calculate ∫_(−∞) ^(+∞) ((cos(ax)ch(bx))/(x^2 +1))dx .


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Question Number 38467 by maxmathsup by imad last updated on 25/Jun/18

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{c o s (a x) c h (b x)}{x^{2} + 1} d x .$
Commented by math khazana by abdo last updated on 28/Jun/18
$let I = ∫_(−∞) ^(+∞) ((cos(ax)ch(bx))/(x^2 +1))dx I =Re( ∫_(−∞) ^(+∞) ((e^(iax) (e^(bx) +e^(−bx) ))/(2(x^2 +1)))dx) =Re( ∫_(−∞) ^(+∞) ((e^((b+ia)x) +e^((−b+ia)x) )/(2(x^2 +1)))dx) let ϕ(z)= ((e^((b+ia)z) +e^((−b+ia)z) )/(2(z^2 +1))) the poles of ϕ are i and −i ∫_(−∞) ^(+∞) ϕ(z)dz=2iπRes(ϕ,i) Re(ϕ,i)= ((e^((b+ia)i) +e^((−b+ia)i) )/(2(2i))) = ((e^(−a +bi) +e^(−a−bi) )/(4i)) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ ((e^(−a +bi) +e^(−a−bi) )/(4i)) =(π/2){ e^(−a) ( cosb +isinb) +e^(−a) (cosb −isinb)} =(π/2) e^(−a) (2cosb) =π e^(−a) cosb ⇒ I =Re( ∫_(−∞) ^(+∞) ϕ(z)dz) = π e^(−a) cos(b).$
$l e t I = \int_{- \infty}^{+ \infty} \frac{c o s (a x) c h (b x)}{x^{2} + 1} d x$ $I = R e (\int_{- \infty}^{+ \infty} \frac{e^{i a x} (e^{b x} + e^{- b x})}{2 (x^{2} + 1)} d x)$ $= R e (\int_{- \infty}^{+ \infty} \frac{e^{(b + i a) x} + e^{(- b + i a) x}}{2 (x^{2} + 1)} d x) l e t$ $φ (z) = \frac{e^{(b + i a) z} + e^{(- b + i a) z}}{2 (z^{2} + 1)} t h e p o l e s o f φ a r e$ $i a n d - i$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e (φ, i) = \frac{e^{(b + i a) i} + e^{(- b + i a) i}}{2 (2 i)} = \frac{e^{- a + b i} + e^{- a - b i}}{4 i}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{e^{- a + b i} + e^{- a - b i}}{4 i}$ $= \frac{π}{2} {e^{- a} (c o s b + i s i n b) + e^{- a} (c o s b - i s i n b)}$ $= \frac{π}{2} e^{- a} (2 c o s b) = π e^{- a} c o s b \Rightarrow$ $I = R e (\int_{- \infty}^{+ \infty} φ (z) d z) = π e^{- a} c o s (b) .$
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