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Question Number 110027 by 150505R last updated on 26/Aug/20

Answered by mnjuly1970 last updated on 26/Aug/20

ans::: (1/π) .(π/e^a )=(1/e^a ) ♣

$${ans}:::\:\:\:\frac{\mathrm{1}}{\pi}\:.\frac{\pi}{{e}^{{a}} }=\frac{\mathrm{1}}{{e}^{{a}} }\:\clubsuit \\ $$

Commented by 150505R last updated on 26/Aug/20

but how ?

$${but}\:{how}\:? \\ $$

Answered by mathmax by abdo last updated on 26/Aug/20

I_a =∫_(−∞) ^(+∞) ((cos(ax))/(π(x^2 +1))) dx ⇒π I_a =Re(∫_(−∞) ^(+∞) (e^(iax) /(x^2 +1))dx) let ϕ(z) =(e^(iaz) /(z^2 +1)) ⇒ϕ(z) =(e^(iaz) /((z−i)(z+i))) rdsidus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) =2iπ×(e^(ia(i)) /(2i)) =π e^(−a) ⇒ πI_a =πe^(−a) ⇒ ★ I_a =e^(−a) ★ (a≥0)

$$\:\mathrm{I}_{\mathrm{a}} =\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{ax}\right)}{\pi\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\mathrm{dx}\:\Rightarrow\pi\:\mathrm{I}_{\mathrm{a}} =\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{iax}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\right)\:\mathrm{let} \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{iaz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{iaz}} }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)}\:\mathrm{rdsidus}\:\mathrm{theorem}\:\mathrm{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:=\mathrm{2i}\pi×\frac{\mathrm{e}^{\mathrm{ia}\left(\mathrm{i}\right)} }{\mathrm{2i}}\:=\pi\:\mathrm{e}^{−\mathrm{a}} \:\Rightarrow \\ $$$$\pi\mathrm{I}_{\mathrm{a}} =\pi\mathrm{e}^{−\mathrm{a}} \:\Rightarrow\:\bigstar\:\mathrm{I}_{\mathrm{a}} =\mathrm{e}^{−\mathrm{a}} \:\:\bigstar\:\:\:\:\left(\mathrm{a}\geqslant\mathrm{0}\right) \\ $$

Commented by mohammad17 last updated on 27/Aug/20

sir whats the mean Res can you tel me

$${sir}\:{whats}\:{the}\:{mean}\:{Res}\:{can}\:{you}\:{tel}\:{me} \\ $$

Commented by mathdave last updated on 27/Aug/20

that is call residue

$${that}\:{is}\:{call}\:{residue} \\ $$

Commented by 1549442205PVT last updated on 27/Aug/20

You want know above it then you need must read books writting above function of complex variable.This knowledge only teached at college level or university

$$\mathrm{You}\:\mathrm{want}\:\mathrm{know}\:\mathrm{above}\:\mathrm{it}\:\mathrm{then}\:\mathrm{you}\:\mathrm{need} \\ $$$$\mathrm{must}\:\mathrm{read}\:\mathrm{books}\:\mathrm{writting}\:\mathrm{above} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{variable}.\mathrm{This}\: \\ $$$$\mathrm{knowledge}\:\mathrm{only}\:\mathrm{teached}\:\mathrm{at}\:\mathrm{college} \\ $$$$\mathrm{level}\:\mathrm{or}\:\mathrm{university} \\ $$

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