find-0-x-2-3-x-4-1-2-dx-

Question Number 35061 by math khazana by abdo last updated on 14/May/18

f i n d \int_{0}^{\infty} \frac{x^{2} + 3}{{(x^{4} + 1)}^{2}} d x

Commented by math khazana by abdo last updated on 15/May/18

l e t p u t I = \int_{0}^{\infty} \frac{x^{2} + 3}{{(x^{4} + 1)}^{2}} d x

2 I = \int_{- \infty}^{+ \infty} \frac{x^{2} + 3}{{(x^{4} + 1)}^{2}} d x l e t c o n s i d e r t h e c o m p l e x

f u n c t i o n f (z) = \frac{z^{2} + 3}{{(z^{4} + 1)}^{2}}

f (z) = \frac{z^{2} + 3}{{(z^{2} - i)}^{2} {(z^{2} + i)}^{2}} = \frac{z^{2} + 3}{{(z - \sqrt{i})}^{2} {(z + \sqrt{i})}^{2} {(z - \sqrt{- i})}^{2} {(z + \sqrt{- i})}^{2}}

= \frac{z^{2} + 3}{{(z - e^{i \frac{π}{4}})}^{2} {(z + e^{i \frac{π}{4}})}^{2} {(z - e^{- i \frac{π}{4}})}^{2} {(z + e^{- i \frac{π}{4}})}^{2}}

a l l p o l e s o f f a r e d o u b l e s

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π {R e s (f, e^{i \frac{π}{4}}) + R e s (f, - e^{- i \frac{π}{4}})}

R e s (f, e^{i \frac{π}{4}}) = {l i m}_{z \to e^{i \frac{π}{4}}} \frac{1}{(2 - 1)!} {{(z - e^{i \frac{π}{4}})}^{2} f (z)}^{^{'}}

= {l i m}_{z \to e^{i \frac{π}{4}}} {(\frac{z^{2} + 3}{{(z + \sqrt{i})}^{2} {(z^{2} + i)}^{2}})}^{^{'}}

= {l i m}_{z \to e^{i \frac{π}{4}}} \frac{2 z {(z + \sqrt{i})}^{2} {(z^{2} + i)}^{2} - (z^{2} + 3) (2 (z + \sqrt{i}) {(z^{2} + i)}^{2} + 4 z (z^{2} + i) {(z + \sqrt{i})}^{2})}{{(z + \sqrt{i})}^{4} {(z^{2} + i)}^{4}}

= {l i m}_{z \to e^{i \frac{π}{4}}} \frac{2 z (z + \sqrt{i}) (z^{2} + i) - 2 (z^{2} + 3) (z^{2} + i + 4 z (z + \sqrt{i}))}{{(z + \sqrt{i})}^{3} {(z^{2} + i)}^{3}}

= \frac{2 e^{i \frac{π}{4}} 2 e^{i \frac{π}{4}} (2 i) - 2 (i + 3) (2 i + 4 e^{i \frac{π}{4}} (2 e^{i \frac{π}{4}})}{{(2 e^{i \frac{π}{4}})}^{2} {(2 i)}^{3}}

= \frac{- 8 - 2 (i + 3) (10 i)}{- 8 i . 4 i} = \frac{8 + 20 i (i + 3)}{- 32}

= - \frac{8 - 20 + 60 i}{32} = - \frac{- 12 + 60 i}{32} = \frac{- 3 + 15 i}{8}

f o r R e s (f, - e^{- i \frac{π}{4}}) w e f o l o w t h e s a m e m e t h o d \dots

b e c o n t i n u e d \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 16/May/18

= \int_{0}^{\infty} \frac{x^{2} + 3}{x^{4} {(x^{2} + \frac{1}{x^{2}})}^{2}}

= \int_{0}^{\infty} \frac{\frac{1}{x^{2}}}{{(x^{2} + \frac{1}{x^{2}})}^{2}} + \int_{0}^{\infty} \frac{\frac{3}{x^{4}}}{{(x^{2} + \frac{1}{x^{2}})}^{2}}

I_{1} = \frac{1}{2} \int_{0}^{\infty} \frac{(1 + \frac{1}{x^{2}}) - (1 - \frac{1}{x^{2}})}{{(x^{2} + \frac{1}{x^{2}})}^{2}} d x

= \frac{1}{2} \int_{0}^{\infty} \frac{(1 + \frac{1}{x^{2}}) d x}{{{(x - \frac{1}{x})}^{2} + 2}^{2}} - \frac{1}{2} \int_{0}^{\infty} \frac{(1 - \frac{1}{x^{2}}) d x}{{{(x + \frac{1}{x})}^{2} - 2}^{2}}

c o n t d

z_{1} = x - \frac{1}{x} z_{2} = x + \frac{1}{x}

\frac{1}{2} \int_{- \infty}^{\infty} \frac{{d z}_{1}}{{(z_{1}^{2} + 2)}^{2}} - \frac{1}{2} \int_{\infty}^{\infty} \frac{{d z}_{2}}{{(z_{2}^{2} - 2)}^{2}} \to i t s v a l u e 0

z_{1} = \sqrt{2} t a n θ

\frac{1}{2} \int_{- Π / 2}^{Π / 2} \frac{\sqrt{2} {s e c}^{2} θ}{2^{2} {s e c}^{4} θ} d θ

