((sin((√2)))/1^3 )+((sin(2(√2)))/2^3 )+((sin(3(√2)))/3^3 )+...=((π^b +1)/( a(√b)))−(π/b) Find a−b


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Question Number 136070 by Dwaipayan Shikari last updated on 18/Mar/21

$\frac{s i n (\sqrt{2})}{1^{3}} + \frac{s i n (2 \sqrt{2})}{2^{3}} + \frac{s i n (3 \sqrt{2})}{3^{3}} + . . . = \frac{π^{b} + 1}{a \sqrt{b}} - \frac{π}{b}$ $F i n d a - b$
	Answered by mnjuly1970 last updated on 18/Mar/21
	$Ω(x)=Σ_(n=1) ^∞ ((sin(nx))/n^3 ) −Im(ln(1−e^(ix) ))=Σ_(n=1) =∞((sin(nx))/n) −Imln((1−cos(x))−isin(x))=Σ_(n=1) ^∞ ((sin(nx))/n) −Im{ln(2sin((x/2))−itan^(−1) (((sin(x))/(1−cos(x))))=Σ_(n=1) ^∞ ((sin(nx))/(n=1)) (π/2)−(x/2)=Σ_(n=1) ^∞ ((sin(nx))/n) (π/2)x−(x^2 /4)=−Σ_(n=1) ^∞ ((cos(nx))/n^2 )+C x=π (π^2 /4)=(π^2 /(12))+C⇒C=(π^2 /6) (π^2 /6)+(x^2 /4)−(π/2)x=Σ_(n=1) ^∞ ((cos(nx))/n^2 ) (π^2 /6)x+(x^3 /(12))−(π/4)x^2 =Σ_(n=1) ^∞ ((sin(nx))/n^3 ) x=(√2) Σ_(n=1) ^∞ ((sin(n(√2)))/n^3 )=(π^2 /6)(√2) +((√2)/6)−(π/2) Ω=((π^2 +1)/(3(√2)))−(π/2) hence :: a−b=3−2=1 please check my solution....$
	$Ω (x) = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n^{3}}$ $- I m (l n (1 - e^{i x})) = \sum_{n = 1} = \infty \frac{s i n (n x)}{n}$ $- I m l n ((1 - c o s (x)) - i s i n (x)) = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n}$ $- I m {l n (2 s i n (\frac{x}{2}) - {i t a n}^{- 1} (\frac{s i n (x)}{1 - c o s (x)}) = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n = 1}$ $\frac{π}{2} - \frac{x}{2} = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n}$ $\frac{π}{2} x - \frac{x^{2}}{4} = - \underset{n = 1}{\sum^{\infty}} \frac{c o s (n x)}{n^{2}} + C$ $x = π$ $\frac{π^{2}}{4} = \frac{π^{2}}{12} + C \Rightarrow C = \frac{π^{2}}{6}$ $\frac{π^{2}}{6} + \frac{x^{2}}{4} - \frac{π}{2} x = \underset{n = 1}{\sum^{\infty}} \frac{c o s (n x)}{n^{2}}$ $\frac{π^{2}}{6} x + \frac{x^{3}}{12} - \frac{π}{4} x^{2} = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n^{3}}$ $x = \sqrt{2}$ $\underset{n = 1}{\sum^{\infty}} \frac{s i n (n \sqrt{2})}{n^{3}} = \frac{π^{2}}{6} \sqrt{2} + \frac{\sqrt{2}}{6} - \frac{π}{2}$ $Ω = \frac{π^{2} + 1}{3 \sqrt{2}} - \frac{π}{2}$ $h e n c e :: a - b = 3 - 2 = 1$ $p l e a s e c h e c k m y s o l u t i o n . . . .$
	Commented by Dwaipayan Shikari last updated on 18/Mar/21

	$G r e a t s i r!$
	Commented by mnjuly1970 last updated on 18/Mar/21

	$t h a n k y o u f o r y o u r n i c e q u e s t i o n s . . .$
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