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# calculate-A-0-e-x-2-x-4-1-dx-with-gt-0-and-find-0-1-A-d-

Question Number 68595 by Abdo msup. last updated on 14/Sep/19
$${calculate}\:\:{A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−\lambda{x}^{\mathrm{2}} } }{{x}^{\mathrm{4}} +\mathrm{1}}{dx}\:\:{with}\:\lambda>\mathrm{0}\:\:{and}\: \\$$$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{A}_{\lambda} \:{d}\lambda \\$$
Commented by mathmax by abdo last updated on 14/Sep/19
$${A}_{\lambda} =\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−\lambda{x}^{\mathrm{2}} } }{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\:\Rightarrow\mathrm{2}{A}_{\lambda} =\int_{−\infty} ^{+\infty} \:\frac{{e}^{−\lambda{x}^{\mathrm{2}} } }{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\:{let}\:{W}\left({z}\right)\:=\frac{{e}^{−\lambda{z}^{\mathrm{2}} } }{{z}^{\mathrm{4}} +\mathrm{1}} \\$$$${poles}\:{of}\:{W}?\:\:{we}\:{have}\:{W}\left({z}\right)\:=\frac{{e}^{−\lambda{x}^{\mathrm{2}} } }{\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} +{i}\right)} \\$$$$=\frac{{e}^{−\lambda{x}^{\mathrm{2}} } }{\left({z}−\sqrt{{i}}\right)\left({z}+\sqrt{{i}}\right)\left({z}−\sqrt{−{i}}\right)\left(\mathrm{2}+\sqrt{−{i}}\right)}\:=\frac{{e}^{−\lambda{x}^{\mathrm{2}} } }{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\$$$${so}\:{the}\:{poles}\:{of}\:{W}\:{are}\:\overset{−} {+}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{and}\:\overset{−} {+}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\$$$${residus}\:{tbeorem}\:{give}\: \\$$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{{Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\$$$${Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{{e}^{−\lambda\left({i}\right)} }{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{−\lambda{i}−\frac{{i}\pi}{\mathrm{4}}} \:=\frac{\mathrm{1}}{\mathrm{4}{i}}{e}^{−\left(\lambda+\frac{\pi}{\mathrm{4}}\right){i}} \\$$$${Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\:\frac{{e}^{−\lambda\left(−{i}\right)} }{\left(−\mathrm{2}{i}\right)\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{\lambda{i}+\frac{{i}\pi}{\mathrm{4}}} =\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{\left(\lambda+\frac{\pi}{\mathrm{4}}\right){i}} \:\Rightarrow \\$$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{\mathrm{1}}{\mathrm{4}{i}}{e}^{−\left(\lambda+\frac{\pi}{\mathrm{4}}\right){i}} \:+\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{\left(\lambda+\frac{\pi}{\mathrm{4}}\right){i}} \right\} \\$$$$=\frac{\pi}{\mathrm{2}}\left\{\:\mathrm{2}{Re}\left({e}^{\left(\lambda+\frac{\pi}{\mathrm{4}}\right){i}} \right)\:=\pi\:\:{cos}\left(\frac{\pi}{\mathrm{4}}\:+\lambda\right)\:\Rightarrow{A}_{\lambda} =\frac{\pi}{\mathrm{2}}{cos}\left(\frac{\pi}{\mathrm{4}}\:+\lambda\right)\right. \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{A}_{\lambda} {d}\lambda\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{cos}\left(\lambda+\frac{\pi}{\mathrm{4}}\right)\:=\left[{sin}\left(\lambda+\frac{\pi}{\mathrm{4}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:={sin}\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}\right)−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\$$