Question-147322

Question Number 147322 by Gbenga last updated on 19/Jul/21

Commented by Mrsof last updated on 19/Jul/21

I=∫_0 ^( ∞) ((cos(ax))/(x^2 +b^2 ))dx=_(x=z) 2I=∫_(−∞) ^( ∞) ((cos(az))/(z^2 +b^2 ))dz 2I=Re(∫_(−∞) ^( ∞) (e^(aiz) /((z−bi)(z+bi)))dz) the function has a resideo of z=bi Res(f,bi)=lim_(z→bi) (z−bi)×(e^(aiz) /((z−bi)(z+bi)))=(e^(−ab) /(2bi)) 2I=2πi(Res(f,bi))⇒I=πi×(e^(−ab) /(2bi))=((πe^(−ab) )/(2b)) ⟨M.T⟩

$${I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({ax}\right)}{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} }{dx}=_{{x}={z}} \mathrm{2}{I}=\int_{−\infty} ^{\:\infty} \frac{{cos}\left({az}\right)}{{z}^{\mathrm{2}} +{b}^{\mathrm{2}} }{dz} \\ $$$$ \\ $$$$\mathrm{2}{I}={Re}\left(\int_{−\infty} ^{\:\infty} \frac{{e}^{{aiz}} }{\left({z}−{bi}\right)\left({z}+{bi}\right)}{dz}\right) \\ $$$$ \\ $$$${the}\:{function}\:{has}\:{a}\:{resideo}\:{of}\:{z}={bi} \\ $$$$ \\ $$$${Res}\left({f},{bi}\right)={lim}_{{z}\rightarrow{bi}} \left({z}−{bi}\right)×\frac{{e}^{{aiz}} }{\left({z}−{bi}\right)\left({z}+{bi}\right)}=\frac{{e}^{−{ab}} }{\mathrm{2}{bi}} \\ $$$$ \\ $$$$\mathrm{2}{I}=\mathrm{2}\pi{i}\left({Res}\left({f},{bi}\right)\right)\Rightarrow{I}=\pi{i}×\frac{{e}^{−{ab}} }{\mathrm{2}{bi}}=\frac{\pi{e}^{−{ab}} }{\mathrm{2}{b}} \\ $$$$ \\ $$$$\langle{M}.{T}\rangle \\ $$

Commented by Gbenga last updated on 19/Jul/21

$${thanks} \\ $$

Commented by Mrsof last updated on 20/Jul/21

$${you}\:{are}\:{welcome} \\ $$

Answered by qaz last updated on 20/Jul/21

∫_0 ^∞ ((L(cos (ax)))/(x^2 +b^2 ))dx =∫_0 ^∞ (s/((x^2 +b^2 )(x^2 +s^2 )))dx =(s/(s^2 −b^2 ))∫_0 ^∞ ((1/(x^2 +b^2 ))−(1/(x^2 +s^2 )))dx =(π/(2b(s+b))) ⇒ ∫_0 ^∞ ((cos (ax))/(x^2 +b^2 ))dx=L^(−1) ((π/(2b(s+b))))=(π/(2b))e^(−ba)

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathscr{L}\left(\mathrm{cos}\:\left(\mathrm{ax}\right)\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{s}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$=\frac{\mathrm{s}}{\mathrm{s}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} }\right)\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{2b}\left(\mathrm{s}+\mathrm{b}\right)} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\left(\mathrm{ax}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\mathrm{dx}=\mathscr{L}^{−\mathrm{1}} \left(\frac{\pi}{\mathrm{2b}\left(\mathrm{s}+\mathrm{b}\right)}\right)=\frac{\pi}{\mathrm{2b}}\mathrm{e}^{−\mathrm{ba}} \\ $$

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