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Question Number 57746 by maxmathsup by imad last updated on 10/Apr/19

l e t f (x) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 x t + 1)}^{2}} w i t h ∣ x ∣< 1 (x r e a l)

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

2) f i n d a l s o g (x) = \int_{- \infty}^{+ \infty} \frac{t d t}{{(t^{2} - 2 x t + 1)}^{3}}

3) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - \sqrt{2} t + 1)}^{2}} a n d \int_{- \infty}^{+ \infty} \frac{t d t}{{(t^{2} - \sqrt{2} t + 1)}^{3}}

4) c a l c u l a t e A (θ) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 c o s θ t + 1)}^{2}} a n d

B (θ) = \int_{- \infty}^{+ \infty} \frac{t d t}{{(t^{2} - 2 c o s θ t + 1)}^{3}} w i t h 0 < θ < \frac{π}{2} .

Commented by maxmathsup by imad last updated on 13/Apr/19

1) 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 φ (z) = \frac{1}{{(z^{2} - 2 x z + 1)}^{2}}

l e t d e t e r m i n e t h e p o l e s o f φ z^{2} - 2 x z + 1 = 0

Δ^{^{'}} = x^{2} - 1 = {(i \sqrt{1 - x^{2}})}^{2} \Rightarrow z_{1} = x + i \sqrt{1 - x^{2}} a n d z_{2} = x - i \sqrt{1 - x^{2}}

φ (z) = \frac{1}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} t h e p o l e s o f φ a r e z_{1} a n d z_{2} (d o u b l e s)

r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1})

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

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

= \frac{- 2}{{(z_{1} - z_{2})}^{3}} = \frac{- 2}{{(2 i \sqrt{1 - x^{2}})}^{3}} = \frac{- 2}{- 8 i (1 - x^{2}) \sqrt{1 - x^{2}}} = \frac{1}{4 i (1 - x^{2}) \sqrt{1 - x^{2}}} \Rightarrow

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

w i t h ∣ x ∣< 1 .

Commented by maxmathsup by imad last updated on 13/Apr/19

2) w e h a v e f^{^{'}} (x) = \int_{- \infty}^{+ \infty} \frac{\partial}{\partial x} (\frac{1}{{(t^{2} - 2 x t + 1)}^{2}}) d t

= \int_{- \infty}^{+ \infty} \frac{2 t (t^{2} - 2 x t + 1)}{{(t^{2} - 2 x t + 1)}^{4}} d t = 2 \int_{- \infty}^{+ \infty} \frac{t d t}{{(t^{2} - 2 x t + 1)}^{3}} = 2 g (x) \Rightarrow g (x) = \frac{1}{2} f^{^{'}} (x)

w e h a v e f (x) = \frac{π}{2 (1 - x^{2}) \sqrt{1 - x^{2}}} \Rightarrow f (x) = \frac{π}{2} {{(1 - x^{2})}^{- 1} {(1 - x^{2})}^{- \frac{1}{2}}} \Rightarrow

f^{^{'}} (x) = \frac{π}{2} {2 x {(1 - x^{2})}^{- 2} {(1 - x^{2})}^{- \frac{1}{2}} + x {(1 - x^{2})}^{- \frac{3}{2}} {(1 - x^{2})}^{- 1}}

= \frac{π}{2} {\frac{2 x}{{(1 - x^{2})}^{2} \sqrt{1 - x^{2}}} + \frac{x}{(1 - x^{2}) (1 - x^{2}) \sqrt{1 - x^{2}}}}

= \frac{π x}{2} {\frac{2}{{(1 - x^{2})}^{2} \sqrt{1 - x^{2}}} + \frac{x}{{(1 - x^{2})}^{2} \sqrt{1 - x^{2}}}}

= \frac{π x (x + 2)}{2 {(1 - x^{2})}^{2} \sqrt{1 - x^{2}}} \Rightarrow g (x) = \frac{π x (x + 2)}{4 {(1 - x^{2})}^{2} \sqrt{1 - x^{2}}} .

Commented by maxmathsup by imad last updated on 13/Apr/19

3) \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - \sqrt{2} t + 1)}^{2}} = f (\frac{\sqrt{2}}{2}) = \frac{π}{2 (1 - {(\frac{\sqrt{2}}{2})}^{2} \sqrt{1 - {(\frac{\sqrt{2}}{2})}^{2}}}

= \frac{π}{2 (1 - \frac{1}{2}) \sqrt{1 - \frac{1}{2}}} = \frac{π}{\sqrt{\frac{1}{2}}} = π \sqrt{2}

\int_{- \infty}^{+ \infty} \frac{t d t}{{(t^{2} - \sqrt{2} t + 1)}^{3}} = g (\frac{\sqrt{2}}{2}) = \frac{π \frac{\sqrt{2}}{2} (\frac{\sqrt{2}}{2} + 2)}{4 {(1 - {(\frac{\sqrt{2}}{2})}^{2})}^{2} \sqrt{1 - {(\frac{\sqrt{2}}{2})}^{2}}}

= \frac{π \sqrt{2} (4 + \sqrt{2})}{16 {(1 - \frac{1}{2})}^{2} \sqrt{1 - \frac{1}{2}}} = \frac{π \sqrt{2} (4 + \sqrt{2})}{4 . \frac{1}{\sqrt{2}}} = \frac{2 π (4 + \sqrt{2})}{4} = \frac{π}{2} (4 + \sqrt{2}) .

Commented by maxmathsup by imad last updated on 13/Apr/19

4) w e h a v e A (θ) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 c o s θ t + 1)}^{2}} = f (c o s θ) = \frac{π}{2 (1 - {c o s}^{2} θ) \sqrt{1 - {c o s}^{2} θ}}

= \frac{π}{2 {s i n}^{2} θ s i n θ} = \frac{π}{2 {s i n}^{3} θ} (0 < θ < \frac{π}{2})

a l s o w e h a v e B (θ) = \int_{- \infty}^{+ \infty} \frac{t d t}{{(t^{2} - 2 c o s θ t + 1)}^{3}} \Rightarrow B (θ) = g (c o s θ)

= \frac{π c o s θ (c o s θ + 2)}{4 {(1 - {c o s}^{2} θ)}^{2} \sqrt{1 - {c o s}^{2} θ}} = \frac{π c o s θ (2 + c o s θ)}{4 {s i n}^{4} θ s i n θ} = \frac{2 π c o s θ + π {c o s}^{2} θ}{4 {s i n}^{5} θ} .