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

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

$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{cos}\left({ax}\right){ch}\left({bx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:. \\ $$

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).

$${let}\:{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({ax}\right){ch}\left({bx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$${I}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{iax}} \:\left({e}^{{bx}} \:+{e}^{−{bx}} \right)}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}\right) \\ $$$$={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{e}^{\left({b}+{ia}\right){x}} \:+{e}^{\left(−{b}+{ia}\right){x}} }{\mathrm{2}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{dx}\right)\:{let} \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{\left({b}+{ia}\right){z}} \:+{e}^{\left(−{b}+{ia}\right){z}} }{\mathrm{2}\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${i}\:{and}\:−{i} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\right) \\ $$$${Re}\left(\varphi,{i}\right)=\:\frac{{e}^{\left({b}+{ia}\right){i}} \:+{e}^{\left(−{b}+{ia}\right){i}} }{\mathrm{2}\left(\mathrm{2}{i}\right)}\:=\:\frac{{e}^{−{a}\:+{bi}} \:+{e}^{−{a}−{bi}} }{\mathrm{4}{i}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:\frac{{e}^{−{a}\:+{bi}} \:+{e}^{−{a}−{bi}} }{\mathrm{4}{i}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{\:{e}^{−{a}} \left(\:{cosb}\:+{isinb}\right)\:+{e}^{−{a}} \left({cosb}\:−{isinb}\right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\:{e}^{−{a}} \left(\mathrm{2}{cosb}\right)\:=\pi\:{e}^{−{a}} {cosb}\:\Rightarrow \\ $$$${I}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\right)\:=\:\pi\:{e}^{−{a}} \:{cos}\left({b}\right). \\ $$

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