Question Number 202017 by sonukgindia last updated on 18/Dec/23
Answered by Mathspace last updated on 18/Dec/23
![f(x)=Σ_(n=1) ^∞ ((sin(nx))/2^n ) ⇒ ∫_0 ^π f(x)dx=∫_0 ^π Σ_(n=1) ^∞ ((sin(nx))/2^n )dx =Σ_(n=1) ^∞ (1/2^n )∫_0 ^π sin(nx)dx =Σ_(n=1) ^∞ (1/2^n )[−(1/n)cos(nx)]_0 ^π =−Σ_(n=1) ^∞ (1/(n2^n )){(−1)^n −1} =2Σ_(p=0) ^∞ (1/((2p+1)2^(2p+1) )) =Σ_(n=0) ^∞ (1/(2n+1))((1/2))^(2n+1) s(x)=Σ_(n=0) ^∞ (x^(2n+1) /(2n+1)) (∣x∣<1) s^′ (x)=Σ_(n=0) ^∞ x^(2n) =(1/(1−x^2 )) ⇒ s(x)=∫(dx/(1−x^2 )) +λ =(1/2)∫((1/(1−x))+(1/(1+x)))dx +λ =(1/2)ln(((1+x)/(1−x))) +λ s(0)=0=λ ⇒ s(x)=(1/2)ln(((1+x)/(1−x))) and Σ_(n=0) ^∞ (1/(2n+1))((1/2))^(2n+1) =s((1/2))=(1/2)ln(((3/2)/(1/2)))=(1/2)ln3 ⇒ ⇒∫_0 ^π f(θ)dθ =((ln3)/2)](https://www.tinkutara.com/question/Q202026.png)
$${f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{sin}\left({nx}\right)}{\mathrm{2}^{{n}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\pi} \sum_{{n}=\mathrm{1}} ^{\infty} \frac{{sin}\left({nx}\right)}{\mathrm{2}^{{n}} }{dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}^{{n}} }\int_{\mathrm{0}} ^{\pi} {sin}\left({nx}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left[−\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}\mathrm{2}^{{n}} }\left\{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right\} \\ $$$$=\mathrm{2}\sum_{{p}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)\mathrm{2}^{\mathrm{2}{p}+\mathrm{1}} } \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{1}} \\ $$$${s}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:\:\:\:\left(\mid{x}\mid<\mathrm{1}\right) \\ $$$${s}^{'} \left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${s}\left({x}\right)=\int\frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} }\:+\lambda \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\mathrm{1}+{x}}\right){dx}\:+\lambda \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:+\lambda \\ $$$${s}\left(\mathrm{0}\right)=\mathrm{0}=\lambda\:\Rightarrow \\ $$$${s}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:{and} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$={s}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}\:\Rightarrow \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} {f}\left(\theta\right){d}\theta\:=\frac{{ln}\mathrm{3}}{\mathrm{2}} \\ $$
Commented by Mathspace last updated on 18/Dec/23
$${sorry}\:\:\:\:\int_{\mathrm{0}} ^{\pi} {f}\left(\theta\right){d}\theta=\mathrm{2}{s}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\bigstar\int_{\mathrm{0}} ^{\pi} {f}\left(\theta\right){d}\theta={ln}\mathrm{3}\bigstar \\ $$
Answered by mr W last updated on 19/Dec/23
![METHOD I ∫_0 ^π sin kθ dθ =(1/k)∫_0 ^π sin kθ d(kθ) =(1/k)[−cos kθ]_0 ^π =((1−cos kπ)/k) = { ((0, if k=2nπ)),(((2/k), if k=2n+1)) :} ∫_0 ^π f(θ)dθ =Σ_(n=0) ^∞ [∫_0 ^π ((sin 2(n+1)θ)/2^(2(n+1)) )dθ+∫_0 ^π ((sin (2n+1)θ)/2^(2n+1) )dθ] =Σ_(n=0) ^∞ [∫_0 ^π ((sin (2n+1)θ)/2^(2n+1) )dθ] =Σ_(n=0) ^∞ [(2/((2n+1)2^(2n+1) ))] =Σ_(n=0) ^∞ (1/((2n+1)2^(2n) ))=S, say Σ_(n=0) ^∞ x^(2n) =(1/(1−x^2 )) Σ_(n=0) ^∞ ∫_0 ^x x^(2n) dx=∫_0 ^x (dx/(1−x^2 ))=(1/2)∫_0 ^x ((1/(1+x))+(1/(1−x)))dx=(1/2) ln ∣((1+x)/(1−x))∣ Σ_(n=0) ^∞ (x^(2n+1) /(2n+1))=(1/2) ln ∣((1+x)/(1−x))∣ ⇒Σ_(n=0) ^∞ (x^(2n) /(2n+1))=(1/(2x)) ln ∣((1+x)/(1−x))∣ with x=(1/2) ⇒Σ_(n=0) ^∞ (1/((2n+1)2^(2n) ))=ln 3=S ⇒∫_0 ^π f(θ)dθ=ln 3](https://www.tinkutara.com/question/Q202036.png)
$$\boldsymbol{{METHOD}}\:\boldsymbol{{I}} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{sin}\:{k}\theta\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{{k}}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}\:{k}\theta\:{d}\left({k}\theta\right) \\ $$$$=\frac{\mathrm{1}}{{k}}\left[−\mathrm{cos}\:{k}\theta\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\:{k}\pi}{{k}} \\ $$$$=\begin{cases}{\mathrm{0},\:{if}\:{k}=\mathrm{2}{n}\pi}\\{\frac{\mathrm{2}}{{k}},\:{if}\:{k}=\mathrm{2}{n}+\mathrm{1}}\end{cases} \\ $$$$\int_{\mathrm{0}} ^{\pi} {f}\left(\theta\right){d}\theta \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\:\mathrm{2}\left({n}+\mathrm{1}\right)\theta}{\mathrm{2}^{\mathrm{2}\left({n}+\mathrm{1}\right)} }{d}\theta+\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\:\left(\mathrm{2}{n}+\mathrm{1}\right)\theta}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }{d}\theta\right] \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\:\left(\mathrm{2}{n}+\mathrm{1}\right)\theta}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }{d}\theta\right] \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\right] \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{2}{n}} }={S},\:{say} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{{x}} {x}^{\mathrm{2}{n}} {dx}=\int_{\mathrm{0}} ^{{x}} \frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ln}\:\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ln}\:\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid \\ $$$$\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}{x}}\:\mathrm{ln}\:\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid \\ $$$${with}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{2}{n}} }=\mathrm{ln}\:\mathrm{3}={S} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} {f}\left(\theta\right){d}\theta=\mathrm{ln}\:\mathrm{3} \\ $$
Answered by mr W last updated on 19/Dec/23
![METHOD II A=((cos θ)/2)+((cos 2θ)/2^2 )+((cos 3θ)/2^3 )+... B=((sin θ)/2)+((sin 2θ)/2^2 )+((sin 3θ)/2^3 )+...=f(θ) A+Bi=Σ_(n=1) ^∞ ((cos nθ+i sin nθ)/2^n ) =Σ_(n=1) ^∞ [(e^(θi) /2)]^n =(e^(iθ) /(2×(1−(e^(iθ) /2))))=((cos θ+i sin θ)/(2−cos θ−i sin θ)) =(((cos θ+i sin θ)(2−cos θ+i sin θ))/((2−cos θ)^2 +sin^2 θ)) =((cos θ−1+i sin θ)/(2((5/4)−cos θ))) ⇒A=((cos θ−1)/(2((5/4)−cos θ))), B=((sin θ)/(2((5/4)−cos θ)))=f(θ) ∫_0 ^π f(θ)dθ=∫_0 ^π ((sin θ)/(2((5/4)−cos θ)))dθ =(1/2)∫_0 ^π ((d(cos θ))/((cos θ−(5/4)))) =(1/2)[ln ((5/4)−cos θ)]_0 ^π =(1/2)ln 9=ln 3](https://www.tinkutara.com/question/Q202038.png)
$$\boldsymbol{{METHOD}}\:\boldsymbol{{II}} \\ $$$${A}=\frac{\mathrm{cos}\:\theta}{\mathrm{2}}+\frac{\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{cos}\:\mathrm{3}\theta}{\mathrm{2}^{\mathrm{3}} }+… \\ $$$${B}=\frac{\mathrm{sin}\:\theta}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{sin}\:\mathrm{3}\theta}{\mathrm{2}^{\mathrm{3}} }+…={f}\left(\theta\right) \\ $$$${A}+{Bi}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{cos}\:{n}\theta+{i}\:\mathrm{sin}\:{n}\theta}{\mathrm{2}^{{n}} } \\ $$$$\:\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{{e}^{\theta{i}} }{\mathrm{2}}\right]^{{n}} =\frac{{e}^{{i}\theta} }{\mathrm{2}×\left(\mathrm{1}−\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)}=\frac{\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta}{\mathrm{2}−\mathrm{cos}\:\theta−{i}\:\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:=\frac{\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)\left(\mathrm{2}−\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)}{\left(\mathrm{2}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\:\:\:\:=\frac{\mathrm{cos}\:\theta−\mathrm{1}+{i}\:\mathrm{sin}\:\theta}{\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{cos}\:\theta\right)} \\ $$$$\Rightarrow{A}=\frac{\mathrm{cos}\:\theta−\mathrm{1}}{\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{cos}\:\theta\right)},\:{B}=\frac{\mathrm{sin}\:\theta}{\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{cos}\:\theta\right)}={f}\left(\theta\right) \\ $$$$\int_{\mathrm{0}} ^{\pi} {f}\left(\theta\right){d}\theta=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\:\theta}{\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{cos}\:\theta\right)}{d}\theta \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{d}\left(\mathrm{cos}\:\theta\right)}{\left(\mathrm{cos}\:\theta−\frac{\mathrm{5}}{\mathrm{4}}\right)} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\left(\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{cos}\:\theta\right)\right]_{\mathrm{0}} ^{\pi} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{9}=\mathrm{ln}\:\mathrm{3} \\ $$
Commented by Mathspace last updated on 19/Dec/23
$${yes}\:{sir}\:{thanks} \\ $$