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Question Number 67008 by mathmax by abdo last updated on 21/Aug/19

l e t f (x) = \int_{0}^{\infty} \frac{d t}{{(x^{2} + t^{2})}^{2}} w i t h x > 0

1) f i n d a e x p l i c i t f o r m o f (x)

2) f i n d a l s o g (x) = \int_{0}^{\infty} \frac{d t}{{(x^{2} + t^{2})}^{3}}

3) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{d t}{{(t^{2} + 3)}^{2}} a n d \int_{0}^{\infty} \frac{d t}{{(t^{2} + 3)}^{3}}

4) c a l c u l a t e U_{θ} = \int_{0}^{\infty} \frac{d t}{{(t^{2} + {c o s}^{2} θ)}^{2}} w i t h 0 < θ < \frac{π}{2}

5) f i n d f^{(n)} (x) a n d f^{(n)} (0)

6) d e v e l o p p f a t i n t e g r s e r i e

Commented by mathmax by abdo last updated on 25/Aug/19

1) f (x) = \int_{0}^{\infty} \frac{d t}{{(x^{2} + t^{2})}^{2}} \Rightarrow 2 f (x) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + x^{2})}^{2}} l e t

=_{t = x z} \int_{- \infty}^{+ \infty} \frac{x d z}{x^{4} {(z^{2} + 1)}^{2}} = \frac{1}{x^{3}} \int_{- \infty}^{+ \infty} \frac{d z}{{(z^{2} + 1)}^{2}} l e t φ (z) = \frac{1}{{(z^{2} + 1)}^{2}}

w e h a v e φ (z) = \frac{1}{{(z - i)}^{2} {(z + i)}^{2}} {r e s i}_{} d u s t b e o r e m h i v e

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

R e s (φ, i) = {l i m}_{z \to i} {{(z - i)}^{2} φ (z)}^{(1)} = {l i m}_{z \to i} {{(z + i)}^{- 2}}^{(1)}

= {l i m}_{z \to i} - 2 {(z + i)}^{- 3} = - 2 {(2 i)}^{- 3} = \frac{- 2}{{(2 i)}^{3}} = \frac{- 2}{- 8 i} = \frac{1}{4 i} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \times \frac{1}{4 i} = \frac{π}{2} \Rightarrow 2 f (x) = \frac{π}{2 x^{3}} \Rightarrow f (x) = \frac{π}{4 x^{3}} (x > 0)

2) w e h a v e f^{^{'}} (x) = - \int_{0}^{\infty} \frac{2 (2 x) (x^{2} + t^{2})}{{(x^{2} + t^{2})}^{4}} d x = - 4 x \int_{0}^{\infty} \frac{d x}{{(x^{2} + t^{2})}^{3}}

= - 4 x g (x) \Rightarrow g (x) = - \frac{1}{4 x} f^{^{'}} (x)

f^{^{'}} (x) = \frac{π}{4} {(x^{- 3})}^{^{'}} = \frac{π}{4} (- 3) x^{- 4} = \frac{- 3 π}{4 x^{4}} \Rightarrow

g (x) = - \frac{1}{4 x} \times \frac{- 3 π}{4 x^{4}} = \frac{3 π}{16 x^{5}}

Commented by mathmax by abdo last updated on 25/Aug/19

3) \int_{0}^{\infty} \frac{d t}{{(t^{2} + 3)}^{2}} = f (\sqrt{3}) = \frac{π}{4 {(\sqrt{3})}^{3}} = \frac{π}{12 \sqrt{3}} = \frac{π \sqrt{3}}{36}

\int_{0}^{\infty} \frac{d t}{{(t^{2} + 3)}^{3}} = g (\sqrt{3}) = \frac{3 π}{16 {(\sqrt{3})}^{5}} = \frac{3 π}{16 \times 9 \times \sqrt{3}} = \frac{π}{48 \sqrt{3}} = \frac{π \sqrt{3}}{144}