let f(x) = ∫_(−∞) ^(+∞) ((cos(xt))/((t−i)^2 )) dt 1) let R =Re(f(x)) and I =Im(f(x)) extract R and I 2) calculate R and I 3) conclude the value of f(x) 4) calculate ∫_(−∞) ^(+∞) ((cos(2t))/((t−i)^2 ))dt 5) let u_n = ∫_(−∞) ^(+∞) ((cos((t/n)))/((t−i)^2 ))dt (n natral integer not o) find lim_(n→+∞) u_n and study the convergence of Σu


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 42305 by maxmathsup by imad last updated on 22/Aug/18

$l e t f (x) = \int_{- \infty}^{+ \infty} \frac{c o s (x t)}{{(t - i)}^{2}} d t$ $1) l e t R = R e (f (x)) a n d I = I m (f (x)) e x t r a c t R a n d I$ $2) c a l c u l a t e R a n d I$ $3) c o n c l u d e t h e v a l u e o f f (x)$ $4) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{c o s (2 t)}{{(t - i)}^{2}} d t$ $5) l e t u_{n} = \int_{- \infty}^{+ \infty} \frac{c o s (\frac{t}{n})}{{(t - i)}^{2}} d t (n n a t r a l i n t e g e r n o t o)$ $f i n d {l i m}_{n \to + \infty} u_{n} a n d s t u d y t h e c o n v e r g e n c e o f Σ u_{n}$
Commented by maxmathsup by imad last updated on 22/Aug/18

$i^{2} = - 1$
Commented by maxmathsup by imad last updated on 23/Aug/18

$1) w e h a v e f (x) = \int_{- \infty}^{+ \infty} \frac{c o s (x t) {(t + i)}^{2}}{{(t - i)}^{2} {(t + i)}^{2}} d t = \int_{- \infty}^{+ \infty} \frac{(t^{2} + 2 i t - 1) c o s (x t)}{{(t^{2} + 1)}^{2}} d t$ $= \int_{- \infty}^{+ \infty} \frac{(t^{2} - 1) c o s (x t)}{{(t^{2} + 1)}^{2}} d t + i \int_{- \infty}^{+ \infty} \frac{2 t c o s (x t)}{{(t^{2} + 1)}^{2}} d t \Rightarrow$ $R e (f (x)) = \int_{- \infty}^{+ \infty} \frac{(t^{2} - 1) c o s (x t)}{{(t^{2} + 1)}^{2}} d t a n d I m (f (x)) = \int_{- \infty}^{+ \infty} \frac{2 t c o s (x t)}{{(t^{2} + 1)}^{2}} d t (= 0)$
Commented by maxmathsup by imad last updated on 23/Aug/18
$2) let caoculate R =Re( ∫_(−∞) ^(+∞) (((x^2 −1)e^(ixt) )/((x^2 +1)^2 ))dt) let consider the complex function ϕ(z) =(((z^2 −1)e^(ixz) )/((z^2 +1)^2 )) we have ϕ(z) =(((z^2 −1)e^(ixz) )/((z−i)^2 (z+i)^2 ))( iand −i double poles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { (((z^2 −1)e^(ixz) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((2z +ix(z^2 −1))e^(ixz) (z+i)^2 −2(z+i)(z^2 −1)e^(ixz) )/((z+i)^4 )) =lim_(z→i) (((2z +ixz^2 −ix)e^(ixz) (z+i)−2(z^2 −1)e^(ixz) )/((z+i)^3 )) = (((2i−ix −ix)e^(−x) (2i) +4 e^(−x) )/((2i)^3 )) =((2i(2i−2ix)e^(−x) +4 e^(−x) )/(−8i)) =((−4(1−x)e^(−x) +4e^(−x) )/(−8i)) = ((4x e^(−x) )/(−8i)) =−(x/(2i)) e^(−x) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(−(x/(2i))e^(−x) ) = −πx e^(−x) ⇒ R =−π x e^(−x) also I =Im(f(x))=0$
$2) l e t c a o c u l a t e R = R e (\int_{- \infty}^{+ \infty} \frac{(x^{2} - 1) e^{i x t}}{{(x^{2} + 1)}^{2}} d t) l e t c o n s i d e r t h e c o m p l e x$ $f u n c t i o n φ (z) = \frac{(z^{2} - 1) e^{i x z}}{{(z^{2} + 1)}^{2}} w e h a v e φ (z) = \frac{(z^{2} - 1) e^{i x z}}{{(z - i)}^{2} {(z + i)}^{2}} (i a n d - i d o u b l e p o l e s)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i) b u t R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{(z^{2} - 1) e^{i x z}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{(2 z + i x (z^{2} - 1)) e^{i x z} {(z + i)}^{2} - 2 (z + i) (z^{2} - 1) e^{i x z}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{(2 z + {i x z}^{2} - i x) e^{i x z} (z + i) - 2 (z^{2} - 1) e^{i x z}}{{(z + i)}^{3}}$ $= \frac{(2 i - i x - i x) e^{- x} (2 i) + 4 e^{- x}}{{(2 i)}^{3}} = \frac{2 i (2 i - 2 i x) e^{- x} + 4 e^{- x}}{- 8 i}$ $= \frac{- 4 (1 - x) e^{- x} + 4 e^{- x}}{- 8 i} = \frac{4 x e^{- x}}{- 8 i} = - \frac{x}{2 i} e^{- x} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (- \frac{x}{2 i} e^{- x})$ $= - π x e^{- x} \Rightarrow R = - π x e^{- x} a l s o I = I m (f (x)) = 0$
Commented by maxmathsup by imad last updated on 23/Aug/18

$3) w e h a v e f (x) = R e (f (x)) + i I m (f (x)) \Rightarrow f (x) = - π x e^{- x} .$
Commented by maxmathsup by imad last updated on 23/Aug/18

$4) \int_{- \infty}^{+ \infty} \frac{c o s (2 t)}{{(t - i)}^{2}} d t = f (2) = - 2 π e^{- 2} = \frac{- 2 π}{e^{2}} .$
Commented by maxmathsup by imad last updated on 23/Aug/18

$5) w e h a v e u_{n} = \int_{- \infty}^{+ \infty} \frac{c o s (\frac{t}{n})}{{(t - i)}^{2}} d t = f (\frac{1}{n}) = - \frac{π}{n} e^{- \frac{1}{n}} \Rightarrow {l i m}_{n \to + \infty} u_{n} = 0$ $a n d \sum_{n = 1}^{\infty} u_{n} = - \sum_{n = 1}^{\infty} \frac{π}{n} e^{- \frac{1}{n}} b u t e^{u} = 1 + u + o (u^{2}) (u \to 0) \Rightarrow$ $e^{- \frac{1}{n}} = 1 - \frac{1}{n} + o (\frac{1}{n^{2}}) \Rightarrow \frac{π}{n} e^{- \frac{1}{n}} = \frac{π}{n} - \frac{π}{n^{2}} + o (\frac{1}{n}) b u t Σ \frac{π}{n} d i v e r g e s$ $Σ \frac{π}{n^{2}} c o n v e r g e s \Rightarrow Σ u_{n} d i v e r g e s .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum