let f(a) =∫_0 ^∞ (dx/((x^2 +1)(x^2 +a))) with a>0 1) determine a explicit form of f(a) 2) calculate g(a) =∫_0 ^∞ (dx/((x^2 +1)(x^2 +a)^2 )) 3)give f^((n)) (a) at form of integral 4)calculate ∫_0 ^∞ (dx/((x^2 +1)(x^2 +3)^2 )) and ∫


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Question Number 67674 by Abdo msup. last updated on 30/Aug/19

$l e t f (a) = \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (x^{2} + a)} w i t h a > 0$ $1) d e t e r m i n e a e x p l i c i t f o r m o f f (a)$ $2) c a l c u l a t e g (a) = \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) {(x^{2} + a)}^{2}}$ $3) g i v e f^{(n)} (a) a t f o r m o f i n t e g r a l$ $4) c a l c u l a t e \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) {(x^{2} + 3)}^{2}} a n d$ $\int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{3}}$
Commented by mathmax by abdo last updated on 30/Aug/19
$1)f(a) =∫_0 ^∞ (dx/((x^2 +1)(x^2 +a))) ⇒2f(a) =∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +a))) let W(z) =(1/((z^2 +1)(z^2 +a))) ⇒W(z) =(1/((z−i)(z+i)(z−i(√a))(z+i(√a)))) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,i) +Res(W,i(√a))} Res(W,i) =lim_(z→i) (z−i)W(z) =(1/(2i(a−1))) ( a≠1) Res(W,i(√a)) =lim_(z→i(√a)) (z−i(√a))W(z) =(1/(2i(√a)(1−a))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{(1/(2i(a−1))) +(1/(2i(√a)(1−a)))} =(π/(a−1)) +(π/((√a)(1−a))) =(π/(a−1))−(π/((a−1)(√a))) =(π/(a−1))(1−(1/(√a))) =((π((√a)−1))/((√a)(a−1))) =(π/((√a)((√a)+1))) ⇒ f(a) =(π/(2(√a)((√a)+1))) =(π/(2(a+(√a)))) another way f(a) =(1/(a−1))∫_0 ^∞ {(1/(x^2 +1))−(1/(x^2 +a))}dx =(1/(a−1))×(π/2)−(1/(a−1)) ∫_0 ^∞ (dx/(x^2 +a)) changement x =(√a)t give ∫_0 ^∞ (dx/(x^2 +a)) =∫_0 ^∞ (((√a)dt)/(a(t^2 +1))) =(1/(√a))×(π/2) ⇒f(a)=(π/(2(a−1)))−(π/(2(√a)(a−1)))$
$1) f (a) = \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (x^{2} + a)} \Rightarrow 2 f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) (x^{2} + a)}$ $l e t W (z) = \frac{1}{(z^{2} + 1) (z^{2} + a)} \Rightarrow W (z) = \frac{1}{(z - i) (z + i) (z - i \sqrt{a}) (z + i \sqrt{a})}$ $r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, i) + R e s (W, i \sqrt{a})}$ $R e s (W, i) = {l i m}_{z \to i} (z - i) W (z) = \frac{1}{2 i (a - 1)} (a \neq 1)$ $R e s (W, i \sqrt{a}) = {l i m}_{z \to i \sqrt{a}} (z - i \sqrt{a}) W (z) = \frac{1}{2 i \sqrt{a} (1 - a)} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {\frac{1}{2 i (a - 1)} + \frac{1}{2 i \sqrt{a} (1 - a)}}$ $= \frac{π}{a - 1} + \frac{π}{\sqrt{a} (1 - a)} = \frac{π}{a - 1} - \frac{π}{(a - 1) \sqrt{a}} = \frac{π}{a - 1} (1 - \frac{1}{\sqrt{a}})$ $= \frac{π (\sqrt{a} - 1)}{\sqrt{a} (a - 1)} = \frac{π}{\sqrt{a} (\sqrt{a} + 1)} \Rightarrow f (a) = \frac{π}{2 \sqrt{a} (\sqrt{a} + 1)} = \frac{π}{2 (a + \sqrt{a})}$ $a n o t h e r w a y f (a) = \frac{1}{a - 1} \int_{0}^{\infty} {\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + a}} d x$ $= \frac{1}{a - 1} \times \frac{π}{2} - \frac{1}{a - 1} \int_{0}^{\infty} \frac{d x}{x^{2} + a} c h a n g e m e n t x = \sqrt{a} t g i v e$ $\int_{0}^{\infty} \frac{d x}{x^{2} + a} = \int_{0}^{\infty} \frac{\sqrt{a} d t}{a (t^{2} + 1)} = \frac{1}{\sqrt{a}} \times \frac{π}{2} \Rightarrow f (a) = \frac{π}{2 (a - 1)} - \frac{π}{2 \sqrt{a} (a - 1)}$
Commented by mathmax by abdo last updated on 30/Aug/19

