help please ∫_1 ^∞ ((1/t)−sin^(−1) ((1/t))) dt I can show that it is equal to −Σ_(n≥0) (−1)^n (((2n−1)!)/(4^n (n!)^2 (2n+1))) but I cant calculate it...


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Question Number 170349 by antoineop last updated on 21/May/22

$h e l p p l e a s e$ ${\int^{\infty}}_{1} (\frac{1}{t} - \sin^{- 1} (\frac{1}{t})) d t$ $I c a n s h o w t h a t i t i s e q u a l t o$ $- \sum_{n ⩾ 0} {(- 1)}^{n} \frac{(2 n - 1)!}{4^{n} {(n!)}^{2} (2 n + 1)}$ $b u t I c a n t c a l c u l a t e i t . . .$
	Answered by aleks041103 last updated on 21/May/22

	$\frac{1}{\sqrt{1 - x^{2}}} = {(1 + (- x^{2}))}^{- 1 / 2}$ ${(1 + z)}^{a} = 1 + a z + \frac{a (a - 1)}{2} z^{2} + \frac{a (a - 1) (a - 2)}{3!} z^{3} + . . . =$ $= 1 + \underset{i = 1}{\sum^{\infty}} \frac{z^{i}}{i!} \underset{j = 0}{\prod^{i - 1}} (a - j)$ $\frac{1}{\sqrt{1 - x^{2}}} = 1 + \underset{i = 1}{\sum^{\infty}} \frac{{(- x^{2})}^{i}}{i!} \underset{j = 0}{\prod^{i - 1}} (- \frac{1}{2} - j) =$ $= 1 + \underset{i = 1}{\sum^{\infty}} \frac{{(- 1)}^{i} x^{2 i}}{i!} \underset{j = 0}{\prod^{i - 1}} (- \frac{1}{2}) (2 j + 1) =$ $\underset{j = 0}{\prod^{i - 1}} (- \frac{1}{2}) (2 j + 1) = {(- \frac{1}{2})}^{i} (1.3 . . . . . (2 i - 1))$ $1.3 . . . . . (2 i - 1) = \frac{(2 i)!}{2^{i} i!}$ $\frac{1}{\sqrt{1 - x^{2}}} = 1 + \underset{i = 1}{\sum^{\infty}} \frac{(2 i)!}{4^{i} {(i!)}^{2}} x^{2 i} = \underset{i = 0}{\sum^{\infty}} \frac{(2 i)!}{4^{i} {(i!)}^{2}} x^{2 i}$ ${s i n}^{- 1} (x) = \int_{0}^{x} \frac{d x}{\sqrt{1 - x^{2}}} =$ $= \underset{i = 0}{\sum^{\infty}} \frac{(2 i)!}{4^{i} {(i!)}^{2}} \int_{0}^{x} x^{2 i} d x = \underset{i = 0}{\sum^{\infty}} \frac{(2 i)!}{4^{i} (2 i + 1) {(i!)}^{2}} x^{2 i + 1}$ $\frac{1}{t} - {s i n}^{- 1} (t) =$ $= \frac{1}{t} - \underset{i = 0}{\sum^{\infty}} \frac{(2 i)!}{4^{i} (2 i + 1) {(i!)}^{2}} \frac{1}{t^{2 i + 1}} =$ $= - \underset{i = 1}{\sum^{\infty}} \frac{(2 i)!}{4^{i} (2 i + 1) {(i!)}^{2}} \frac{1}{t^{2 i + 1}}$ $\Rightarrow \int_{1}^{\infty} (\frac{1}{t} - a s i n (\frac{1}{t})) d t =$ $= - \underset{i = 1}{\sum^{\infty}} \frac{(2 i)!}{4^{i} (2 i + 1) {(i!)}^{2}} \int_{1}^{\infty} t^{- 2 i - 1} d t =$ $= - \underset{i = 1}{\sum^{\infty}} \frac{(2 i)!}{4^{i} (2 i + 1) {(i!)}^{2}} {(\frac{t^{- 2 i}}{- 2 i}]}_{1}^{\infty} =$ $= - \underset{i = 1}{\sum^{\infty}} \frac{(2 i - 1)!}{4^{i} (2 i + 1) {(i!)}^{2}}$
	Commented by aleks041103 last updated on 21/May/22

	Commented by antoineop last updated on 21/May/22
	Yep thats what I did, thanks ! (I'm sorry my English is really bad) but I'm supposed to calculate this integral, so, I'm supposed to calculate this sum and I don't know how..
	Commented by aleks041103 last updated on 21/May/22

	$T h e m a i n p a r t o f t h e s o l u t i o n i s$ $f i n d i n g a s e r i e s r e p r e s e n t a t i o n f o r$ $t h e i n v e r s e s i n e f u n c t i o n .$ $A f t e r t h a t {i t}^{'} s e a s y .$
	Commented by antoineop last updated on 21/May/22
	My exercise just tell me "Justify the existence of this integral and then, calculate it" I dont know how to calculate this sum, I've also tried IBP but .. still impossible
	Commented by aleks041103 last updated on 21/May/22

	$I d o n t t h i n k t h a t t h e r e i s a c l o s e d f o r m .$
	Commented by antoineop last updated on 21/May/22
	That's what I think too, I'll ask my teacher next week and share it if he has a way (of not) to calculate that ! thanks :)
	Commented by aleks041103 last updated on 21/May/22

	$Y o u r t a s k w a s t o p r o v e t h a t t h e i m t e g r a l$ $c o n v e r g e s a n d t h e n c a l c u l a t e i t, w h i c h$ ${y o u}^{'} v e a l r e a d y d o n e b y f i n d i n g t h i s s u m$
	Commented by aleks041103 last updated on 21/May/22

	$f (x) = {s i n}^{- 1} (x) - x$ $w e a n a l y z e x \in [0, 1]$ $w e k n o w t h a t$ $x ⩾ s i n (x), x ⩾ 0$ $\Rightarrow {s i n}^{- 1} x ⩾ x ⩾ 0$ $\Rightarrow {s i n}^{- 1} x - x ⩾ 0$ $\Rightarrow \int_{1}^{\infty} - f (\frac{1}{t}) d t ⩽ 0$
	Commented by antoineop last updated on 22/May/22
	yeah but in France in 99% of analysis exercises we are able to calculate directly the sum
	Commented by Tawa11 last updated on 08/Oct/22

	$Great sir$
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