calculate f^′ (x) 1) f(x) =∫_0 ^∞ ((cos(xt))/(t^2 +x^2 ))dt 2)f(x)=∫


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Question Number 120025 by mathmax by abdo last updated on 28/Oct/20

$calculate f^{^{'}} (x)$ $1) f (x) = \int_{0}^{\infty} \frac{\cos (xt)}{t^{2} + x^{2}} dt$ $2) f (x) = \int_{0}^{\infty} \frac{\sin ({xt}^{2} + \sqrt{2})}{t^{2} + x^{2} + 3} dt$
	Answered by Bird last updated on 29/Oct/20
	$1)f(x)=_(t=xu) ∫_0 ^∞ ((cos(x^2 u))/(x^2 (u^2 +1)))xdu =(1/x)∫_0 ^∞ ((cos(x^2 u))/(u^2 +1))du (we tske x>0) =(1/(2x)) Re(∫_(−∞) ^∞ (e^(ix^2 u) /(u^2 +1))du) ∫_(−∞) ^∞ (e^(ix^2 z) /(z^2 +1))dz =2iπ Res(f,i) f(z)=(e^(ix^2 z) /((z−i)(z+i))) ⇒Res(f,i)=(e^(−x^2 ) /(2i)) ⇒∫_(−∞) ^(+∞) (e^(ix^2 z) /(z^2 +1))dz =2iπ.(e^(−x^2 ) /(2i)) =π e^(−x^2 ) ⇒f(x) =(π/(2x))e^(−x^2 ) ⇒ f^′ (x) =(π/2){−(1/x^2 )e^(−x^2 ) +(1/x)(−2x)e^(−x^2 ) } =(π/2){−(1/x^2 )e^(−x^2 ) −2e^(−x^2 ) } =−(π/2){(1/x^2 )+2}e^(−x^2 ) =−π×((1+2x^2 )/(2x^2 )) e^(−x^2 )$
	$1) f (x) =_{t = x u} \int_{0}^{\infty} \frac{c o s (x^{2} u)}{x^{2} (u^{2} + 1)} x d u$ $= \frac{1}{x} \int_{0}^{\infty} \frac{c o s (x^{2} u)}{u^{2} + 1} d u (w e t s k e x > 0)$ $= \frac{1}{2 x} R e (\int_{- \infty}^{\infty} \frac{e^{{i x}^{2} u}}{u^{2} + 1} d u)$ $\int_{- \infty}^{\infty} \frac{e^{{i x}^{2} z}}{z^{2} + 1} d z = 2 i π R e s (f, i)$ $f (z) = \frac{e^{{i x}^{2} z}}{(z - i) (z + i)} \Rightarrow R e s (f, i) = \frac{e^{- x^{2}}}{2 i}$ $\Rightarrow \int_{- \infty}^{+ \infty} \frac{e^{{i x}^{2} z}}{z^{2} + 1} d z = 2 i π . \frac{e^{- x^{2}}}{2 i}$ $= π e^{- x^{2}} \Rightarrow f (x) = \frac{π}{2 x} e^{- x^{2}} \Rightarrow$ $f^{^{'}} (x) = \frac{π}{2} {- \frac{1}{x^{2}} e^{- x^{2}} + \frac{1}{x} (- 2 x) e^{- x^{2}}}$ $= \frac{π}{2} {- \frac{1}{x^{2}} e^{- x^{2}} - 2 e^{- x^{2}}}$ $= - \frac{π}{2} {\frac{1}{x^{2}} + 2} e^{- x^{2}}$ $= - π \times \frac{1 + 2 x^{2}}{2 x^{2}} e^{- x^{2}}$
	Answered by Bird last updated on 29/Oct/20
