...nice calculus.. prove that:: lim_(x→0) (((2φ(x))/x^2 ) +(π^2 /(3x))) =^(???) ζ(3) where φ(x)=∫


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Question Number 123687 by mnjuly1970 last updated on 27/Nov/20

$. . . n i c e c a l c u l u s . .$ $p r o v e t h a t :: {l i m}_{x \to 0} (\frac{2 ϕ (x)}{x^{2}} + \frac{π^{2}}{3 x}) \overset{? ? ?}{=} ζ (3)$ $w h e r e$ $ϕ (x) = \int_{0}^{1} \frac{(t^{x} - 1) (l n (1 - t))}{t l n (t)} d t$
	Answered by mnjuly1970 last updated on 28/Nov/20
	$solution: φ(x)=−∫_0 ^( 1) {((t^x −1)/(ln(t)))Σ_(n=1) ^∞ (t^(n−1) /n)}dx =−Σ_(n=1 ) ^∞ (1/n)∫_0 ^( 1) ((t^(x+n−1) −t^(n−1) )/(ln(t)))dt =−Σ_(n≥1) (1/n)[ln(x+n)−ln(n)] =−Σ_(n≥1) (1/n)ln(((x+n)/n))=−Σ_(n≥1) (1/n)ln(1+(x/n)) =−Σ_(n≥1) (1/n)Σ_(m≥1) (((((−x)/n))^m )/m)=Σ_(m≥1) (((−x)^m )/m)Σ_(n≥1) (1/n^(m+1) ) ∴ φ(x)=Σ_(m≥1) (((−x)^m ζ(m+1))/m) =(((−x)^1 )/1) ζ(2)+(x^2 /2)ζ(3)−(x^3 /3)ζ(4)+... ⇒((2φ(x))/x^2 )=ζ(3)−((2ζ(2))/x)−((2x)/3)ζ(4)+... ⇒2((φ(x))/x^2 )+(π^2 /(3x))=ζ(3)−((2x)/3)φζ(4)+... limit from both sides as x→0... lim_(x→0) (((2φ(x))/x^2 )+(π^2 /(3x)))=ζ(3) ✓✓ m.n.$
	$s o l u t i o n : ϕ (x) = - \int_{0}^{1} {\frac{t^{x} - 1}{l n (t)} \underset{n = 1}{\sum^{\infty}} \frac{t^{n - 1}}{n}} d x$ $= - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} \frac{t^{x + n - 1} - t^{n - 1}}{l n (t)} d t$ $= - \sum_{n ⩾ 1} \frac{1}{n} [l n (x + n) - l n (n)]$ $= - \sum_{n ⩾ 1} \frac{1}{n} l n (\frac{x + n}{n}) = - \sum_{n ⩾ 1} \frac{1}{n} l n (1 + \frac{x}{n})$ $= - \sum_{n ⩾ 1} \frac{1}{n} \sum_{m ⩾ 1} \frac{{(\frac{- x}{n})}^{m}}{m} = \sum_{m ⩾ 1} \frac{{(- x)}^{m}}{m} \sum_{n ⩾ 1} \frac{1}{n^{m + 1}}$ $∴ ϕ (x) = \sum_{m ⩾ 1} \frac{{(- x)}^{m} ζ (m + 1)}{m}$ $= \frac{{(- x)}^{1}}{1} ζ (2) + \frac{x^{2}}{2} ζ (3) - \frac{x^{3}}{3} ζ (4) + . . .$ $\Rightarrow \frac{2 ϕ (x)}{x^{2}} = ζ (3) - \frac{2 ζ (2)}{x} - \frac{2 x}{3} ζ (4) + . . .$ $\Rightarrow 2 \frac{ϕ (x)}{x^{2}} + \frac{π^{2}}{3 x} = ζ (3) - \frac{2 x}{3} ϕ ζ (4) + . . .$ $l i m i t f r o m b o t h s i d e s a s x \to 0 . . .$ ${l i m}_{x \to 0} (\frac{2 ϕ (x)}{x^{2}} + \frac{π^{2}}{3 x}) = ζ (3) ✓ ✓$ $m . n .$
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