1) let f(x) = ∫_0 ^∞ (dt/(1+x^2 t^4 )) with x >0 find a simple form of f(x) 2) calculate ∫_0 ^(+∞) (dt/(1+t^4 )) 3) calculate ∫


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Question Number 39030 by maxmathsup by imad last updated on 01/Jul/18

$1) l e t f (x) = \int_{0}^{\infty} \frac{d t}{1 + x^{2} t^{4}} w i t h x > 0$ $f i n d a s i m p l e f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{+ \infty} \frac{d t}{1 + t^{4}}$ $3) c a l c u l a t e \int_{0}^{\infty} \frac{d t}{1 + 3 t^{4}}$
Commented bymaxmathsup by imad last updated on 02/Jul/18
$1)we have 2f(x)= ∫_(−∞) ^(+∞) (dt/(1+x^2 t^4 )) let ϕ(z)= (1/(x^2 z^4 +1)) ϕ(z)= (1/(x^2 (z^4 +(1/x^2 )))) = (1/(x^2 (z^2 −(i/x))(z^2 +(i/x)))) = (1/(x^2 ( z −(1/(√x))e^((iπ)/4) )(z+(1/(√x))e^((iπ)/4) )(z−(1/(√x))e^(−((iπ)/4)) )(z+(1/(√x))e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,(1/(√x))e^((iπ)/4) ) +Res(ϕ,−(1/(√x))e^(−((iπ)/4)) ) but Res(ϕ,z_i ) = (1/(4x^2 z_i ^3 )) wehave z_i ^4 =((−1)/x^2 ) ⇒Res(ϕ,z_i )= (z_i /(4x^2 (−(1/x^2 )))) =−(z_i /4) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {−(1/(4(√x))) e^((iπ)/4) +(1/(4(√x))) e^(−((iπ)/4)) } =−((iπ)/(2(√x))){ e^((iπ)/4) −e^(−((iπ)/4)) } =−((iπ)/(2(√x))){2i ((√2)/2)} =((π(√2))/(√x)) ⇒ f(x) = ((π(√2))/(2(√x))) . 2) ∫_0 ^∞ (dt/(1+t^4 )) = f(1) = ((π(√2))/2) 3) ∫_0 ^∞ (dt/(1+3t^4 )) =f((√3)) = ((π(√2))/(2(^4 (√3)))) .$
$1) w e h a v e 2 f (x) = \int_{- \infty}^{+ \infty} \frac{d t}{1 + x^{2} t^{4}} l e t φ (z) = \frac{1}{x^{2} z^{4} + 1}$ $φ (z) = \frac{1}{x^{2} (z^{4} + \frac{1}{x^{2}})} = \frac{1}{x^{2} (z^{2} - \frac{i}{x}) (z^{2} + \frac{i}{x})}$ $= \frac{1}{x^{2} (z - \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) + R e s (φ, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) b u t$ $R e s (φ, z_{i}) = \frac{1}{4 x^{2} z_{i}^{3}} w e h a v e z_{i}^{4} = \frac{- 1}{x^{2}} \Rightarrow R e s (φ, z_{i}) = \frac{z_{i}}{4 x^{2} (- \frac{1}{x^{2}})} = - \frac{z_{i}}{4}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {- \frac{1}{4 \sqrt{x}} e^{\frac{i π}{4}} + \frac{1}{4 \sqrt{x}} e^{- \frac{i π}{4}}}$ $= - \frac{i π}{2 \sqrt{x}} {e^{\frac{i π}{4}} - e^{- \frac{i π}{4}}} = - \frac{i π}{2 \sqrt{x}} {2 i \frac{\sqrt{2}}{2}} = \frac{π \sqrt{2}}{\sqrt{x}} \Rightarrow$ $f (x) = \frac{π \sqrt{2}}{2 \sqrt{x}} .$ $2) \int_{0}^{\infty} \frac{d t}{1 + t^{4}} = f (1) = \frac{π \sqrt{2}}{2}$ $3) \int_{0}^{\infty} \frac{d t}{1 + 3 t^{4}} = f (\sqrt{3}) = \frac{π \sqrt{2}}{2 (^{4} \sqrt{3})} .$
Commented byajfour last updated on 02/Jul/18

$S i r, h a v e y o u f o u n d 2 f (x) o r f (x);$ $p l e a s e c h e c k . .$
Commented bymaxmathsup by imad last updated on 02/Jul/18

$e r r o r f r o m l i n e 7 \int_{- \infty}^{+ \infty} φ (z) d z = \frac{π \sqrt{2}}{2 \sqrt{x}} \Rightarrow f (x) = \frac{π \sqrt{2}}{4 \sqrt{x}}$ $2) \int_{0}^{\infty} \frac{d t}{1 + t^{4}} = f (1) = \frac{π \sqrt{2}}{4}$ $3) \int_{0}^{\infty} \frac{d t}{1 + 3 t^{4}} = f (\sqrt{3}) = \frac{π \sqrt{2}}{4 (^{4} \sqrt{3})} .$
	Answered by behi83417@gmail.com last updated on 02/Jul/18

