let-a-gt-0-b-from-C-and-Re-b-gt-0-1-calculate-b-cos-ax-x-2-b-2-dx-2-find-the-value-of-x-sin-ax-x-2-b-2-dx-

Question Number 37357 by math khazana by abdo last updated on 12/Jun/18

l e t a > 0 b f r o m C a n d R e (b) > 0

1) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{b c o s (a x)}{x^{2} + b^{2}} d x

2) f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{x s i n (a x)}{x^{2} + b^{2}} d x .

Commented by abdo.msup.com last updated on 13/Jun/18

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

I = R e (\int_{- \infty}^{+ \infty} \frac{b e^{i a x}}{x^{2} + b^{2}} d x) l e t c o n s i d e r

φ (z) = \frac{b e^{i a z}}{z^{2} + b^{2}}

φ (z) = \frac{{b e}^{i a x}}{(z - i b) (z + i b)} s o t h e p o l e s o f φ

a r e i b a n d - i b w e h a v e R e (b) > 0 \Rightarrow

I m (i b) > 0 s o

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

R e s (φ, i b) = {l i m}_{z \to i b} (z - i b) φ (z)

= \frac{b e^{i a (i b)}}{2 i b} = \frac{e^{- a b}}{2 i} \Rightarrow

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

I = π e^{- a b} .

Commented by abdo.msup.com last updated on 13/Jun/18

2) l e t J = \int_{- \infty}^{+ \infty} \frac{x s i n (a x)}{x^{2} + b^{2}} d x

J = I m (\int_{- \infty}^{+ \infty} \frac{x e^{i a x}}{x^{2} + b^{2}} d x) l e t

ψ (z) = \frac{z e^{i a z}}{z^{2} + b^{2}} t h e p o l e s o f ψ a r e

i b a n d - i b

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

= 2 i π \frac{i b e^{i a (i b)}}{2 i b} = i π e^{- a b} \Rightarrow J = π e^{- a b}