calculate f(a) =∫ (dx/((√(1+ax))−(√(1−ax)))) with a>0 . 2) calculate U_n =∫_(−(1/(na))) ^(1/(na)) (dx/((√(1+ax))−(√(1−ax)))) with n from N and n>1 find lim_(n→+∞) U_n and study the convergence of Σ U


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Question Number 54011 by maxmathsup by imad last updated on 27/Jan/19

$c a l c u l a t e f (a) = \int \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}} w i t h a > 0 .$ $2) c a l c u l a t e U_{n} = \int_{- \frac{1}{n a}}^{\frac{1}{n a}} \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}} w i t h n f r o m N a n d n > 1$ $f i n d {l i m}_{n \to + \infty} U_{n} a n d s t u d y t h e c o n v e r g e n c e o f Σ U_{n}$
Commented by maxmathsup by imad last updated on 28/Jan/19
$1) we have f(a) =_(ax =t) ∫ (dt/(a((√(1+t))−(√(1−t))))) ⇒ af(a) =∫ (((√(1+t))+(√(1−t)))/(2t)) dt = (1/2) ∫ ((√(1+t))/t)dt +(1/2)∫ ((√(1−t))/t) dt ⇒ 2af(a) = ∫ ((√(1+t))/t) dt +∫ ((√(1−t))/t) dt but ∫ ((√(1+t))/t)dt =_(1+t=u^2 ) ∫ (u/(u^2 −1)) (2u)du =2 ∫ (u^2 /(u^2 −1)) du =2 ∫ ((u^2 −1+1)/(u^2 −1)) =2u +2 ∫ (du/(u^2 −1)) =2u + ∫ ((1/(1−u)) +(1/(1+u)))du =2u +ln∣((1+u)/(1−u))∣ +c_0 =2(√(1+t))+ln∣((1+(√(1+t)))/(1−(√(1−t))))∣ +c_0 =2 (√(1+ax)) +ln∣((1+(√(1+ax)))/(1−(√(1−ax))))∣ +c_0 also we have ∫ ((√(1−t))/t) dt =_(1−t =u^2 ) ∫ (u/(1−u^2 )) (−2u)du = 2 ∫ (u^2 /(u^2 −1)) ⇒ 2af(a) =2 ∫ ((√(1+t))/t) dt ⇒f(a) =(1/a){2(√(1+ax)) +ln∣((1+(√(1+ax)))/(1−(√(1−ax))))∣} . f(a) =(1/a){2(√(1+ax))+ln∣((1+(√(1+ax)))/(1−(√(1−ax))))∣} +C .$
$1) w e h a v e f (a) =_{a x = t} \int \frac{d t}{a (\sqrt{1 + t} - \sqrt{1 - t})} \Rightarrow$ $a f (a) = \int \frac{\sqrt{1 + t} + \sqrt{1 - t}}{2 t} d t = \frac{1}{2} \int \frac{\sqrt{1 + t}}{t} d t + \frac{1}{2} \int \frac{\sqrt{1 - t}}{t} d t \Rightarrow$ $2 a f (a) = \int \frac{\sqrt{1 + t}}{t} d t + \int \frac{\sqrt{1 - t}}{t} d t b u t$ $\int \frac{\sqrt{1 + t}}{t} d t =_{1 + t = u^{2}} \int \frac{u}{u^{2} - 1} (2 u) d u = 2 \int \frac{u^{2}}{u^{2} - 1} d u$ $= 2 \int \frac{u^{2} - 1 + 1}{u^{2} - 1} = 2 u + 2 \int \frac{d u}{u^{2} - 1} = 2 u + \int (\frac{1}{1 - u} + \frac{1}{1 + u}) d u$ $= 2 u + l n ∣ \frac{1 + u}{1 - u} ∣ + c_{0} = 2 \sqrt{1 + t} + l n ∣ \frac{1 + \sqrt{1 + t}}{1 - \sqrt{1 - t}} ∣ + c_{0}$ $= 2 \sqrt{1 + a x} + l n ∣ \frac{1 + \sqrt{1 + a x}}{1 - \sqrt{1 - a x}} ∣ + c_{0} a l s o w e h a v e$ $\int \frac{\sqrt{1 - t}}{t} d t =_{1 - t = u^{2}} \int \frac{u}{1 - u^{2}} (- 2 u) d u = 2 \int \frac{u^{2}}{u^{2} - 1} \Rightarrow$ $2 a f (a) = 2 \int \frac{\sqrt{1 + t}}{t} d t \Rightarrow f (a) = \frac{1}{a} {2 \sqrt{1 + a x} + l n ∣ \frac{1 + \sqrt{1 + a x}}{1 - \sqrt{1 - a x}} ∣} .