let-f-xsin-x-1-x-2-2-dx-calculate-f-and-f-from-R-

Question Number 48040 by maxmathsup by imad last updated on 18/Nov/18

l e t f (α) = \int_{- \infty}^{+ \infty} \frac{x s i n (α x)}{{(1 + x^{2})}^{2}} d x

c a l c u l a t e f (α) a n d f^{^{'}} (α) . (α f r o m R) .

Commented by Abdo msup. last updated on 20/Nov/18

1) w e h a v e ? f (α) = I m (\int_{- \infty}^{+ \infty} \frac{x^{} e^{i α x}}{{(1 + x^{2})}^{2}} d x) l e t

φ (z) = \frac{z e^{i α z}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{z e^{i α z}}{{(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) \Rightarrow \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 e^{i α z}}{{(z + i)}^{2}}}^{(1)}

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

= {l i m}_{z \to i} \frac{(1 + i α z) e^{i α z} (z + i) - 2 z e^{i α z}}{{(z + i)}^{3}}

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

= \frac{α}{4} e^{- α} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{α}{4} e^{- α} = i \frac{α π}{2} e^{- α} \Rightarrow

f (α) = \frac{π α}{2} e^{- α}

2) f^{^{'}} (α) = \frac{π}{2} {e^{- α} - α e^{- α}} = \frac{π}{2} {1 - α} e^{- α} b u t

f^{^{'}} (α) = \int_{- \infty}^{+ \infty} \frac{\partial}{\partial α} (\frac{x s i n (α x)}{{(1 + x^{2})}^{2}}) d x

= \int_{- \infty}^{+ \infty} \frac{x^{2} c o s (α x)}{{(1 + x^{2})}^{2}} d x \Rightarrow

\int_{- \infty}^{+ \infty} \frac{x^{2} c o s (α x)}{{(1 + x^{2})}^{2}} d x = \frac{π}{2} (1 - α) e^{- α} .