Question-45885

Question Number 45885 by Meritguide1234 last updated on 17/Oct/18

Commented by maxmathsup by imad last updated on 18/Oct/18

let A_n =∫_0 ^(π/2) n(1−(sinx)^(1/n) )dx =∫_0 ^(π/2) f_n (x)dx with f_n (x)=n(1−(sinx^(1/n) )) but we have (sinx)^(1/n) =e^((1/n)ln(sinx)) =1+((ln(sinx))/n) +o((1/n)) ⇒ 1−e^((1/n)ln(sinx)) =−((ln(sinx))/n) +o((1/n)) ⇒n(1−e^((ln(sinx))/n) )=−ln(sinx) +o(1) ⇒ f_n (x)=−ln(sinx)+o(1) ⇒lim_(n→+∞) A_n = ∫_0 ^(π/2) −ln(sinx)dx =−(−(π/2)ln(2))=(π/2)ln(2).

$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{n}\left(\mathrm{1}−\left({sinx}\right)^{\frac{\mathrm{1}}{{n}}} \right){dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{f}_{{n}} \left({x}\right){dx}\:{with}\:{f}_{{n}} \left({x}\right)={n}\left(\mathrm{1}−\left({sinx}^{\frac{\mathrm{1}}{{n}}} \right)\right) \\ $$$${but}\:{we}\:{have}\:\:\left({sinx}\right)^{\frac{\mathrm{1}}{{n}}} \:={e}^{\frac{\mathrm{1}}{{n}}{ln}\left({sinx}\right)} =\mathrm{1}+\frac{{ln}\left({sinx}\right)}{{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow \\ $$$$\mathrm{1}−{e}^{\frac{\mathrm{1}}{{n}}{ln}\left({sinx}\right)} =−\frac{{ln}\left({sinx}\right)}{{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{n}\left(\mathrm{1}−{e}^{\frac{{ln}\left({sinx}\right)}{{n}}} \right)=−{ln}\left({sinx}\right)\:+{o}\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${f}_{{n}} \left({x}\right)=−{ln}\left({sinx}\right)+{o}\left(\mathrm{1}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −{ln}\left({sinx}\right){dx} \\ $$$$=−\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right)=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right). \\ $$

Commented by Meritguide1234 last updated on 19/Oct/18

$${nice}\:{approach} \\ $$

Commented by maxmathsup by imad last updated on 20/Oct/18

$${thank}\:{you}\:{sir}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18

lim_(n→∞) n∫_0 ^(π/2) {1−(sinx)^(1/n) }dx lim_(n→∞) n[∣x∣_0 ^(π/2) −∫_0 ^(π/2) sin^(2p−1) xcoz^(2q−1) x dx] here 2p−1=(1/n) 2p=1+(1/n) p=((n+1)/(2n)) 2q−1=0 q=(1/2) using gamma beta function formula ∫_0 ^(π/2) sin^(2p−1) xcos^(2a−1) xdx=((⌈(p)⌈(q))/(2⌈(p+q))) lim_(n→∞) n[(π/2)−((⌈(p)⌈(q))/(2⌈(p+q)))] lim_(n→∞) n[(π/2)−((⌈(((n+1)/(2n)))⌈((1/2)))/(2⌈(((n+1)/(2n))+(1/2))))] lim_(n→∞) n[(π/2)−((⌈(1+((1−n)/(2n)))×(√π))/(2⌈(1+((n+1)/(2n))−(1/2)))) ] lim_(n→∞) n[(π/2)−(((√π) ×⌈(1+((1−n)/(2n))))/(2⌈(1+(1/(2n)))))]

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{\mathrm{1}−\left({sinx}\right)^{\frac{\mathrm{1}}{{n}}} \right\}{dx} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\left[\mid{x}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} {xcoz}^{\mathrm{2}{q}−\mathrm{1}} {x}\:{dx}\right] \\ $$$${here}\:\mathrm{2}{p}−\mathrm{1}=\frac{\mathrm{1}}{{n}}\:\:\:\:\:\mathrm{2}{p}=\mathrm{1}+\frac{\mathrm{1}}{{n}}\:\:\:\:{p}=\frac{{n}+\mathrm{1}}{\mathrm{2}{n}} \\ $$$$\:\:\mathrm{2}{q}−\mathrm{1}=\mathrm{0}\:\:\:\:{q}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:{using}\:{gamma}\:{beta}\:{function} \\ $$$${formula}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} {xcos}^{\mathrm{2}{a}−\mathrm{1}} {xdx}=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\mathrm{2}\lceil\left({p}+{q}\right)} \\ $$$$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\left[\frac{\pi}{\mathrm{2}}−\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\mathrm{2}\lceil\left({p}+{q}\right)}\right] \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\left[\frac{\pi}{\mathrm{2}}−\frac{\lceil\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\lceil\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\right] \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\left[\frac{\pi}{\mathrm{2}}−\frac{\lceil\left(\mathrm{1}+\frac{\mathrm{1}−{n}}{\mathrm{2}{n}}\right)×\sqrt{\pi}}{\mathrm{2}\lceil\left(\mathrm{1}+\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:\:\:\right] \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\left[\frac{\pi}{\mathrm{2}}−\frac{\sqrt{\pi}\:×\lceil\left(\mathrm{1}+\frac{\mathrm{1}−{n}}{\mathrm{2}{n}}\right)}{\mathrm{2}\lceil\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)}\right] \\ $$

Commented by Meritguide1234 last updated on 18/Oct/18