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-

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) 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 .