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Question Number 33352 by caravan msup abdo. last updated on 15/Apr/18

l e t g i v e S (x) = \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{\sqrt{x + n}}, x > 0

1) s t u d y t h e c o n t n u i t y, d e r i v s b i l i t y, l i m i t s

a t 0^{+} a n d + \infty

2) w e g i v e \int_{0}^{\infty} e^{- t^{2}} d t = \frac{\sqrt{π}}{2} . p r o v e t h a t

\forall x > 0 S (x) = \frac{1}{\sqrt{π}} \int_{0}^{\infty} \frac{e^{- t x}}{\sqrt{t} (1 + e^{- t})} d t .

Commented by math khazana by abdo last updated on 20/Apr/18

1) l e t p u t f_{n} (x) = \frac{{(- 1)}^{n}}{\sqrt{x + n}} t h e s e q u e n c e (f_{n}) a r e c o n t i n u e

n d t h e s e r i e Σ f_{n} (x) i s c o n v e r g e n t (i f w e t a k e

φ (t) = \frac{1}{\sqrt{x + t}} . φ i s d e c r e s z i n g s o Σ f_{n} i s a l t e r n a t i n g)

i t s s u m S (x) i s c o n t i n u e f o r x > 0 a l s o t h e (f_{n}) a r e

d e r i v a b l e s a n d f_{n}^{^{'}} (x) = {(- 1)}^{n} \frac{- 1}{2 (x + n) \sqrt{x + n}}

\sum^{} f_{n}^{^{'}} (x) i s c o n v e r g e n t a s a a l t e r n a t i n g s e r i e s o

S i s d e r i v a b l e a n d S^{^{'}} (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{2 (x + n) \sqrt{x + n}} .

2) w e h a v e \int_{0}^{\infty} \frac{e^{- t x}}{\sqrt{t} (1 + e^{- t})} d t

= \int_{0}^{\infty} \frac{e^{- t x}}{\sqrt{t}} (\sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- n t}) d t

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} \frac{e^{- (n + x) t}}{\sqrt{t}} d t (c h . \sqrt{t} = u)

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} \frac{e^{- (n + x) u^{2}}}{u} 2 u d u

= 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} e^{- (n + x) u^{2}} d u

=_{\sqrt{n + x} u = t} 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} e^{- t^{2}} \frac{d t}{\sqrt{n + x}}

= 2 . \frac{\sqrt{π}}{2} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{\sqrt{n + x}} = \sqrt{π} S (x) \Rightarrow

S (x) = \frac{1}{\sqrt{π}} \int_{0}^{\infty} \frac{e^{- t x}}{\sqrt{t} (1 + e^{- t})} d t .