$2) w e h a v e f^{^{'}} (a) = - \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) {(x^{2} + a)}^{2}} = - g (a) \Rightarrow$ $g (a) = - f^{^{'}} (a) b u t f (a) = \frac{π}{2 (a + \sqrt{a})} \Rightarrow f^{^{'}} (a) = - \frac{π}{2} \times \frac{{(a + \sqrt{a})}^{^{'}}}{{(a + \sqrt{a})}^{2}}$ $= - \frac{π}{2} \times \frac{1 + \frac{1}{2 \sqrt{a}}}{{(a + \sqrt{a})}^{2}} = - \frac{π}{2} \frac{2 \sqrt{a} + 1}{2 \sqrt{a} {(a + \sqrt{a})}^{2}} = - \frac{π}{4} \times \frac{2 \sqrt{a} + 1}{\sqrt{a} {(a + \sqrt{a})}^{2}} \Rightarrow$ $g (a) = \frac{π (2 \sqrt{a} + 1)}{4 \sqrt{a} {(a + \sqrt{a})}^{2}}$
Commented by mathmax by abdo last updated on 30/Aug/19
$4)∫_0 ^∞ (dx/((x^2 +1)(x^2 +3)^2 )) =g(3) =((π(2(√3)+1))/(4(√3)(3+(√3))^2 )) let find I =∫_0 ^∞ (dx/((x^2 +1)^3 )) ⇒ 2I =∫_(−∞) ^(+∞) (dx/((x^2 +1)^3 )) let w(z)=(1/((z^2 +1)^3 )) ⇒w(z) =(1/((z−i)^3 (z+i)^3 )) the poles of w are i and −i(triples) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,i) Res(w,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 w(z)}^((2)) =lim_(z→i) (1/2){(z+i)^(−3) }^((2)) =lim_(z→i) (1/2){−3(z+i)^(−4) }^((1)) =lim_(z→i) ((−3)/2){−4(z+i)^(−5) } =6 (2i)^(−5) =(6/(2^5 i^5 )) =(6/(32i)) =((3.2)/(16.2i)) =(3/(16i)) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ×(3/(16i)) =((3π)/8) =2I ⇒ I =((3π)/(16))$
$4) \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) {(x^{2} + 3)}^{2}} = g (3) = \frac{π (2 \sqrt{3} + 1)}{4 \sqrt{3} {(3 + \sqrt{3})}^{2}}$ $l e t f i n d I = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{3}} \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 1)}^{3}} l e t w (z) = \frac{1}{{(z^{2} + 1)}^{3}}$ $\Rightarrow w (z) = \frac{1}{{(z - i)}^{3} {(z + i)}^{3}} t h e p o l e s o f w a r e i a n d - i (t r i p l e s)$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} w (z) d z = 2 i π R e s (w, i)$ $R e s (w, i) = {l i m}_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} w (z)}^{(2)}$ $= {l i m}_{z \to i} \frac{1}{2} {{(z + i)}^{- 3}}^{(2)} = {l i m}_{z \to i} \frac{1}{2} {- 3 {(z + i)}^{- 4}}^{(1)}$ $= {l i m}_{z \to i} \frac{- 3}{2} {- 4 {(z + i)}^{- 5}} = 6 {(2 i)}^{- 5} = \frac{6}{2^{5} i^{5}} = \frac{6}{32 i} = \frac{3.2}{16.2 i} = \frac{3}{16 i}$ $\Rightarrow \int_{- \infty}^{+ \infty} w (z) d z = 2 i π \times \frac{3}{16 i} = \frac{3 π}{8} = 2 I \Rightarrow I = \frac{3 π}{16}$
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