calculate-I-x-1-x-2-1-2-dx-

Question Number 36432 by prof Abdo imad last updated on 02/Jun/18

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

Commented by abdo mathsup 649 cc last updated on 03/Jun/18

l e t i n t r o d u c e t h e c o m p l e x f u n c t i o n

φ (z) = \frac{z + 1}{{(z^{2} + 1)}^{2}} w e h a v e

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

(d o u b l e s)

\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} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{^{'}}

= {l i m}_{z \to i} {\frac{z + 1}{{(z + i)}^{2}}}^{^{'}}

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

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

= \frac{- 2}{{(2 i)}^{3}} = \frac{- 2}{- 8 i} = \frac{1}{4 i}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{4 i} = \frac{π}{2} s o

★ I = \frac{π}{2} ★

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

x = t a n (θ

d x = {s e c}^{2} θ d θ

\int_{\frac{- Π}{2}}^{\frac{Π}{2}} \frac{(t a n θ + 1) \times {s e c}^{2} θ}{{s e c}^{2} θ \times {s e c}^{2} θ} d θ

\int_{- \frac{Π}{2}}^{\frac{Π}{2}} {\frac{s i n θ}{c o s θ} \times {c o s}^{2} θ + {c o s}^{2} θ} d θ

\int_{- \frac{Π}{2}}^{\frac{Π}{2}} s i n θ d (s i n θ) + \int_{- \frac{Π}{2}}^{\frac{Π}{2}} \frac{1 + c o s 2 θ}{2} d θ

=∣ \frac{{s i n}^{2} θ}{2} ∣_{- \frac{Π}{2}}^{\frac{Π}{2}} + \frac{1}{2} ∣ θ ∣_{- \frac{Π}{2}}^{\frac{Π}{2}} + \frac{1}{2} \times \frac{1}{2} ∣ s i n 2 θ ∣_{- \frac{Π}{2}}^{\frac{Π}{2}}

= 0 + \frac{1}{2} (Π) + \frac{1}{4} \times 0

= \frac{Π}{2}

Commented by abdo mathsup 649 cc last updated on 03/Jun/18

c o r r e c t a n s w e r s i r T a n m a y t h a n k s . .