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Question Number 85160 by mathmax by abdo last updated on 19/Mar/20

1) f i n d f (a) = \int_{0}^{\infty} \frac{d x}{x^{4} + a} w i t h a > 0

2) f i n d g (a) = \int_{0}^{\infty} \frac{d x}{{(x^{4} + a)}^{2}}

3) f i n d v a l u e o f i n t e g r a l s \int_{0}^{\infty} \frac{d x}{x^{4} + 1}, \int_{0}^{\infty} \frac{d x}{2 x^{4} + 8}

\int_{0}^{\infty} \frac{d x}{{(x^{4} + 1)}^{2}} a n d \int_{0}^{\infty} \frac{d x}{{(2 x^{4} + 8)}^{2}}

Commented by mathmax by abdo last updated on 19/Mar/20

1) f (a) = \int_{0}^{\infty} \frac{d x}{x^{4} + a} =_{x =^{4} \sqrt{a} t} \int_{0}^{\infty} \frac{(^{4} \sqrt{a}) d t}{a (1 + t^{4})} = \frac{1}{a^{1 - \frac{1}{4}}} \int_{0}^{\infty} \frac{d t}{t^{4} + 1}

= \frac{1}{a^{\frac{3}{4}}} \int_{0}^{\infty} \frac{d t}{t^{4} + 1} c h a n g e m e n t t = u^{\frac{1}{4}} g i v e

\int_{0}^{\infty} \frac{d t}{t^{4} + 1} = \frac{1}{4} \int_{0}^{\infty} \frac{u^{\frac{1}{4} - 1}}{1 + u} d u = \frac{1}{4} \times \frac{π}{s i n (\frac{π}{4})} = \frac{π}{4 \times \frac{\sqrt{2}}{2}} = \frac{π}{2 \sqrt{2}} \Rightarrow

f (a) = \frac{π}{2 \sqrt{2}} a^{- \frac{3}{4}}

2) w e h a v e f^{^{'}} (a) = - \int_{0}^{\infty} \frac{d x}{{(x^{4} + a)}^{2}} = - g (a) \Rightarrow g (a) = - f^{^{'}} (a)

g (a) = - \frac{π}{2 \sqrt{2}} \times (\frac{- 3}{4}) a^{\frac{- 3}{4} - 1} = \frac{3 π}{8 \sqrt{2}} a^{- \frac{7}{4}}

Commented by mathmax by abdo last updated on 19/Mar/20

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

\int_{0}^{\infty} \frac{d x}{2 x^{4} + 8} = \frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4} + 4} = \frac{1}{2} f (4) = \frac{π}{4 \sqrt{2}} {(4)}^{- \frac{3}{4}}

= \frac{π}{4 \sqrt{2} 4^{\frac{3}{4}}} = \frac{π}{4^{\frac{7}{4}} \sqrt{2}} = \frac{π}{2^{\frac{7}{2}} \sqrt{2}} = \frac{π}{8 \times 2} = \frac{π}{16}

\int_{0}^{\infty} \frac{d x}{{(x^{4} + 1)}^{2}} = g (1) = \frac{3 π \sqrt{2}}{16}

Answered by mind is power last updated on 19/Mar/20

x = \sqrt{\sqrt{a} t g (t)} \Rightarrow

f (a) = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sqrt{a}} (1 + {t g}^{2} (t))}{2 \sqrt{t g (t)}} . \frac{d t}{a (1 + {t g}^{2} (t))}

= \frac{1}{2 a^{\frac{3}{4}}} . \int_{0}^{\frac{π}{2}} {c o s}^{\frac{1}{2}} (t) {s i n}^{\frac{- 1}{2}} (t) d t

= \frac{1}{2 a^{\frac{3}{4}}} . \frac{β (\frac{3}{4}, \frac{1}{4})}{2} = \frac{1}{4 a^{\frac{3}{4}}} . \frac{Γ (\frac{1}{4}) Γ (\frac{3}{4})}{Γ (1)} = \frac{π}{4 a^{\frac{3}{4}} s i n (\frac{π}{4})}

= \frac{π \sqrt{2}}{4 a^{\frac{3}{4}}} = f (a)

g (a) = - f^{'} (a) = \frac{3 π \sqrt{2}}{16 a^{\frac{7}{4}}} .

\int_{0}^{+ \infty} \frac{d x}{x^{4} + 1} = f (1) = \frac{π \sqrt{2}}{4} = \frac{π}{2 \sqrt{2}}

\int_{0}^{+ \infty} \frac{d x}{2 x^{4} + 8} = \frac{f (4)}{2} = \frac{π \sqrt{2}}{8 (2^{\frac{3}{2}})} = \frac{π}{16}

\int \frac{d x}{{(1 + x^{4})}^{2}} = g (1) = - f^{'} (1) = \frac{3 π \sqrt{2}}{16}

\int_{0}^{+ \infty} \frac{d x}{{(2 x^{4} + 8)}^{2}} = \frac{1}{4} g (4) = - \frac{f^{'} (4)}{4} = \frac{3 π \sqrt{2}}{64 . 4^{\frac{7}{4}}}

= \frac{3 π}{512}

Commented by mathmax by abdo last updated on 19/Mar/20

t h a n k y o u s i r .

Commented by mind is power last updated on 19/Mar/20

w i t h e p l e a s u r