Menu Close

L-tsin-t-

Tinku Tara 0 Comments
Pin




Question Number 223006 by mnjuly1970 last updated on 12/Jul/25
L { tsin((√t) )}=?
$$ \\ $$$$\:\:\:\:\:\:\:\:\mathscr{L}\:\:\left\{\:{tsin}\left(\sqrt{{t}}\:\right)\right\}=? \\ $$
Answered by MrGaster last updated on 12/Jul/25
L { tsin((√t) )}=∫_0 ^∞ e^(−st) t sin((√t))dt sin((√t))=Σ_(n=0) ^∞ (((−1)^n ((√t))^(2n+1) )/((2n+1)!))=Σ_(n=0) ^∞ (((−1)^n t^(n+(1/2)) )/((2n+1)!)) t sin((√t))=t∙Σ_(n=0) ^∞ (((−1)^n t^(n+(1/2)) )/((2n+1)!))=Σ_(n=0) ^∞ (((−1)^n t^(n+(3/2)) )/((2n−1)!)) ∫_0 ^∞ e^(−st) t^(n+(3/2)) dt=Γ(n+(5/2))s^(−(n+(5/2))) ((Γ(n+(5/2)))/((2n+1)!))=(((√π)(2n+3))/(2^(2n+2) n!)) Σ_(n=0) ^∞ ((Γ(n+(5/2)))/((2n+1)!))Γ(n+(5/2))s^(−(n+(5/2))) =((√π)/(4s^(5/2) ))Σ_(n=0) ^∞ (((−1)^n (2n+3))/(n!))((1/(4s)))^n Σ_(n=0) ^∞ (((2n+3))/(n!))z^n =(3+2z)e^z ,z=−(1/(4s)) ((√π)/(4s^(5/2) ))(3−(1/(2s)))e^(−(1/(4s))) =(((√π)(6s−1))/(8s^(7/2) ))e^(−(1/(4s)))
$$\mathscr{L}\:\:\left\{\:{tsin}\left(\sqrt{{t}}\:\right)\right\}=\int_{\mathrm{0}} ^{\infty} {e}^{−{st}} {t}\:\mathrm{sin}\left(\sqrt{{t}}\right){dt} \\ $$$$\mathrm{sin}\left(\sqrt{{t}}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\sqrt{{t}}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${t}\:\mathrm{sin}\left(\sqrt{{t}}\right)={t}\centerdot\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{{n}+\frac{\mathrm{3}}{\mathrm{2}}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)!} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{st}} {t}^{{n}+\frac{\mathrm{3}}{\mathrm{2}}} {dt}=\Gamma\left({n}+\frac{\mathrm{5}}{\mathrm{2}}\right){s}^{−\left({n}+\frac{\mathrm{5}}{\mathrm{2}}\right)} \\ $$$$\frac{\Gamma\left({n}+\frac{\mathrm{5}}{\mathrm{2}}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\frac{\sqrt{\pi}\left(\mathrm{2}{n}+\mathrm{3}\right)}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} {n}!} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\frac{\mathrm{5}}{\mathrm{2}}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\Gamma\left({n}+\frac{\mathrm{5}}{\mathrm{2}}\right){s}^{−\left({n}+\frac{\mathrm{5}}{\mathrm{2}}\right)} =\frac{\sqrt{\pi}}{\mathrm{4}{s}^{\frac{\mathrm{5}}{\mathrm{2}}} }\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}+\mathrm{3}\right)}{{n}!}\left(\frac{\mathrm{1}}{\mathrm{4}{s}}\right)^{{n}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)}{{n}!}{z}^{{n}} =\left(\mathrm{3}+\mathrm{2}{z}\right){e}^{{z}} ,{z}=−\frac{\mathrm{1}}{\mathrm{4}{s}} \\ $$$$\frac{\sqrt{\pi}}{\mathrm{4}{s}^{\frac{\mathrm{5}}{\mathrm{2}}} }\left(\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}{s}}\right){e}^{−\frac{\mathrm{1}}{\mathrm{4}{s}}} =\frac{\sqrt{\pi}\left(\mathrm{6}{s}−\mathrm{1}\right)}{\mathrm{8}{s}^{\frac{\mathrm{7}}{\mathrm{2}}} }{e}^{−\frac{\mathrm{1}}{\mathrm{4}{s}}} \\ $$
Commented by mnjuly1970 last updated on 12/Jul/25
$$\: \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *