find-the-value-of-f-a-0-x-2-1-x-4-a-4-2-dx-2-calculate-0-x-2-1-x-4-1-2-dx-

Question Number 37602 by prof Abdo imad last updated on 15/Jun/18

f i n d t h e v a l u e o f

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

2) c a l c u l a t e \int_{0}^{\infty} \frac{x^{2} - 1}{{(x^{4} + 1)}^{2}} d x

Commented by prof Abdo imad last updated on 16/Jun/18

1) w e h a v e f (a) = \int_{- \infty}^{+ \infty} \frac{x^{2} - 1}{{(x^{4} + a^{4})}^{2}} d x

l e t φ (z) = \frac{z^{2} - 1}{{(z^{4} + a^{4})}^{2}}

φ (z) = \frac{z^{2} - 1}{{{(z^{2})}^{2} - {({i a}^{2})}^{2}}^{2}}

= \frac{z^{2} - 1}{{(z^{2} - {i a}^{2})}^{2} {(z^{2} + {i a}^{2})}^{2}}

= \frac{z^{2} - 1}{{(z - a e^{i \frac{π}{4}})}^{2} {(z + {a e}^{\frac{i π}{4}})}^{2} {(z - a e^{- \frac{i π}{4}})}^{2} {(z + a e^{- \frac{i π}{4}})}^{2}}

t h e p o l e s o f φ a r e \bar{+} e^{i \frac{π}{4}} a n d \bar{+} e^{- \frac{i π}{4}} (d o u b l e s)

l e t s u p p o s e a > 0

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

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

= {l i m}_{z \to {a e}^{\frac{i π}{4}}} {\frac{z^{2} - 1}{{(z + a e^{\frac{i π}{4}})}^{2} {(z^{2} + {i a}^{2})}^{2}}}^{(1)}

= {l i m}_{z \to {a e}^{\frac{i π}{4}}} \frac{2 z (z + {a e}^{\frac{i π}{4}}) {(z^{2} + {i a}^{2})}^{2} - (z^{2} - 1) (2 (z + {a e}^{\frac{i π}{4}}) {(z^{2} + {i a}^{2})}^{2} + 4 z (z^{2} + {ia}^{2}) {(z + {ae}^{\frac{i π}{4}})}^{2}}{{(z + {a e}^{\frac{i π}{4}})}^{4} {(z^{2} + {i a}^{2})}^{4}}

= {l i m}_{z \to a e^{\frac{i π}{4}}} \frac{2 z (z^{2} + {i a}^{2}) - (z^{2} - 1) {2 (z^{2} + {i a}^{2}) + 4 z (z + {a e}^{\frac{i π}{4}})}}{{(z + a e^{\frac{i π}{4}})}^{3} {(z^{2} + {i a}^{2})}^{3}}

= \frac{2 a e^{\frac{i π}{4}} (2 {i a}^{2}) - ({i a}^{2} - 1) {4 {i a}^{2} + 4 {a e}^{\frac{i π}{4}} (2 {a e}^{\frac{i π}{4}})}}{{(2 {a e}^{i \frac{π}{4}})}^{3} {(2 {i a}^{2})}^{3}}

= \frac{4 {i a}^{3} e^{\frac{i π}{4}} - ({i a}^{2} - 1) {4 {i a}^{2} + 8 a^{2} i)}{8 a^{3} i e^{\frac{i π}{4}} (- 8 {i a}^{6})}

= \frac{4 {i a}^{3} e^{\frac{i π}{4}} - 12 {i a}^{2} ({i a}^{2} - 1)}{64 a^{9} e^{\frac{i π}{4}}}

= \frac{4 {i a}^{3} e^{\frac{i π}{4}} + 12 a^{4} + 12 {i a}^{2}}{64 a^{9} e^{\frac{i π}{4}}} = \dots . .

