calculate ∫_0 ^1 (([nx])/(2x+1))dx


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Question Number 51186 by Abdo msup. last updated on 24/Dec/18

$c a l c u l a t e \int_{0}^{1} \frac{[n x]}{2 x + 1} d x$
Commented by Abdo msup. last updated on 25/Dec/18
$let A_n =∫_0 ^1 (([nx])/(2x+1))dx ⇒A_n =Σ_(k=0) ^(n−1) ∫_(k/n) ^((k+1)/n) (k/(2x+1)) ([nx]=k) ⇒A_n =(1/2) Σ_(k=0) ^(n−1) k [ln(2x+1)]_(k/n) ^((k+1)/n) =(1/2) Σ_(k=0) ^(n−1) k {ln(((2(k+1))/n)+1)−ln(((2k)/n)+1)} =(1/2)Σ_(k=0) ^(n−1) k{ ln(2k+n+2)+ln(2k+n)} =(1/2)Σ_(k=0) ^(n−1) k ln(((2k+n +2)/(2k+n))) ⇒ A_n =(1/2){ln(((n+4)/(n+3)))+2ln(((n+6)/(n+4)))+3ln(((n+8)/(n+6)))+... +(n−1)ln(((3n)/(3n−2)))}$
$l e t A_{n} = \int_{0}^{1} \frac{[n x]}{2 x + 1} d x \Rightarrow A_{n} = \sum_{k = 0}^{n - 1} \int_{\frac{k}{n}}^{\frac{k + 1}{n}} \frac{k}{2 x + 1}$ $([n x] = k) \Rightarrow A_{n} = \frac{1}{2} \sum_{k = 0}^{n - 1} k {[l n (2 x + 1)]}_{\frac{k}{n}}^{\frac{k + 1}{n}}$ $= \frac{1}{2} \sum_{k = 0}^{n - 1} k {l n (\frac{2 (k + 1)}{n} + 1) - l n (\frac{2 k}{n} + 1)}$ $= \frac{1}{2} \sum_{k = 0}^{n - 1} k {l n (2 k + n + 2) + l n (2 k + n)}$ $= \frac{1}{2} \sum_{k = 0}^{n - 1} k l n (\frac{2 k + n + 2}{2 k + n}) \Rightarrow$ $A_{n} = \frac{1}{2} {l n (\frac{n + 4}{n + 3}) + 2 l n (\frac{n + 6}{n + 4}) + 3 l n (\frac{n + 8}{n + 6}) + . . .$ $+ (n - 1) l n (\frac{3 n}{3 n - 2})}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18
	$∫_0 ^(1/n) (([nx])/(2(x+(1/2))))dx+∫_(1/n) ^1 (([nx])/(2(x+(1/2))))dx ∫_0 ^(1/n) (0/(2(x+(1/2))))dx+∫_(1/n) ^1 (1/(2(x+(1/2))))dx (1/2)∣ln(x+(1/2))∣_(1/n) ^1 =(1/2){ln((3/2))−ln((1/n)+(1/2))} =(1/2)ln(((3/2)/((2+n)/(2n)))) (1/2)ln(((3×2n)/(2(2+n)))) =(1/2)ln(((3n)/(2+n)))$
	$\int_{0}^{\frac{1}{n}} \frac{[n x]}{2 (x + \frac{1}{2})} d x + \int_{\frac{1}{n}}^{1} \frac{[n x]}{2 (x + \frac{1}{2})} d x$ $\int_{0}^{\frac{1}{n}} \frac{0}{2 (x + \frac{1}{2})} d x + \int_{\frac{1}{n}}^{1} \frac{1}{2 (x + \frac{1}{2})} d x$ $\frac{1}{2} ∣ l n (x + \frac{1}{2}) ∣_{\frac{1}{n}}^{1}$ $= \frac{1}{2} {l n (\frac{3}{2}) - l n (\frac{1}{n} + \frac{1}{2})}$ $= \frac{1}{2} l n (\frac{\frac{3}{2}}{\frac{2 + n}{2 n}})$ $\frac{1}{2} l n (\frac{3 \times 2 n}{2 (2 + n)})$ $= \frac{1}{2} l n (\frac{3 n}{2 + n})$
	Commented by Abdo msup. last updated on 25/Dec/18

	$s i r T a n m a y i f \frac{1}{n} ⩽ x ⩽ 1 \Rightarrow 1 ⩽ n x ⩽ n s o [n x] i s$ $n o t e q u a l t o 1!!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18

	$w h a t i s t h e m e s n i n g o f 1 ⩽ n x$
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