Question Number 110027 by 150505R last updated on 26/Aug/20
Answered by mnjuly1970 last updated on 26/Aug/20
$${ans}:::\:\:\:\frac{\mathrm{1}}{\pi}\:.\frac{\pi}{{e}^{{a}} }=\frac{\mathrm{1}}{{e}^{{a}} }\:\clubsuit \\ $$
Commented by 150505R last updated on 26/Aug/20
$${but}\:{how}\:? \\ $$
Answered by mathmax by abdo last updated on 26/Aug/20
$$\:\mathrm{I}_{\mathrm{a}} =\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{ax}\right)}{\pi\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\mathrm{dx}\:\Rightarrow\pi\:\mathrm{I}_{\mathrm{a}} =\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{iax}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\right)\:\mathrm{let} \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{iaz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{iaz}} }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)}\:\mathrm{rdsidus}\:\mathrm{theorem}\:\mathrm{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:=\mathrm{2i}\pi×\frac{\mathrm{e}^{\mathrm{ia}\left(\mathrm{i}\right)} }{\mathrm{2i}}\:=\pi\:\mathrm{e}^{−\mathrm{a}} \:\Rightarrow \\ $$$$\pi\mathrm{I}_{\mathrm{a}} =\pi\mathrm{e}^{−\mathrm{a}} \:\Rightarrow\:\bigstar\:\mathrm{I}_{\mathrm{a}} =\mathrm{e}^{−\mathrm{a}} \:\:\bigstar\:\:\:\:\left(\mathrm{a}\geqslant\mathrm{0}\right) \\ $$
Commented by mohammad17 last updated on 27/Aug/20
$${sir}\:{whats}\:{the}\:{mean}\:{Res}\:{can}\:{you}\:{tel}\:{me} \\ $$
Commented by mathdave last updated on 27/Aug/20
$${that}\:{is}\:{call}\:{residue} \\ $$
Commented by 1549442205PVT last updated on 27/Aug/20
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