Commented by behi83417@gmail.com last updated on 16/Jun/18

x = a \sqrt{t g t} \Rightarrow d x = \frac{a (1 + {t g}^{2} t) d t}{2 \sqrt{t g t}}

I = \int \frac{a^{2} t g t - 1}{{(a^{4} {t g}^{2} t + a^{4})}^{2}} . \frac{a (1 + {t g}^{2} t) d t}{2 \sqrt{t g t}}

\Rightarrow I = \int \frac{a^{2} t g t - 1}{a^{8} {({t g}^{2} t + 1)}^{2}} . \frac{a (1 + {t g}^{2} t) d t}{2 \sqrt{t g t}} =

= \frac{1}{a^{7}} \int \frac{a^{2} t g t - 1}{2 \sqrt{t g t} (1 + {t g}^{2} t)} d t, t g t = u^{2}

(1 + {t g}^{2} t) d t = 2 u d u \Rightarrow d t = \frac{2 u d u}{1 + u^{4}}

\Rightarrow I = \frac{1}{a^{7}} \int \frac{a^{2} u^{2} - 1}{2 u (1 + u^{4})} . \frac{2 u d u}{1 + u^{4}} = \frac{1}{a^{7}} \int \frac{a^{2} u^{2} - 1}{{(1 + u^{4})}^{2}} d u =

= \frac{1}{a^{7}} [\int \frac{a^{2} u^{2}}{{(1 + u^{4})}^{2}} d u - \int \frac{d u}{{(1 + u^{4})}^{2}}] = \dots

You can't use 'macro parameter character #' in math mode

+ \frac{\sqrt{2}}{16} \boldsymbol l n \frac{\boldsymbol u^{2} - \boldsymbol u \sqrt{2} + 1}{\sqrt{1 + \boldsymbol u^{4}}} + \boldsymbol c o n s t

You can't use 'macro parameter character #' in math mode

- \frac{3 \sqrt{2}}{16} \boldsymbol l n \frac{\boldsymbol u^{2} - \boldsymbol u \sqrt{2} + 1}{\sqrt{1 + \boldsymbol u^{4}}} + \boldsymbol c o n s t

\boldsymbol b y \boldsymbol s y m p l i f i n g :

You can't use 'macro parameter character #' in math mode

c h a n g e \Rightarrow \boldsymbol u \to \frac{\boldsymbol x}{\boldsymbol a}

I = \frac{1}{\boldsymbol a^{5}} [\frac{\boldsymbol {a x}^{3}}{4 (\boldsymbol a^{4} + \boldsymbol x^{4})} + \frac{\sqrt{2}}{16} \boldsymbol {t g}^{- 1} (\frac{\boldsymbol a x \sqrt{2}}{\boldsymbol a^{2} - \boldsymbol x^{2}}) -

- \frac{3 \sqrt{2}}{16} \boldsymbol l n \frac{\boldsymbol x^{2} - \boldsymbol a x \sqrt{2} + \boldsymbol a^{2}}{\sqrt{\boldsymbol a^{4} + \boldsymbol x^{4}}}] - \frac{1}{\boldsymbol a^{7}} [\frac{\boldsymbol a^{3} \boldsymbol x}{4 (\boldsymbol a^{4} + \boldsymbol x^{4})} +

+ \frac{3 \sqrt{2}}{16} \boldsymbol {t g}^{- 1} \frac{\boldsymbol a x \sqrt{2}}{\boldsymbol a^{2} - \boldsymbol x^{2}} - \frac{3 \sqrt{2}}{16} \boldsymbol l n \frac{\boldsymbol x^{2} - \boldsymbol a x \sqrt{2} + \boldsymbol a^{2}}{\sqrt{\boldsymbol a^{4} + \boldsymbol x^{4}}}] + \boldsymbol c o n s t

3) \boldsymbol I = \boldsymbol F (\infty) - \boldsymbol F (0) = 0

4) \boldsymbol a = 1 \Rightarrow \boldsymbol I = \frac{1}{8} [\frac{2 \boldsymbol x}{1 + \boldsymbol x^{4}} + 2 \sqrt{2} \boldsymbol {t g}^{- 1} \frac{\boldsymbol x \sqrt{2}}{1 - \boldsymbol x^{2}} -

- \sqrt{2} \boldsymbol l n \frac{\boldsymbol x^{2} - \boldsymbol x \sqrt{2} + 1}{\sqrt{1 + \boldsymbol x^{4}}}] + \boldsymbol c o n s t .

Commented by behi83417@gmail.com last updated on 16/Jun/18

d e a r p r o f . a b d o! i s i t o k ?

Commented by prof Abdo imad last updated on 16/Jun/18

s i r B e h i y o u h a v e u s i n g a g o o d c h a n g e m e n t

a t t h e b e g i n i n g a n d i t s s e e m s t h a t y o u r m e t h o d

i s c o r r e c t a n d i w i l l g i v e y o u t h e f i n a l a n s w e r

a f t e r f i n i s h i n g t h e c a l c u l u s \dots