Question-184753

Question Number 184753 by cortano1 last updated on 11/Jan/23

Answered by qaz last updated on 11/Jan/23

lim_(r→∞) ((r^c ∫_0 ^(π/2) x^r sin xdx)/(∫_0 ^(π/2) x^r cos xdx))=lim_(r→∞) ((r^c ∫_0 ^(π/2) e^(rln((π/2)−x)) cos xdx)/(∫_0 ^(π/2) e^(rln((π/2)−x)) sin xdx)) =lim_(r→∞) ((r^c ∫_0 ^(π/2) e^(rln(1−(2/π)x)) cos xdx)/(∫_0 ^(π/2) e^(rln(1−(2/π)x)) sin xdx))=lim_(r→∞) ((r^c ∫_0 ^∞ e^(−(2/π)rx) cos xdx)/(∫_0 ^∞ e^(−(2/π)rx) sin xdx)).......by laplace method =lim_(r→∞) ((r^c (((2/π)r)/(((2/π)r)^2 +1)))/(1/(((2/π)r)^2 +1)))=lim_(r→∞) ((2r^(c+1) )/π)=L ⇒c=−1 , L=(2/π)

$$\underset{{r}\rightarrow\infty} {{lim}}\frac{{r}^{{c}} \int_{\mathrm{0}} ^{\pi/\mathrm{2}} {x}^{{r}} \mathrm{sin}\:{xdx}}{\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {x}^{{r}} \mathrm{cos}\:{xdx}}=\underset{{r}\rightarrow\infty} {{lim}}\frac{{r}^{{c}} \int_{\mathrm{0}} ^{\pi/\mathrm{2}} {e}^{{rln}\left(\frac{\pi}{\mathrm{2}}−{x}\right)} {cos}\:{xdx}}{\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {e}^{{rln}\left(\frac{\pi}{\mathrm{2}}−{x}\right)} {sin}\:{xdx}} \\ $$$$=\underset{{r}\rightarrow\infty} {{lim}}\frac{{r}^{{c}} \int_{\mathrm{0}} ^{\pi/\mathrm{2}} {e}^{{rln}\left(\mathrm{1}−\frac{\mathrm{2}}{\pi}{x}\right)} {cos}\:{xdx}}{\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {e}^{{rln}\left(\mathrm{1}−\frac{\mathrm{2}}{\pi}{x}\right)} {sin}\:{xdx}}=\underset{{r}\rightarrow\infty} {{lim}}\frac{{r}^{{c}} \int_{\mathrm{0}} ^{\infty} {e}^{−\frac{\mathrm{2}}{\pi}{rx}} {cos}\:{xdx}}{\int_{\mathrm{0}} ^{\infty} {e}^{−\frac{\mathrm{2}}{\pi}{rx}} \mathrm{sin}\:{xdx}}…….{by}\:{laplace}\:{method} \\ $$$$=\underset{{r}\rightarrow\infty} {{lim}}\frac{{r}^{{c}} \frac{\frac{\mathrm{2}}{\pi}{r}}{\left(\frac{\mathrm{2}}{\pi}{r}\right)^{\mathrm{2}} +\mathrm{1}}}{\frac{\mathrm{1}}{\left(\frac{\mathrm{2}}{\pi}{r}\right)^{\mathrm{2}} +\mathrm{1}}}=\underset{{r}\rightarrow\infty} {{lim}}\frac{\mathrm{2}{r}^{{c}+\mathrm{1}} }{\pi}={L}\:\:\:\:\:\Rightarrow{c}=−\mathrm{1}\:\:,\:{L}=\frac{\mathrm{2}}{\pi} \\ $$

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