	$(1 - {i x t}^{2}) (1 + {i x t}^{2}) = 0$ $\Rightarrow t^{2} = \frac{1}{i x}, - \frac{1}{i x} \Rightarrow t = \pm \frac{1}{\sqrt{i x}}, \frac{\pm i}{\sqrt{i x}}, \frac{1}{\sqrt{i x}} = m$ $m = \frac{1}{\sqrt{i x}} = \frac{\sqrt{i}}{i \sqrt{x}} = - \frac{{i e}^{\frac{i π}{4}}}{\sqrt{x}} = - \frac{i (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2})}{\sqrt{x}} = \frac{1 - i}{\sqrt{2 x}}$ $I = \int \frac{d t}{(t - m) (t + m) (t - i m) (t + i m)} =$ $\frac{1}{(t - m) (t + m) (t - i m) (t + i m)} =$ $= \frac{a}{t - m} + \frac{b}{t + m} + \frac{c}{t - i m} + \frac{d}{t + i m}$ $a = \frac{1}{2 m . m^{2} (1 - i^{2})} = \frac{1}{4 m^{3}}$ $b = \frac{1}{- 2 m . m^{2} (1 - i^{2})} = \frac{- 1}{4 m^{3}}$ $c = \frac{1}{m (i - 1) m (i + 1) (2 i m)} = \frac{i}{4 m^{3}}$ $d = \frac{1}{- m (i + 1) m (i - 1) (2 i m)} = \frac{- i}{4 m^{3}}$ $I = \frac{1}{4 m^{3}} \int [\frac{1}{t - m} - \frac{1}{t + m} + \frac{i}{t - i m} - \frac{i}{t + i m}] d t$ $= \frac{1}{4 m^{3}} [l n \frac{t - m}{t + m} + i . l n \frac{t - i m}{t + i m}] =$ $= \frac{1}{4 m^{3}} [l n \frac{t^{2} - 2 m t + m^{2}}{t^{2} - m^{2}} + i . l n \frac{t^{2} - 2 m i t - m^{2}}{t^{2} + m^{2}}] =$ $= \frac{1}{4 . \frac{{(1 - i)}^{3}}{2 x \sqrt{2 x}}} [l n \frac{t^{2} - 2 t \frac{1 - i}{\sqrt{2 x}} + \frac{i}{x}}{t^{2} - \frac{i}{x}} + i . l n \frac{t^{2} - 2 t \frac{1 + i}{\sqrt{2 x}} - \frac{i}{x}}{t^{2} + \frac{i}{x}}] =$ $= - \frac{x \sqrt{2 x}}{4 (2 i + 1)} [l n (\frac{x \sqrt{2 x} . t^{2} - 2 (1 - i) x t + i . \sqrt{2 x}}{{x t}^{2} - i}) +$ $+ i . l n (\frac{x \sqrt{2 x} . t^{2} - 2 (1 + i) x t - i \sqrt{2 x}}{{x t}^{2} + i})] + c o n s t .$ $F (x) = F (\infty) - F (0) = - \frac{x \sqrt{2 x}}{4 (2 i + 1)} [l n \sqrt{2 x} + i . l n \sqrt{2 x} -$ $- l n (- \sqrt{2 x}) - i . l n (- \sqrt{2 x})] =$ $= \frac{- x \sqrt{2 x}}{4 (2 i + 1)} [l n (- 1) + i l n (- 1)] =$ $= \frac{- x \sqrt{2 x}}{4 (- 4 - 1)} (2 i - 1) (1 + i) . i π = \frac{- π . x \sqrt{2 x}}{20} (3 i + 1)$ $F (1) = \frac{- π \sqrt{2}}{20} (3 i + 1)$ $F (\sqrt{3}) = \frac{- π \sqrt{6 \sqrt{3}}}{20} (3 i + 1) .$
	Answered by ajfour last updated on 02/Jul/18

	$f (x) = \frac{1}{x^{2}} \int_{0}^{\infty} \frac{d t}{t^{4} + {(\frac{1}{x})}^{2}}$ $l e t \frac{1}{x} = y$ $f (x) = y^{2} \int \frac{\frac{d t}{t^{2}}}{t^{2} + \frac{y^{2}}{t^{2}}} = \frac{y}{2} \int \frac{[(1 + \frac{y}{t^{2}}) - (1 - \frac{y}{t^{2}})] d t}{{(t - \frac{y}{t})}^{2} + 2 y}$ $= \frac{y}{2} [\int \frac{d u}{u^{2} + {(\sqrt{2 y})}^{2}} - \int^{} \frac{d v}{v^{2} - {(\sqrt{2 y})}^{2}}]$ $= \frac{y}{2 \sqrt{2 y}} \tan^{- 1} (\frac{u}{\sqrt{2 y}}) - \frac{y}{4 \sqrt{2 y}} \ln ∣ \frac{v - \sqrt{2 y}}{v + \sqrt{2 y}} ∣ + c$ $= \frac{1}{2 \sqrt{2 x}} \tan^{- 1} (\frac{t - \frac{1}{t x}}{\sqrt{\frac{2}{x}}}) ∣_{0}^{\infty}$ $- \frac{1}{4 \sqrt{2 x}} \ln ∣ \frac{t + \frac{1}{t x} - \sqrt{\frac{2}{x}}}{t + \frac{1}{t x} + \sqrt{\frac{2}{x}}} ∣_{0}^{\infty}$ $\Rightarrow f (x) = \frac{π}{2 \sqrt{2 x}}$
	Commented byajfour last updated on 02/Jul/18

	$w i s h i c o u l d p l a y w i t h t h e r e s i d u e$ $t h e o r e m e v e n . .$
	Commented bymaxmathsup by imad last updated on 02/Jul/18

	$s i r A j f o u r i t s e e m s t h a t y o u h a v e p l a y e d a g a m e w i t h t h i s i n t e g r a l . . .$
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