	$2)f(x)=∫_0 ^∞ ((sin(xt^2 +(√2)))/(t^2 +x^2 +3))dt ⇒ f(x)=(1/2)∫_(−∞) ^(+∞) ((sin(xt^2 +(√2)))/(t^2 +x^2 +3))dt =(1/2)Im(∫_(−∞) ^(+∞) (e^(i(xt^2 +(√2))) /(t^2 +x^2 +3))dt) g(z)=(e^(i(xz^2 +(√2))) /(z^2 +x^2 +3)) ⇒ g(z) =(e^(i(xz^2 +(√2))) /((z−i(√(x^2 +3)))(z+i(√(x^2 +3))))) ∫_(−∞) ^(+∞) g(x)dz =2iπ Res(g,i(√(x^2 +3))) =2iπ×(e^(i(x(−x^2 −3)+(√2))) /(2i(√(x^2 +3)))) =(π/( (√(x^2 +3)))).e^(i(−x^3 −3x+(√2))) =(π/( (√(x^2 +3)))){cos(−x^3 −3x+(√2)) +isin(−x^3 −3x+(√2))} ⇒ f(x)=(π/(2(√(x^2 +3))))sin(−x^3 −3x+(√2)) =−(π/(2(√(x^2 +3))))sin(x^3 +3x−(√2)) f(x)=−(π/2)(x^2 +3)^(−(1/2)) sin(x^3 +3x−(√2)) ⇒ f^′ (x)=−(π/2){−x(x^2 +3)^(−(3/2)) sin(x^3 +3x−(√2))+(x^2 +3)^(−(1/2)) (3x^2 +3)cos(x^3 +3x−(√2))} f^′ (x) =−(π/2){−(x/((x^2 +3)(√(x^2 +3))))sin(x^3 +3x−(√2))+((3x^2 +3)/( (√(x^2 +3))))cos(x^3 +3x+(√2))}$
	$2) f (x) = \int_{0}^{\infty} \frac{s i n ({x t}^{2} + \sqrt{2})}{t^{2} + x^{2} + 3} d t \Rightarrow$ $f (x) = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{s i n ({x t}^{2} + \sqrt{2})}{t^{2} + x^{2} + 3} d t$ $= \frac{1}{2} I m (\int_{- \infty}^{+ \infty} \frac{e^{i ({x t}^{2} + \sqrt{2})}}{t^{2} + x^{2} + 3} d t)$ $g (z) = \frac{e^{i ({x z}^{2} + \sqrt{2})}}{z^{2} + x^{2} + 3} \Rightarrow$ $g (z) = \frac{e^{i ({x z}^{2} + \sqrt{2})}}{(z - i \sqrt{x^{2} + 3}) (z + i \sqrt{x^{2} + 3})}$ $\int_{- \infty}^{+ \infty} g (x) d z = 2 i π R e s (g, i \sqrt{x^{2} + 3})$ $= 2 i π \times \frac{e^{i (x (- x^{2} - 3) + \sqrt{2})}}{2 i \sqrt{x^{2} + 3}}$ $= \frac{π}{\sqrt{x^{2} + 3}} . e^{i (- x^{3} - 3 x + \sqrt{2})}$ $= \frac{π}{\sqrt{x^{2} + 3}} {c o s (- x^{3} - 3 x + \sqrt{2})$ $+ i s i n (- x^{3} - 3 x + \sqrt{2})} \Rightarrow$ $f (x) = \frac{π}{2 \sqrt{x^{2} + 3}} s i n (- x^{3} - 3 x + \sqrt{2})$ $= - \frac{π}{2 \sqrt{x^{2} + 3}} s i n (x^{3} + 3 x - \sqrt{2})$ $f (x) = - \frac{π}{2} {(x^{2} + 3)}^{- \frac{1}{2}} s i n (x^{3} + 3 x - \sqrt{2}) \Rightarrow$ $f^{^{'}} (x) = - \frac{π}{2} {- x {(x^{2} + 3)}^{- \frac{3}{2}} s i n (x^{3} + 3 x - \sqrt{2}) + {(x^{2} + 3)}^{- \frac{1}{2}} (3 x^{2} + 3) c o s (x^{3} + 3 x - \sqrt{2})}$ $f^{^{'}} (x) = - \frac{π}{2} {- \frac{x}{(x^{2} + 3) \sqrt{x^{2} + 3}} s i n (x^{3} + 3 x - \sqrt{2}) + \frac{3 x^{2} + 3}{\sqrt{x^{2} + 3}} c o s (x^{3} + 3 x + \sqrt{2})}$
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