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find-n-1-1-n-2-by-use-of-integral-0-pi-2-ln-2cos-d-




Question Number 60976 by maxmathsup by imad last updated on 28/May/19
find Σ_(n=1) ^∞  (1/n^2 ) by use of integral ∫_0 ^(π/2) ln(2cosθ)dθ  .
$${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:{by}\:{use}\:{of}\:{integral}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{2}{cos}\theta\right){d}\theta\:\:. \\ $$
Commented by maxmathsup by imad last updated on 31/May/19
we have ∫_0 ^(π/2) ln(2cosθ)dθ =∫_0 ^(π/2) ln(e^(iθ)  +e^(−iθ) )dθ  =∫_0 ^(π/2) ln{e^(iθ) (1+e^(−2iθ) )}dθ =∫_0 ^(π/2) iθ dθ +∫_0 ^(π/2)  ln(1+e^(−2iθ) )dθ  =i[(θ^2 /2)]_0 ^(π/2)  +∫_0 ^(π/2) ln(1+e^(−2iθ) )dθ =i(π^2 /8) +∫_0 ^(π/2) ln(1+e^(−2iθ) )dθ  we have for ∣z∣≤1 ln^′ (1+z) =(1/(1+z)) =Σ_(n=0) ^∞ (−1)^n z^n  ⇒  ln(1+z) =Σ_(n=0) ^∞  (((−1)^n  z^(n+1) )/(n+1)) +c        (c=0)  =Σ_(n=1) ^∞  (((−1)^(n−1)  z^n )/n)  ⇒∫_0 ^(π/2) ln(1+e^(−2iθ) ) =∫_0 ^(π/2) (Σ_(n=1) ^∞  (((−1)^(n−1) e^(−2inθ) )/n))dθ  =Σ_(n=1) ^∞  (((−1)^(n−1) )/n) ∫_0 ^(π/2)  e^(−2inθ)  dθ  =Σ_(n=1) ^∞   (((−1)^(n−1) )/n) [−(1/(2in)) e^(−2inθ) ]_0 ^(π/2)   =−(1/(2i)) Σ_(n=1) ^∞   (((−1)^(n−1) )/n^2 ){  (−1)^n −1}  =(1/i) Σ_(n=0) ^∞   (1/((2n+1)^2 )) =−i Σ_(n=0) ^∞   (1/((2n+1)^2 )) ⇒  ∫_0 ^(π/2) ln(2cosθ)dθ =i((π^2 /8) − Σ_(n=0) ^∞  (1/((2n+1)^2 )))  but this integral is  real ⇒(π^2 /8) −Σ_(n=0) ^∞  (1/((2n+1)^2 )) =0 ⇒Σ_(n=0) ^∞   (1/((2n+1)^2 )) =(π^2 /8)  we have Σ_(n=1) ^∞  (1/n^2 ) =(1/4) Σ_(n=1) ^∞  (1/n^2 ) +Σ_(n=0) ^∞  (1/((2n+1)^2 )) ⇒  (3/4) Σ_(n=1) ^∞  (1/n^2 ) =(π^2 /8) ⇒ Σ_(n=1) ^∞  (1/n^2 ) =(4/3)(π^2 /8) =(π^2 /6) .
$${we}\:{have}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{2}{cos}\theta\right){d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({e}^{{i}\theta} \:+{e}^{−{i}\theta} \right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left\{{e}^{{i}\theta} \left(\mathrm{1}+{e}^{−\mathrm{2}{i}\theta} \right)\right\}{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {i}\theta\:{d}\theta\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}\theta} \right){d}\theta \\ $$$$={i}\left[\frac{\theta^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}\theta} \right){d}\theta\:={i}\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}\theta} \right){d}\theta \\ $$$${we}\:{have}\:{for}\:\mid{z}\mid\leqslant\mathrm{1}\:{ln}^{'} \left(\mathrm{1}+{z}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{z}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {z}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{z}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{z}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:+{c}\:\:\:\:\:\:\:\:\left({c}=\mathrm{0}\right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{z}^{{n}} }{{n}}\:\:\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}\theta} \right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {e}^{−\mathrm{2}{in}\theta} }{{n}}\right){d}\theta \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{e}^{−\mathrm{2}{in}\theta} \:{d}\theta \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\left[−\frac{\mathrm{1}}{\mathrm{2}{in}}\:{e}^{−\mathrm{2}{in}\theta} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{i}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\left\{\:\:\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{1}}{{i}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=−{i}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{2}{cos}\theta\right){d}\theta\:={i}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:−\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:\:{but}\:{this}\:{integral}\:{is} \\ $$$${real}\:\Rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\mathrm{0}\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{4}}{\mathrm{3}}\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$

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