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Question Number 114777 by mathdave last updated on 21/Sep/20

	Answered by maths mind last updated on 21/Sep/20
	$(√x)=t⇒dx=2tdt =∫_0 ^(π/2) 2tcl_2 (t)dt =∫_0 ^(π/2) 2tΣ_(k≥1) ((sin(kt))/k^2 )dt =2Σ_(k≥1) ∫_0 ^(π/2) ((sin(kt))/k^2 )tdt ∫_0 ^(π/2) ((sin(kt))/k^2 )tdt=(1/k^2 ){[((−cos(kt))/k)t]_0 ^(π/2) +∫_0 ^(π/2) ((cos(kt))/k)dt} =(π/2)[((−cos(((kπ)/2)))/k^3 )]+(1/k^4 )(sin(k(π/2))) sin(((kπ)/2))=0 if k=2m sin(((kπ)/2))=(−1)^m if k=2m+1 cos(((kπ)/2)) =1 si k=4m,cos(((kπ)/2))=−1 ifk=4m+2 cos(((kπ)/2))=0 if k=4m+1 or 4m+3 so we get Σ_(k≥1) (π/2).[((−cos(((kπ)/2)))/k^3 )]=Σ_(m≥0) (π/2).[((−cos((4m+2)(π/2)))/((4m+2)^3 ))]Σ_(m≥1) −(π/2)((cos(4m.(π/2)))/((4m)^3 )) +Σ_(m≥0) ((sin((π/2)(2m+1)))/((2m+1)^4 )) =Σ_(m≥0) (π/2).(1/(8(2m+1)^3 ))+−(π/2).Σ_(m≥1) (1/(64m^3 ))+Σ_(m≥0) (((−1)^m )/((2m+1)^4 )) =(π/(16))Σ_(m≥0) (1/((2m+1)^3 ))+Σ_(m≥0) (((−1)^m )/((2m+1)^4 )) β(s)=Σ_(n≥0) (((−1)^n )/((2n+1)^s )) Dirichlet Betta function Σ(((−1)^m )/((2m+1)^4 ))=β(4),Σ(1/((2m+1)^3 ))=(ζ(3)−(1/8)ζ(3))=(7/8)ζ(3) we get =(π/(16)).(7/( 8))ζ(3)−(π/(128))ζ(3)+β(4)=((6π)/(128))ζ(3)+β(4)= ∫_0 ^(π/2) cl_2 ((√t))dt=2∫_0 ^1 tcl_2 (t)dt=2.[((6π)/(128))+β(4)] =((12π)/(128))ζ(3)+2β(4)=((3π)/(32))ζ(3)+2β(4)$
	$\sqrt{x} = t \Rightarrow d x = 2 t d t$ $= \int_{0}^{\frac{π}{2}} 2 {t c l}_{2} (t) d t$ $= \int_{0}^{\frac{π}{2}} 2 t \sum_{k ⩾ 1} \frac{s i n (k t)}{k^{2}} d t$ $= 2 \sum_{k ⩾ 1} \int_{0}^{\frac{π}{2}} \frac{s i n (k t)}{k^{2}} t d t$ $\int_{0}^{\frac{π}{2}} \frac{s i n (k t)}{k^{2}} t d t = \frac{1}{k^{2}} {{[\frac{- c o s (k t)}{k} t]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} \frac{c o s (k t)}{k} d t}$ $= \frac{π}{2} [\frac{- c o s (\frac{k π}{2})}{k^{3}}] + \frac{1}{k^{4}} (s i n (k \frac{π}{2}))$ $s i n (\frac{k π}{2}) = 0 i f k = 2 m$ $s i n (\frac{k π}{2}) = {(- 1)}^{m} i f k = 2 m + 1$ $c o s (\frac{k π}{2}) = 1 s i k = 4 m, c o s (\frac{k π}{2}) = - 1 i f k = 4 m + 2$ $c o s (\frac{k π}{2}) = 0 i f k = 4 m + 1 o r 4 m + 3$ $s o w e g e t \sum_{k ⩾ 1} \frac{π}{2} . [\frac{- c o s (\frac{k π}{2})}{k^{3}}] = \sum_{m ⩾ 0} \frac{π}{2} . [\frac{- c o s ((4 m + 2) \frac{π}{2})}{{(4 m + 2)}^{3}}] \sum_{m ⩾ 1} - \frac{π}{2} \frac{c o s (4 m . \frac{π}{2})}{{(4 m)}^{3}}$ $+ \sum_{m ⩾ 0} \frac{s i n (\frac{π}{2} (2 m + 1))}{{(2 m + 1)}^{4}}$ $= \sum_{m ⩾ 0} \frac{π}{2} . \frac{1}{8 {(2 m + 1)}^{3}} + - \frac{π}{2} . \sum_{m ⩾ 1} \frac{1}{64 m^{3}} + \sum_{m ⩾ 0} \frac{{(- 1)}^{m}}{{(2 m + 1)}^{4}}$ $= \frac{π}{16} \sum_{m ⩾ 0} \frac{1}{{(2 m + 1)}^{3}} + \sum_{m ⩾ 0} \frac{{(- 1)}^{m}}{{(2 m + 1)}^{4}}$ $β (s) = \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{s}} D i r i c h l e t B e t t a f u n c t i o n$ $Σ \frac{{(- 1)}^{m}}{{(2 m + 1)}^{4}} = β (4), Σ \frac{1}{{(2 m + 1)}^{3}} = (ζ (3) - \frac{1}{8} ζ (3)) = \frac{7}{8} ζ (3)$ $w e g e t$ $= \frac{π}{16} . \frac{7}{8} ζ (3) - \frac{π}{128} ζ (3) + β (4) = \frac{6 π}{128} ζ (3) + β (4) =$ $\int_{0}^{\frac{π}{2}} {c l}_{2} (\sqrt{t}) d t = 2 \int_{0}^{1} {t c l}_{2} (t) d t = 2 . [\frac{6 π}{128} + β (4)]$ $= \frac{12 π}{128} ζ (3) + 2 β (4) = \frac{3 π}{32} ζ (3) + 2 β (4)$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
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