\frac{\sqrt{2}}{8} \int_{- Π / 2}^{Π / 2} \frac{1 + c o s 2 θ}{2} d θ

= \frac{\sqrt{2}}{16} (Π) + \frac{\sqrt{2}}{16} \times ∣ \frac{s i n 2 θ}{2} ∣_{- Π / 2}^{Π / 2}

= \frac{\sqrt{2} (Π)}{16} + \frac{\sqrt{2}}{32} {s i n Π - s i n (- Π)}

= \sqrt{2} \frac{Π}{16}

p l s c h e c k

Answered by MJS last updated on 15/May/18

\int \frac{x^{2} + 3}{{(x^{4} + 1)}^{2}} d x = i

[\begin{matrix} {Ostrogradski}^{'} s Method \\ \int \frac{p (x)}{q (x)} d x = \frac{p_{1} (x)}{q_{1} (x)} + \int \frac{p_{2} (x)}{q_{2} (x)} d x \\ q_{1} (x) = \gcd (q (x), q^{'} (x)) \\ q_{2} (x) = \frac{q (x)}{q_{1} (x)} \\ p (x) = x^{2} + 3; q (x) = {(x^{4} + 1)}^{2} \\ q^{'} (x) = 8 x^{3} (x^{4} + 1); q_{1} (x) = q_{2} (x) = x^{4} + 1 \\ now we calculate the factors of \\ p_{1} (x) and p_{2} (x) with degree (p_{n}) < degree (q_{n}) \\ using this equation (comparing factors) \\ \frac{p (x)}{q (x)} = \frac{d}{d x} [\frac{p_{1} (x)}{q_{1} (x)}] + \frac{p_{2} (x)}{q_{2} (x)} \\ p_{1} (x) = a_{1} x^{3} + b_{1} x^{2} + c_{1} x + d_{1} \\ p_{2} (x) = a_{2} x^{3} + b_{2} x^{2} + c_{2} x + d_{2} \\ \Rightarrow p_{1} (x) = \frac{1}{4} (x^{3} + 3 x); p_{2} (x) = \frac{1}{4} (x^{2} + 9) \end{matrix}]

= \frac{x (x^{2} + 3)}{4 (x^{4} + 1)} + \frac{1}{4} \int \frac{x^{2} + 9}{x^{4} + 1} d x =

\int \frac{x^{2} + 9}{x^{4} + 1} d x = \int \frac{x^{2} + 9}{(x^{2} + \sqrt{2} x + 1) (x^{2} - \sqrt{2} x + 1)} d x =

= \frac{\sqrt{2}}{4} \int (\frac{8 x + 9 \sqrt{2}}{x^{2} + \sqrt{2} x + 1} - \frac{8 x - 9 \sqrt{2}}{x^{2} - \sqrt{2} x + 1}) d x =

= \frac{\sqrt{2}}{4} \int (4 \frac{2 x + \sqrt{2}}{x^{2} + \sqrt{2} x + 1} + 5 \sqrt{2} \frac{1}{x^{2} + \sqrt{2} x + 1} - 4 \frac{2 x - \sqrt{2}}{x^{2} - \sqrt{2} x + 1} + 5 \sqrt{2} \frac{1}{x^{2} - \sqrt{2} x + 1}) d x =

[\begin{matrix} the 1^{st} and 3^{rd} terms are of the form \\ F (x) = \int a \frac{f^{'} (x)}{f (x)} d x \Rightarrow F (x) = a \ln (f (x)) \\ the 2^{nd} and 4^{th} terms can be solved using \\ u = \sqrt{2} x \pm 1 \to d x = \frac{d u}{\sqrt{2}} and lead to \\ F (x) = \sqrt{2} \int \frac{d u}{u^{2} + 2} \Rightarrow F (x) = \sqrt{2} \arctan (u) \end{matrix}]

= \sqrt{2} (\ln (x^{2} + \sqrt{2} x + 1) - \ln (x^{2} - \sqrt{2} x + 1)) + \frac{5 \sqrt{2}}{2} (\arctan (\sqrt{2} x + 1) + \arctan (\sqrt{2} x - 1))

= \frac{x (x^{2} + 3)}{4 (x^{4} + 1)} + \frac{\sqrt{2}}{4} \ln (\frac{∣ x^{2} + \sqrt{2} x + 1 ∣}{∣ x^{2} - \sqrt{2} x + 1 ∣}) + \frac{5 \sqrt{2}}{8} (\arctan (\sqrt{2} x + 1) + \arctan (\sqrt{2} x - 1)) + C

\underset{0}{\int^{\infty}} \frac{x^{2} + 3}{{(x^{4} + 1)}^{2}} d x = \frac{5 \sqrt{2}}{8} π

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