mathrm{2}} +\left(\mathrm{2}{k}−{n}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{Re}\left(....\right)=\frac{{x}}{{x}^{\mathrm{2}} +\left(\mathrm{2}{k}−{n}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${L}\left({sin}^{{n}} {x}\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} \:\frac{{x}}{{x}^{\mathrm{2}} +\left(\mathrm{2}{k}−{n}\right)^{\mathrm{2}} } \\ $$$$+{i}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} \:\frac{\mathrm{2}{k}−{n}}{{x}^{\mathrm{2}} +\left(\mathrm{2}{k}−{n}\right)^{\mathrm{2}} } \\ $$
Commented by Gbenga last updated on 03/Aug/22
$${thanks}\:{sk} \\ $$
Commented by Tawa11 last updated on 03/Aug/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Mathspace last updated on 03/Aug/22
$${thanks} \\ $$