....nice mathematics... evaluate :::: Ω=∫_0 ^( ∞) ((√(x+1)) −(√x) )^(10) dx=?


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Question Number 123550 by mnjuly1970 last updated on 26/Nov/20

$. . . . n i c e m a t h e m a t i c s . . .$ $e v a l u a t e ::::$ $Ω = \int_{0}^{\infty} {(\sqrt{x + 1} - \sqrt{x})}^{10} d x = ?$
	Answered by Olaf last updated on 26/Nov/20

	$Let x = \sinh^{2} u$ $d x = 2 \sinh u \cosh u d u = \sinh (2 u) d u$ $Ω = \int_{0}^{\infty} {(\sqrt{1 + \sinh^{2} u} - \sqrt{\sinh^{2} u})}^{10} \sinh (2 u) d u$ $Ω = \int_{0}^{\infty} {(\cosh u - \sinh u)}^{10} \sinh (2 u) d u$ $Ω = \int_{0}^{\infty} {(e^{- u})}^{10} \frac{e^{2 u} - e^{- 2 u}}{2} d u$ $Ω = \int_{0}^{\infty} \frac{e^{- 8 u} - e^{- 12 u}}{2} d u$ $Ω = {[\frac{- \frac{1}{8} e^{- 8 u} + \frac{1}{12} e^{- 12 u}}{2}]}_{0}^{\infty}$ $Ω = - \frac{1}{2} (- \frac{1}{8} + \frac{1}{12})$ $Ω = \frac{1}{48}$
	Commented by mnjuly1970 last updated on 26/Nov/20

	$e x c e l l e n t m r o l a f . t h a n k y o u . .$
	Answered by MJS_new last updated on 26/Nov/20

	$\int {(\sqrt{x + 1} - \sqrt{x})}^{10} d x =$ $[t = {(\sqrt{x + 1} - \sqrt{x})}^{10} \to d x = - \frac{\sqrt{x} \sqrt{x + 1}}{5 {(\sqrt{x + 1} - \sqrt{x})}^{10}} d t]$ $= \frac{1}{20} \int \frac{t^{2 / 5} - 1}{t^{1 / 5}} d t = \frac{t^{6 / 5}}{24} - \frac{t^{4 / 5}}{16}$ $borders for x : [0; \infty) \Rightarrow borders for t : [1; 0]$ $\Rightarrow answer is \frac{1}{48}$
	Commented by MJS_new last updated on 26/Nov/20

	$\underset{0}{\int^{\infty}} {(\sqrt{x + 1} - \sqrt{x})}^{n} d x = \frac{2}{n^{2} - 4} at least for n \in N \land n > 2$
	Commented by MJS_new last updated on 26/Nov/20

	$. . . it holds for n \in R \land n > 2$
	Commented by mnjuly1970 last updated on 26/Nov/20

	$g r a t e f u l m r M J S . .$
	Commented by mnjuly1970 last updated on 26/Nov/20


	Answered by mathmax by abdo last updated on 26/Nov/20
	$let I_n =∫_0 ^∞ ((√(x+1))−(√x))^n dx changement x=sh^2 t give I_n =∫_0 ^∞ (ch(t)−sh(t))^n 2sh(t)cht dt =∫_0 ^∞ (((e^t +e^(−t) )/2)−((e^t −e^(−t) )/2))^n sh(2t)dt =∫_0 ^∞ e^(−nt) ×((e^(2t) −e^(−2t) )/2) dt =(1/2)∫_0 ^∞ (e^(−(n−2)t) −e^(−(n+2)t) )dt =(1/2)[−(1/(n−2))e^(−(n−2)t) +(1/(n+2))e^(−(n+2)t) ]_0 ^∞ =(1/2){(1/(n−2))−(1/(n+2))}=(1/2)×((n+2−n+2)/(n^2 −4))=(2/(n^2 −4)) n=10 ⇒ ∫_0 ^∞ ((√(x+1))−(√x))^(10) dx =(2/(100−4))=(2/(96))=(1/(48))$
	$let I_{n} = \int_{0}^{\infty} {(\sqrt{x + 1} - \sqrt{x})}^{n} dx changement x = {sh}^{2} t give$ $I_{n} = \int_{0}^{\infty} {(ch (t) - sh (t))}^{n} 2 sh (t) cht dt$ $= \int_{0}^{\infty} {(\frac{e^{t} + e^{- t}}{2} - \frac{e^{t} - e^{- t}}{2})}^{n} sh (2 t) dt$ $= \int_{0}^{\infty} e^{- nt} \times \frac{e^{2 t} - e^{- 2 t}}{2} dt = \frac{1}{2} \int_{0}^{\infty} (e^{- (n - 2) t} - e^{- (n + 2) t}) dt$ $= \frac{1}{2} {[- \frac{1}{n - 2} e^{- (n - 2) t} + \frac{1}{n + 2} e^{- (n + 2) t}]}_{0}^{\infty}$ $= \frac{1}{2} {\frac{1}{n - 2} - \frac{1}{n + 2}} = \frac{1}{2} \times \frac{n + 2 - n + 2}{n^{2} - 4} = \frac{2}{n^{2} - 4}$ $n = 10 \Rightarrow \int_{0}^{\infty} {(\sqrt{x + 1} - \sqrt{x})}^{10} dx = \frac{2}{100 - 4} = \frac{2}{96} = \frac{1}{48}$
	Commented by mnjuly1970 last updated on 27/Nov/20

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