let a>0 b from C and Re(b)>0 1) calculate ∫_(−∞) ^(+∞) ((b cos(ax))/(x^2 +b^2 ))dx 2) find the value of ∫


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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 byabdo.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 byabdo.msup.com last updated on 13/Jun/18

$2) l e t J = \int_{- \infty}^{+ \infty} \frac{x s i n (a x)}{x^{2} + b^{2}} d x$ $J = I m (\int_{- \infty}^{+ \infty} \frac{x e^{i a x}}{x^{2} + b^{2}} d x) l e t$ $ψ (z) = \frac{z e^{i a z}}{z^{2} + b^{2}} t h e p o l e s o f ψ a r e$ $i b a n d - i b$ $\int_{- \infty}^{+ \infty} ψ (z) d z = 2 i π R e s (ψ, i b)$ $= 2 i π \frac{i b e^{i a (i b)}}{2 i b} = i π e^{- a b} \Rightarrow J = π e^{- a b}$
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