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Question Number 117806 by mnjuly1970 last updated on 13/Oct/20

                ... nice  calculus...      i ::   1 +(4/9)+(9/(36))+((16)/(100))+...= ??       ii::  ∫_0 ^( (π/2)) x^2 cot(x) dx=??                m.n.1970

...nicecalculus...i::1+49+936+16100+...=??ii::0π2x2cot(x)dx=??m.n.1970

Commented by TANMAY PANACEA last updated on 13/Oct/20

ii)it is not solution but reflection of idea  ∫_0 ^(π/2) x^2 cotx dx  f(x)=x^2 cotx  f(0)=0  f((π/2))=0  which means x^2 cotx make a loop in interval  x=0 and x=(π/2)  let (x^2 cotx)_(max)  value=h in that interval  ∫_0 ^(π/2) h dx≥∫_0 ^(π/2) x^2 cotx dx≥0  h×(π/2)≥I≥0

ii)itisnotsolutionbutreflectionofidea0π2x2cotxdxf(x)=x2cotxf(0)=0f(π2)=0whichmeansx2cotxmakealoopinintervalx=0andx=π2let(x2cotx)maxvalue=hinthatinterval0π2hdx0π2x2cotxdx0h×π2I0

Commented by Lordose last updated on 13/Oct/20

having trilogarithm Li_3 (z).

havingtrilogarithmLi3(z).

Answered by Dwaipayan Shikari last updated on 13/Oct/20

1+((2/3))^2 +((3/6))^2 +((4/(10)))^2 +....  T_n =((4n^2 )/((n(n+1))^2 ))  Σ_(n=1) ^∞ ((4n^2 )/(n^2 (n+1)^2 ))=4Σ_(n=1) ^∞ (1/((n+1)^2 ))=4((π^2 /6)−1)=((2π^2 )/3)−4

1+(23)2+(36)2+(410)2+....Tn=4n2(n(n+1))2n=14n2n2(n+1)2=4n=11(n+1)2=4(π261)=2π234

Commented by mnjuly1970 last updated on 13/Oct/20

mercey mr dwaipayan..

merceymrdwaipayan..

Answered by mnjuly1970 last updated on 13/Oct/20

Answered by mindispower last updated on 14/Oct/20

∫x^2 cot(x) dx =[x^2 ln(sin(x))dx]−∫2xlnsin(x)dx  =2∫x(−ln(sin(x))dx  byfourier −ln(sin(x))=ln(2)+Σ_(n≥1) ((cos(2nx))/n)  we get 2∫_0 ^(π/2) (xln(2)+xΣ_(n≥1) ((cos(2nx))/n))dx  =(π^2 /4)ln(2)+2Σ_(n≥1) (1/n)∫_0 ^(π/2) xcos(2nx)dx  ∫xcos(2nx)dx=((xsin(2nx))/(2n))+((cos(2nx))/(4n^2 ))+c   we get  (π^2 /4)ln(2)+2Σ_(n≥1) (1/n)((((−1)^n −1)/(4n^2 )))  =((π^2 ln(2))/4)−Σ_(n≥0) (1/((2n+1)^3 ))=((π^2 ln(2))/4)−(7/8)ζ(3)

x2cot(x)dx=[x2ln(sin(x))dx]2xlnsin(x)dx=2x(ln(sin(x))dxbyfourierln(sin(x))=ln(2)+n1cos(2nx)nweget20π2(xln(2)+xn1cos(2nx)n)dx=π24ln(2)+2n11n0π2xcos(2nx)dxxcos(2nx)dx=xsin(2nx)2n+cos(2nx)4n2+cwegetπ24ln(2)+2n11n((1)n14n2)=π2ln(2)4n01(2n+1)3=π2ln(2)478ζ(3)

Commented by mnjuly1970 last updated on 14/Oct/20

very nice .thank you mr  power.

verynice.thankyoumrpower.

Commented by mindispower last updated on 14/Oct/20

withe pleasur

withepleasur

Answered by mnjuly1970 last updated on 14/Oct/20

solution  (ii)    note::  we know that ::   ∫_0 ^( (π/2)) xln(sin(x))dx[=_(ln(sin(x))=−ln(2)−Σ_(n=1) ^∞ ((cos(2nx))/n)) ^(fourier series) ] −ln(2)(π^2 /8)+(7/(16))ζ(3) ✓  therfore::   Ω=∫_0 ^(π/2) x^2 cot(x)dx=^(i.b.p) {x^2 lnsin(x)}_0 ^(π/2) −2∫_0 ^( (π/2)) xln(sin(x))dx      =^(note) −2[−ln(2)(π^2 /8)+(7/(16))ζ(3)]       =ln(2)(π^2 /4) −(7/8)ζ(3) ✓             ...  m.n.1970...

solution(ii)note::weknowthat::0π2xln(sin(x))dx[=fourierseriesln(sin(x))=ln(2)n=1cos(2nx)n]ln(2)π28+716ζ(3)therfore::Ω=0π2x2cot(x)dx=i.b.p{x2lnsin(x)}0π220π2xln(sin(x))dx=note2[ln(2)π28+716ζ(3)]=ln(2)π2478ζ(3)...m.n.1970...

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