Question Number 38467 by maxmathsup by imad last updated on 25/Jun/18
$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{cos}\left({ax}\right){ch}\left({bx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:. \\ $$
Commented by math khazana by abdo last updated on 28/Jun/18
$${let}\:{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({ax}\right){ch}\left({bx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$${I}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{iax}} \:\left({e}^{{bx}} \:+{e}^{−{bx}} \right)}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}\right) \\ $$$$={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{e}^{\left({b}+{ia}\right){x}} \:+{e}^{\left(−{b}+{ia}\right){x}} }{\mathrm{2}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{dx}\right)\:{let} \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{\left({b}+{ia}\right){z}} \:+{e}^{\left(−{b}+{ia}\right){z}} }{\mathrm{2}\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${i}\:{and}\:−{i} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\right) \\ $$$${Re}\left(\varphi,{i}\right)=\:\frac{{e}^{\left({b}+{ia}\right){i}} \:+{e}^{\left(−{b}+{ia}\right){i}} }{\mathrm{2}\left(\mathrm{2}{i}\right)}\:=\:\frac{{e}^{−{a}\:+{bi}} \:+{e}^{−{a}−{bi}} }{\mathrm{4}{i}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:\frac{{e}^{−{a}\:+{bi}} \:+{e}^{−{a}−{bi}} }{\mathrm{4}{i}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{\:{e}^{−{a}} \left(\:{cosb}\:+{isinb}\right)\:+{e}^{−{a}} \left({cosb}\:−{isinb}\right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\:{e}^{−{a}} \left(\mathrm{2}{cosb}\right)\:=\pi\:{e}^{−{a}} {cosb}\:\Rightarrow \\ $$$${I}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\right)\:=\:\pi\:{e}^{−{a}} \:{cos}\left({b}\right). \\ $$