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 | ||
![]() | ||
$${sin}^{{n}} {x}=\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\right)^{{n}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \left({e}^{{ix}} \right)^{{k}} \left(−{e}^{−{ix}} \right)^{{n}−{k}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{n}−{k}} {e}^{{ikx}−{i}\left({n}−{k}\right){x}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} {e}^{{i}\left(\mathrm{2}{k}−{n}\right){x}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} \left\{{cos}\left(\mathrm{2}{k}−{n}\right){x}\:+{isin}\left(\mathrm{2}{k}−{n}\right){x}\right\} \\ $$$$\Rightarrow{L}\left({sin}^{{n}} {x}\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} {L}\left({cos}\left(\mathrm{2}{k}−{n}\right){x}\right. \\ $$$$+{i}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} {L}\left({sin}\left(\mathrm{2}{k}−{n}\right){x}\right. \\ $$$${L}\left({cos}\left(\mathrm{2}{k}−{n}\right){x}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{cos}\left(\mathrm{2}{k}−{n}\right){t}\:{e}^{−{xt}} {dt} \\ $$$$={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}+{i}\left(\mathrm{2}{k}−{n}\right){t}} {dt}\right) \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{x}+{i}\left(\mathrm{2}{k}−{n}\right)\right){t}} {dt} \\ $$$$=\left[\frac{\mathrm{1}}{−{x}+{i}\left(\mathrm{2}{k}−{n}\right)}{e}^{\left(−{x}+{i}\left(\mathrm{2}{k}−{n}\right)\right){t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=−\frac{\mathrm{1}}{−{x}+{i}\left(\mathrm{2}{k}−{n}\right)} \\ $$$$=\frac{\mathrm{1}}{{x}−{i}\left(\mathrm{2}{k}−{n}\right)}=\frac{{x}+{i}\left(\mathrm{2}{k}−{n}\right)}{{x}^{\mathrm{2}} +\left(\mathrm{2}{k}−{n}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{Re}\left(....\right)=\frac{{x}}{{x}^{\mathrm{2}} +\left(\mathrm{2}{k}−{n}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${L}\left({sin}^{{n}} {x}\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} \:\frac{{x}}{{x}^{\mathrm{2}} +\left(\mathrm{2}{k}−{n}\right)^{\mathrm{2}} } \\ $$$$+{i}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{i}\right)^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} \:\frac{\mathrm{2}{k}−{n}}{{x}^{\mathrm{2}} +\left(\mathrm{2}{k}−{n}\right)^{\mathrm{2}} } \\ $$ | ||
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 | ||
![]() | ||
$${thanks} \\ $$ | ||