$ $f (a) = \frac{1}{a} {2 \sqrt{1 + a x} + l n ∣ \frac{1 + \sqrt{1 + a x}}{1 - \sqrt{1 - a x}} ∣} + C .$
Commented by maxmathsup by imad last updated on 28/Jan/19
$2) we have U_n =∫_(−(1/(na))) ^(1/(na)) (dx/((√(1+ax))−(√(1−ax)))) =2lim_(ξ→0) ∫_ξ ^(1/(na)) (dx/((√(1+ax))−(√(1−ax)))) but ∫_ξ ^(1/(na)) (dx/((√(1+ax))−(√(1−ax)))) =[(1/a)(2(√(1+ax))+ln∣((1+(√(1+ax)))/(1−(√(1−ax))))∣)]_(x=ξ) ^(1/(na)) =(1/a){2(√(1+(1/n)))+ln∣((1+(√(1+(1/n))))/(1−(√(1−(1/n)))))∣−2(√(1+aξ))−ln∣((1+(√(1+aξ)))/(1−(√(1−aξ))))∣} but we cant find lim U_n from this quantity be continued...$
$2) w e h a v e U_{n} = \int_{- \frac{1}{n a}}^{\frac{1}{n a}} \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}} = 2 {l i m}_{ξ \to 0} \int_{ξ}^{\frac{1}{n a}} \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}}$ $b u t \int_{ξ}^{\frac{1}{n a}} \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}} = {[\frac{1}{a} (2 \sqrt{1 + a x} + l n ∣ \frac{1 + \sqrt{1 + a x}}{1 - \sqrt{1 - a x}} ∣)]}_{x = ξ}^{\frac{1}{n a}}$ $= \frac{1}{a} {2 \sqrt{1 + \frac{1}{n}} + l n ∣ \frac{1 + \sqrt{1 + \frac{1}{n}}}{1 - \sqrt{1 - \frac{1}{n}}} ∣ - 2 \sqrt{1 + a ξ} - l n ∣ \frac{1 + \sqrt{1 + a ξ}}{1 - \sqrt{1 - a ξ}} ∣}$ $b u t w e c a n t f i n d l i m U_{n} f r o m t h i s q u a n t i t y b e c o n t i n u e d . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19

	$\int \frac{\sqrt{1 + a x} + \sqrt{1 - a x}}{2 a x} d x$ $\int \frac{\sqrt{1 + a x}}{2 a x} d x + \int \frac{\sqrt{1 - a x}}{2 a x} d x$ $t_{1}^{2} = 1 + a x 2 t_{1} d t = a d x$ $t_{2}^{2} = 1 - a x 2 t_{2} d t = - a d x$ $\int \frac{t_{1} \times 2 t_{1} d t}{2 a (t_{1}^{2} - 1)} + \int \frac{t_{2} \times - 2 t_{2}}{2 a \times (1 - t_{1}^{2})} {d t}_{2}$ $\frac{1}{a} \int \frac{t_{1}^{2} - 1 + 1}{t_{1}^{2} - 1} {d t}_{1} + \frac{1}{a} \int \frac{1 - t_{2}^{2} - 1}{1 - t_{2}^{2}} {d t}_{2}$ $\frac{1}{a} [\int {d t}_{1} + \int \frac{{d t}_{1}}{t_{2}^{2} - 1} + \int {d t}_{2} - \int \frac{{d t}_{2}}{1 - t_{2}^{2}}]$ $n o w u s e f o r m u l a . . .$
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