let-f-x-dt-t-2-2t-x-2-4-with-x-gt-1-and-n-integr-natural-1-find-a-explicit-form-for-f-x-2-determine-also-g-x-dt-t-2-2t-x-2-5-3-find-the-values

Question Number 70870 by Abdo msup. last updated on 09/Oct/19

l e t f (x) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + x^{2})}^{4}} w i t h ∣ x ∣> 1

a n d n i n t e g r n a t u r a l

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

2) d e t e r m i n e a l s o g (x) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + x^{2})}^{5}}

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_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + 3)}^{4}} a n d \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + 3)}^{5}}

Commented by mathmax by abdo last updated on 13/Oct/19

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

z^{2} - 2 z + x^{2} = 0 \to Δ^{^{'}} = 1 - x^{2} < 0 \Rightarrow Δ^{^{'}} = {(i \sqrt{x^{2} - 1})}^{2}

z_{1} = 1 + i \sqrt{x^{2} - 1} a n d z_{2} = 1 - i \sqrt{x^{2} - 1}

∣ z_{1} ∣= \sqrt{1 + x^{2} - 1} =∣ x ∣ \Rightarrow z_{1} =∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})}

z_{2} =∣ x ∣ e^{- i a r c t a n (\sqrt{x^{2} - 1})}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, ∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})})

R e s (φ, ∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})}

= {l i m}_{z \to∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})}} \frac{1}{(4 - 1)!} {{(z - ∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})})}^{4} φ (z)}^{(3)}

= {l i m}_{z \to (\dots)} \frac{1}{3!} {\frac{1}{{(z - z_{2})}^{4}}}^{(3)} \dots . b e c o n t i n u e d \dots

Commented by mathmax by abdo last updated on 13/Oct/19

{{(z - z_{2})}^{- 4}}^{(3)} = {- 4 {(z - z_{2})}^{- 5}}^{(2)} = {20 {(z - z_{2})}^{- 6}}^{(1)}

= {- 120 {(z - z_{2})}^{- 7}} \Rightarrow

R e s (φ, ∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1}}) = \frac{1}{6} (- 120) {(z_{1} - z_{2})}^{- 7}

= - 20 {(2 i I m (z_{1}))}^{- 7} = - 20 (2 i s i n {(∣ x ∣ a r c t a n (\sqrt{x^{2} - 1}))}^{- 7}

= \frac{- 20}{{(2 i)}^{7} {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1}))} = \frac{- 20}{2^{7} (- i) {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1})}

i^{7} = {(i^{2})}^{3} i = - i

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{2^{2} \times 5}{2^{3} \times 2^{4} (i) {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1}))}

= \frac{5 π}{16 {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1})} \Rightarrow

f (x) = \frac{5 π}{16 {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1})}

Commented by mathmax by abdo last updated on 13/Oct/19

2) w e h a v e f^{^{'}} (x) = - \int_{- \infty}^{+ \infty} \frac{4 (2 x) {(t^{2} - 2 t + x^{2})}^{3}}{{(t^{2} - 2 t + x^{2})}^{8}} d t

= - 8 x \int_{- \infty}^{+ \infty} \frac{d x}{{(t^{2} - 2 t + x^{2})}^{5}} = - 8 x g (x) \Rightarrow g (x) = - \frac{1}{8 x} f^{^{'}} (x)

r e s t t h e c a l c u l u s o f f^{^{'}} (x) \dots .

Commented by mathmax by abdo last updated on 13/Oct/19

\int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + 3)}^{4}} = f (\sqrt{3}) = \frac{5 π}{16 {s i n}^{7} (\sqrt{3} a r c t a n (\sqrt{3 - 1})}

= \frac{5 π}{16 {s i n}^{7} (\sqrt{3} a r c t a n (2))}