1) calculate f(x,y) =∫_0 ^∞ ((e^(−xt) cos(yt))/(√t)) dt and g(x,y) =∫_0 ^∞ ((e^(−xt) sin(yt))/(√t)) dt with x>0 and y>0 2) find the values of ∫_0 ^∞ ((e^(−2t) cos(t))/(√t)) dt and ∫


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Question Number 62335 by maxmathsup by imad last updated on 19/Jun/19

$1) c a l c u l a t e f (x, y) = \int_{0}^{\infty} \frac{e^{- x t} c o s (y t)}{\sqrt{t}} d t a n d g (x, y) = \int_{0}^{\infty} \frac{e^{- x t} s i n (y t)}{\sqrt{t}} d t$ $w i t h x > 0 a n d y > 0$ $2) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{e^{- 2 t} c o s (t)}{\sqrt{t}} d t a n d \int_{0}^{\infty} \frac{e^{- t} c o s (2 t)}{\sqrt{t}} d t$
Commented bymaxmathsup by imad last updated on 20/Jun/19
$1) we have f(x,y)+ig(x,y) =∫_0 ^∞ (e^(−xt) /(√t))e^(iyt) dt =∫_0 ^∞ (e^(−(x+iy)t) /(√t)) dt changement (√(x+iy))(√t)=u give f(x,y)+ig(x,y) =(√(x+iy))∫_0 ^∞ (e^(−u^2 ) /u) (√t)=(u/(√(x+iy))) ⇒t =(u^2 /(x+iy)) ⇒dt =((2udu)/(x+iy)) ⇒f(x,y)+ig(x,y)=(√(x+iy))∫_0 ^∞ (e^(−u^2 ) /u) ((2udu)/(x+iy)) =(2/(√(x+iy)))∫_0 ^∞ e^(−u^2 ) du =(2/(√(x+iy))) ((√π)/2)=((√π)/(√(x+iy))) we have x+iy =(√(x^2 +y^2 )){(x/(√(x^2 +y^2 ))) +i(y/(√(x^2 +y^2 )))} =r e^(iθ) ⇒r =(√(x^2 +y^2 )) cosθ =(x/(√(x^2 +y^2 ))) and sinθ =(y/(√(x^2 +y^2 ))) ⇒tanθ =(y/x) ⇒ θ =arctan((y/x)) ⇒x+iy =(x^2 +y^2 )^(1/2) e^(iarctan((y/x))) ⇒(√(x+iy)) =(x^2 +y^2 )^(1/4) e^((i/2)arctan((y/x))) ⇒ f(x,y) =(x^2 +y^2 )^(1/4) cos((1/2)arctan((y/x))) and g(x,y) =(x^2 +y^2 )^(1/4) sin((1/2)arctan((y/x)))$
$1) w e h a v e f (x, y) + i g (x, y) = \int_{0}^{\infty} \frac{e^{- x t}}{\sqrt{t}} e^{i y t} d t = \int_{0}^{\infty} \frac{e^{- (x + i y) t}}{\sqrt{t}} d t$ $c h a n g e m e n t \sqrt{x + i y} \sqrt{t} = u g i v e f (x, y) + i g (x, y) = \sqrt{x + i y} \int_{0}^{\infty} \frac{e^{- u^{2}}}{u}$ $\sqrt{t} = \frac{u}{\sqrt{x + i y}} \Rightarrow t = \frac{u^{2}}{x + i y} \Rightarrow d t = \frac{2 u d u}{x + i y} \Rightarrow f (x, y) + i g (x, y) = \sqrt{x + i y} \int_{0}^{\infty} \frac{e^{- u^{2}}}{u} \frac{2 u d u}{x + i y}$ $= \frac{2}{\sqrt{x + i y}} \int_{0}^{\infty} e^{- u^{2}} d u = \frac{2}{\sqrt{x + i y}} \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{\sqrt{x + i y}}$ $w e h a v e x + i y = \sqrt{x^{2} + y^{2}} {\frac{x}{\sqrt{x^{2} + y^{2}}} + i \frac{y}{\sqrt{x^{2} + y^{2}}}} = r e^{i θ} \Rightarrow r = \sqrt{x^{2} + y^{2}}$ $c o s θ = \frac{x}{\sqrt{x^{2} + y^{2}}} a n d s i n θ = \frac{y}{\sqrt{x^{2} + y^{2}}} \Rightarrow t a n θ = \frac{y}{x} \Rightarrow$ $θ = a r c t a n (\frac{y}{x}) \Rightarrow x + i y = {(x^{2} + y^{2})}^{\frac{1}{2}} e^{i a r c t a n (\frac{y}{x})} \Rightarrow \sqrt{x + i y} = {(x^{2} + y^{2})}^{\frac{1}{4}} e^{\frac{i}{2} a r c t a n (\frac{y}{x})} \Rightarrow$ $f (x, y) = {(x^{2} + y^{2})}^{\frac{1}{4}} c o s (\frac{1}{2} a r c t a n (\frac{y}{x})) a n d g (x, y) = {(x^{2} + y^{2})}^{\frac{1}{4}} s i n (\frac{1}{2} a r c t a n (\frac{y}{x}))$
Commented bymaxmathsup by imad last updated on 20/Jun/19

$2) \int_{0}^{\infty} \frac{e^{- 2 t} c o s (t)}{\sqrt{t}} d t = f (2, 1) =^{4} \sqrt{5} c o s (\frac{1}{2} a r c t a n (\frac{1}{2}))$ $\int_{0}^{\infty} \frac{e^{- t} c o s (2 t)}{\sqrt{t}} d t = f (1, 2) =^{4} \sqrt{5} c o s (\frac{1}{2} a r c t a n (2))$
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