find the value of I =∫_0 ^∞ ((e^(−t) sint)/(√t))dt and J =∫_0 ^∞ ((e^(−t) cos(t))/(√t))dt ,study first the convergence.


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Question Number 62262 by maxmathsup by imad last updated on 18/Jun/19

$f i n d t h e v a l u e o f$ $I = \int_{0}^{\infty} \frac{e^{- t} s i n t}{\sqrt{t}} d t a n d J = \int_{0}^{\infty} \frac{e^{- t} c o s (t)}{\sqrt{t}} d t, s t u d y f i r s t t h e c o n v e r g e n c e .$
Commented by maxmathsup by imad last updated on 19/Jun/19
$the convergence o I and J is proved by sir smail let passe to the values we have J−iI =∫_0 ^∞ (e^(−t) /(√t))(cost −isint)dt =∫_0 ^∞ ((e^(−t ) e^(−it) )/(√t)) dt =∫_0 ^∞ (e^(−(i+1)t) /(√t)) dt changement (√t)=x give J−iI =∫_0 ^∞ (e^(−(1+i)x^2 ) /x) (2x)dx =2 ∫_0 ^∞ e^(−(1+i)x^2 ) dx =∫_(−∞) ^(+∞) e^(−(1+i)x^2 ) dx =_((√(1+i))x =u) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√(1+i))) =(1/(√(1+i))) ∫_(−∞) ^(+∞) e^(−u^2 ) du =((√π)/(√(1+i))) 1+i =(√2)e^((iπ)/4) ⇒(√(1+i))=^4 (√2)e^((iπ)/8) ⇒J−iI =((√π)/((^4 (√2)))) e^(−((iπ)/8)) =((√π)/((^4 (√2)))){cos((π/8))−isin((π/8))} ⇒ I =((√π)/((^4 (√2)))) sin((π/8)) and J =((√π)/((^4 (√2)))) cos((π/8)) ⇒ I =((√π)/((^4 (√2))))((√(2−(√2)))/2) and J =((√π)/((^4 (√2)))) ((√(2+(√2)))/2) .$
$t h e c o n v e r g e n c e o I a n d J i s p r o v e d b y s i r s m a i l l e t p a s s e t o t h e v a l u e s$ $w e h a v e J - i I = \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} (c o s t - i s i n t) d t$ $= \int_{0}^{\infty} \frac{e^{- t} e^{- i t}}{\sqrt{t}} d t = \int_{0}^{\infty} \frac{e^{- (i + 1) t}}{\sqrt{t}} d t c h a n g e m e n t \sqrt{t} = x g i v e$ $J - i I = \int_{0}^{\infty} \frac{e^{- (1 + i) x^{2}}}{x} (2 x) d x = 2 \int_{0}^{\infty} e^{- (1 + i) x^{2}} d x = \int_{- \infty}^{+ \infty} e^{- (1 + i) x^{2}} d x$ $=_{\sqrt{1 + i} x = u} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{1 + i}} = \frac{1}{\sqrt{1 + i}} \int_{- \infty}^{+ \infty} e^{- u^{2}} d u = \frac{\sqrt{π}}{\sqrt{1 + i}}$ $1 + i = \sqrt{2} e^{\frac{i π}{4}} \Rightarrow \sqrt{1 + i} =^{4} \sqrt{2} e^{\frac{i π}{8}} \Rightarrow J - i I = \frac{\sqrt{π}}{(^{4} \sqrt{2})} e^{- \frac{i π}{8}}$ $= \frac{\sqrt{π}}{(^{4} \sqrt{2})} {c o s (\frac{π}{8}) - i s i n (\frac{π}{8})} \Rightarrow I = \frac{\sqrt{π}}{(^{4} \sqrt{2})} s i n (\frac{π}{8}) a n d J = \frac{\sqrt{π}}{(^{4} \sqrt{2})} c o s (\frac{π}{8}) \Rightarrow$ $I = \frac{\sqrt{π}}{(^{4} \sqrt{2})} \frac{\sqrt{2 - \sqrt{2}}}{2} a n d J = \frac{\sqrt{π}}{(^{4} \sqrt{2})} \frac{\sqrt{2 + \sqrt{2}}}{2} .$
	Answered by Smail last updated on 18/Jun/19

	$s i n t ⩽ 1 \Leftrightarrow e^{- t} s i n t ⩽ e^{- t}$ $\frac{e^{- t} s i n t}{\sqrt{t}} ⩽ \frac{e^{- t}}{\sqrt{t}} \Leftrightarrow \int_{0}^{\infty} \frac{e^{- t} s i n t}{\sqrt{t}} d t ⩽ \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t$ $K n o w i n g t h a t \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t = 2 \int_{0}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ $w h i c h m e a n s \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t c o n v e r g e s .$ $T h a t m e a n s I = \int_{0}^{\infty} \frac{e^{- t} s i n t}{\sqrt{t}} d t c o n v e r g e s$ $b e c a u s e I ⩽ \sqrt{π}$ $A l s o J ⩽ \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t m e a n i n g J a l s o c o n v e r g e s$ $J - i I = \int_{0}^{\infty} \frac{e^{- t} e^{- i t}}{\sqrt{t}} d t$ $= \int_{0}^{\infty} \frac{e^{- t (1 + i)}}{\sqrt{t}} d t$ $L e t x = \sqrt{t} (\sqrt{1 + i}) \Rightarrow \frac{d t}{\sqrt{t}} = \frac{2 d x}{\sqrt{1 + i}}$ $J - i I = \frac{2}{\sqrt[4]{2} e^{i \frac{π}{8}}} \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{\sqrt[4]{2} e^{i \frac{π}{8}}}$ $= \frac{\sqrt{π}}{\sqrt[4]{2}} (c o s (\frac{π}{8}) - i s i n (\frac{π}{8}))$ $J = \frac{\sqrt{π}}{\sqrt[4]{2}} c o s (\frac{π}{8}) = \frac{\sqrt{π}}{\sqrt[4]{2}} \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}} =$ $J = \frac{\sqrt{π (\sqrt{2} + 1)}}{2}$ $I = \frac{\sqrt{π (\sqrt{2} - 1)}}{2}$
	Commented by maxmathsup by imad last updated on 18/Jun/19

	$t h a n k s s i r$
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