calculate-2x-x-2-mx-1-2-dx-with-m-lt-2-

Question Number 36056 by abdo mathsup 649 cc last updated on 28/May/18

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

Commented by abdo mathsup 649 cc last updated on 30/May/18

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{2 z}{{(z^{2} + m z + 1)}^{2}} p o l e s o f φ!

r o o t s o f z^{2} + m z + 1

Δ = m^{2} - 4 < 0 b e c a u s e ∣ m ∣< 2 \Rightarrow

Δ = - (4 - m^{2}) = {i \sqrt{4 - m^{2}}}^{2}

z_{1} = \frac{- m + i \sqrt{4 - m^{2}}}{2} a n d z_{2} = \frac{- m - i \sqrt{4 - m^{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)

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1})

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

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

Commented by abdo mathsup 649 cc last updated on 30/May/18

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

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

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

= 2 \frac{i \sqrt{4 - m^{2}} - 2 \frac{- m + i \sqrt{4 - m^{2}}}{2}}{{(i \sqrt{4 - m^{2}})}^{3}}

= 2 \frac{m}{- i (4 - m^{2}) \sqrt{4 - m^{(}}} = \frac{- 2 m}{i (4 - m^{2}) \sqrt{4 - m^{2}}}

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

= \frac{- 4 m π}{(4 - m^{2}) \sqrt{4 - m^{2}}} s o

I = \frac{- 4 m π}{(4 - m^{2}) \sqrt{4 - m^{2}}} .

Answered by sma3l2996 last updated on 28/May/18

I = \int_{- \infty}^{+ \infty} \frac{2 x}{{(x^{2} + m x + 1)}^{2}} d x = \int_{- \infty}^{+ \infty} \frac{2 x + m - m}{{(x^{2} + m x + 1)}^{2}} d x

= - {[\frac{1}{x^{2} + m x + 1}]}_{- \infty}^{+ \infty} - m \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 2 \times \frac{m}{2} \times x + \frac{m^{2}}{4} - \frac{m^{2}}{4} + 1)}^{2}}

= 0 - m \int_{- \infty}^{+ \infty} \frac{d x}{{({(x + \frac{m}{2})}^{2} + \frac{4 - m^{2}}{4})}^{2}}

= - m \int_{- \infty}^{+ \infty} \frac{d x}{{(\frac{4 - m^{2}}{4} ({(\frac{2 x + m}{\sqrt{4 - m^{2}}})}^{2} + 1))}^{2}} / ∣ m ∣< 2

l e t u = \frac{2 x + m}{\sqrt{4 - m^{2}}} \Rightarrow d x = \frac{\sqrt{4 - m^{2}}}{2} d u

I = - \frac{8 m \sqrt{4 - m^{2}}}{{(4 - m^{2})}^{2}} \int_{- \infty}^{+ \infty} \frac{d u}{{(u^{2} + 1)}^{2}}

t a n t = u \Rightarrow d t = \frac{d u}{1 + u^{2}}

\frac{d u}{{(u^{2} + 1)}^{2}} = \frac{d t}{1 + {t a n}^{2} t} = {c o s}^{2} (t) d t = (\frac{1 + c o s (2 t)}{2}) d t

\int_{- π / 2}^{π / 2} (1 + c o s (2 t)) d t = {[t + \frac{1}{2} s i n 2 t]}_{- π / 2}^{π / 2} = π

I = - \frac{4 π m \sqrt{4 - m^{2}}}{{(4 - m^{2})}^{2}}

Commented by abdo mathsup 649 cc last updated on 30/May/18

c o r r e c t s i r s m a 3 l t h a n k s a l o t s .