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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) = \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) \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}