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Question Number 137873 by mnjuly1970 last updated on 07/Apr/21

.......nice ............calculus....... 𝛗=Ξ£_(n=1) ^∞ ((sin(nx))/n) =(Ο€/2)βˆ’(x/2) 𝛗=βˆ’Im(1βˆ’e^(ix) )=βˆ’Imln{(1βˆ’cos(x)βˆ’isin(x))} =βˆ’Im{ln((√((1βˆ’cos(x))^2 +sin^2 (x))) +itan^(βˆ’1) (((βˆ’sin(x))/(1βˆ’cos(x))))} =βˆ’Im{(√(2βˆ’2cos(x)_ )) βˆ’itan^(βˆ’1) (tan((Ο€/2)βˆ’(x/2)))} =tan^(βˆ’1) (tan((Ο€/2)βˆ’(x/2))) ........ 𝛗=(Ο€/2)βˆ’(x/2) ...βœ“.....βŸ¨βˆ—βŸ© ^(β€²β€²) ∫ ^(β€²β€²) both sides of βŸ¨βˆ—βŸ© βˆ’Ξ£_(n=1) ^∞ ((cos(nx))/n^2 )dx+C=((Ο€x)/2)βˆ’(x^2 /4) x=0 β‡’ C=(Ο€^2 /6) ...βœ“ Ξ£_(n=1) ^∞ ((cos(nx))/n^2 )=βˆ’((Ο€x)/2)+(x^2 /4)+(Ο€^2 /6) ....βœ“ x=Ο€(1+(√(1/3)) ) .... Ξ£((cos(nΟ€(1+(√(1/3)) )))/n^2 )=(1/2)(βˆ’Ο€^2 (1+(√(1/3)) )+((Ο€^2 (1+(√(1/3)) )^2 )/2)+(Ο€^2 /3)) =(Ο€^2 /2)(βˆ’1βˆ’((√3)/3) +((1+2.((√3)/3) +(1/3))/2)+(1/3)) =(Ο€^2 /2)(βˆ’1βˆ’(((√3) )/3)+(4/6)+(((√3) )/3)+(1/3))=0..βœ“βœ“ ......prepared by mr (Dwaipayan)..... m.n#

$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:.......{nice}\:\:............{calculus}....... \\ $$$$\:\:\:\:\boldsymbol{\phi}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi}{\mathrm{2}}βˆ’\frac{{x}}{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\phi}=βˆ’{Im}\left(\mathrm{1}βˆ’{e}^{{ix}} \right)=βˆ’{Imln}\left\{\left(\mathrm{1}βˆ’{cos}\left({x}\right)βˆ’{isin}\left({x}\right)\right)\right\} \\ $$$$\:\:\:\:=βˆ’{Im}\left\{{ln}\left(\sqrt{\left(\mathrm{1}βˆ’{cos}\left({x}\right)\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \left({x}\right)}\:+{itan}^{βˆ’\mathrm{1}} \left(\frac{βˆ’{sin}\left({x}\right)}{\mathrm{1}βˆ’{cos}\left({x}\right)}\right)\right\}\right. \\ $$$$=βˆ’{Im}\left\{\sqrt{\mathrm{2}βˆ’\mathrm{2}{cos}\left({x}\right)_{\:} }\:βˆ’{itan}^{βˆ’\mathrm{1}} \left({tan}\left(\frac{\pi}{\mathrm{2}}βˆ’\frac{{x}}{\mathrm{2}}\right)\right)\right\} \\ $$$$\:\:={tan}^{βˆ’\mathrm{1}} \left({tan}\left(\frac{\pi}{\mathrm{2}}βˆ’\frac{{x}}{\mathrm{2}}\right)\right)\:\: \\ $$$$\:\:\:\:\:\:\:........\:\:\:\boldsymbol{\phi}=\frac{\pi}{\mathrm{2}}βˆ’\frac{{x}}{\mathrm{2}}\:\:...\checkmark.....\langle\ast\rangle \\ $$$$\:\:\:\:\:^{''} \:\:\int\:\:^{''} \:\:{both}\:{sides}\:{of}\:\langle\ast\rangle \\ $$$$\:\:\:\:\:\:βˆ’\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }{dx}+{C}=\frac{\pi{x}}{\mathrm{2}}βˆ’\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:{x}=\mathrm{0}\:\Rightarrow\:{C}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\:\:...\checkmark \\ $$$$\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }=βˆ’\frac{\pi{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:....\checkmark \\ $$$$\:\:\:\:\:\:\:\:{x}=\pi\left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)\:.... \\ $$$$\:\:\:\:\:\:\:\:\Sigma\frac{{cos}\left({n}\pi\left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)\right)}{{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\left(βˆ’\pi^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)+\frac{\pi^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\left(βˆ’\mathrm{1}βˆ’\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\frac{\mathrm{1}+\mathrm{2}.\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\left(βˆ’\mathrm{1}βˆ’\frac{\sqrt{\mathrm{3}}\:}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{6}}+\frac{\sqrt{\mathrm{3}}\:}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0}..\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:......{prepared}\:{by}\:{mr}\:\left({Dwaipayan}\right)..... \\ $$$$\:\:{m}.{n}# \\ $$$$\:\:\:\: \\ $$

Commented by Dwaipayan Shikari last updated on 07/Apr/21

Have you seen that sir?

$${Have}\:{you}\:{seen}\:{that}\:{sir}? \\ $$

Commented by mnjuly1970 last updated on 07/Apr/21

yes mr payan:: 1:Q:: 137637 ..... 2 :: in brilliant math: this problem was belong to you in brilliant...⇓⇓⇓

$$\:{yes}\:{mr}\:{payan}::\:\mathrm{1}:{Q}::\:\mathrm{137637}\:..... \\ $$$$\:\mathrm{2}\:::\:{in}\:{brilliant}\:{math}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{this}\:{problem}\:{was}\:{belong}\:\:{to} \\ $$$${you}\:{in}\:{brilliant}...\Downarrow\Downarrow\Downarrow \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\: \\ $$$$\:\:\:\: \\ $$

Commented by mnjuly1970 last updated on 07/Apr/21

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