calculate A_λ =∫_0 ^∞ (e^(−λx^2 ) /(x^4 +1))dx with λ>0 and find ∫_0 ^1 A


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Question Number 68595 by Abdo msup. last updated on 14/Sep/19

$c a l c u l a t e A_{λ} = \int_{0}^{\infty} \frac{e^{- λ x^{2}}}{x^{4} + 1} d x w i t h λ > 0 a n d$ $f i n d \int_{0}^{1} A_{λ} d λ$
Commented by mathmax by abdo last updated on 14/Sep/19
$A_λ = ∫_0 ^∞ (e^(−λx^2 ) /(x^4 +1))dx ⇒2A_λ =∫_(−∞) ^(+∞) (e^(−λx^2 ) /(x^4 +1))dx let W(z) =(e^(−λz^2 ) /(z^4 +1)) poles of W? we have W(z) =(e^(−λx^2 ) /((z^2 −i)(z^2 +i))) =(e^(−λx^2 ) /((z−(√i))(z+(√i))(z−(√(−i)))(2+(√(−i))))) =(e^(−λx^2 ) /((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of W are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) residus tbeorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ{Res(W,e^((iπ)/4) )+Res(W,−e^(−((iπ)/4)) )} Res(W,e^((iπ)/4) ) =(e^(−λ(i)) /(2e^((iπ)/4) (2i))) =(1/(4i)) e^(−λi−((iπ)/4)) =(1/(4i))e^(−(λ+(π/4))i) Res(W,−e^(−((iπ)/4)) ) = (e^(−λ(−i)) /((−2i)(−2e^(−((iπ)/4)) ))) =(1/(4i)) e^(λi+((iπ)/4)) =(1/(4i)) e^((λ+(π/4))i) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{ (1/(4i))e^(−(λ+(π/4))i) +(1/(4i)) e^((λ+(π/4))i) } =(π/2){ 2Re(e^((λ+(π/4))i) ) =π cos((π/4) +λ) ⇒A_λ =(π/2)cos((π/4) +λ) ∫_0 ^1 A_λ dλ =(π/2) ∫_0 ^1 cos(λ+(π/4)) =[sin(λ+(π/4))]_0 ^1 =sin(1+(π/4))−((√2)/2)$
$A_{λ} = \int_{0}^{\infty} \frac{e^{- λ x^{2}}}{x^{4} + 1} d x \Rightarrow 2 A_{λ} = \int_{- \infty}^{+ \infty} \frac{e^{- λ x^{2}}}{x^{4} + 1} d x l e t W (z) = \frac{e^{- λ z^{2}}}{z^{4} + 1}$ $p o l e s o f W ? w e h a v e W (z) = \frac{e^{- λ x^{2}}}{(z^{2} - i) (z^{2} + i)}$ $= \frac{e^{- λ x^{2}}}{(z - \sqrt{i}) (z + \sqrt{i}) (z - \sqrt{- i}) (2 + \sqrt{- i})} = \frac{e^{- λ x^{2}}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $s o t h e p o l e s o f W a r e \bar{+} e^{\frac{i π}{4}} a n d \bar{+} e^{- \frac{i π}{4}}$ $r e s i d u s t b e o r e m g i v e$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, e^{\frac{i π}{4}}) + R e s (W, - e^{- \frac{i π}{4}})}$ $R e s (W, e^{\frac{i π}{4}}) = \frac{e^{- λ (i)}}{2 e^{\frac{i π}{4}} (2 i)} = \frac{1}{4 i} e^{- λ i - \frac{i π}{4}} = \frac{1}{4 i} e^{- (λ + \frac{π}{4}) i}$ $R e s (W, - e^{- \frac{i π}{4}}) = \frac{e^{- λ (- i)}}{(- 2 i) (- 2 e^{- \frac{i π}{4}})} = \frac{1}{4 i} e^{λ i + \frac{i π}{4}} = \frac{1}{4 i} e^{(λ + \frac{π}{4}) i} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {\frac{1}{4 i} e^{- (λ + \frac{π}{4}) i} + \frac{1}{4 i} e^{(λ + \frac{π}{4}) i}}$ $= \frac{π}{2} {2 R e (e^{(λ + \frac{π}{4}) i}) = π c o s (\frac{π}{4} + λ) \Rightarrow A_{λ} = \frac{π}{2} c o s (\frac{π}{4} + λ)$ $\int_{0}^{1} A_{λ} d λ = \frac{π}{2} \int_{0}^{1} c o s (λ + \frac{π}{4}) = {[s i n (λ + \frac{π}{4})]}_{0}^{1} = s i n (1 + \frac{π}{4}) - \frac{\sqrt{2}}{2}$
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