1-calculate-dx-1-ix-and-dx-1-ix-2-deduce-the-value-of-dx-1-x-2-3-calculate-dx-1-ix-2-and-dx-1-ix-2-4-deduce-the-value

Question Number 65837 by mathmax by abdo last updated on 04/Aug/19
$$\left.\mathrm{1}\right)\:{calculate}\:\int_{−\infty} ^{\infty} \:\frac{{dx}}{\mathrm{1}+{ix}}\:\:{and}\:\int_{−\infty} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}−{ix}} \\$$$$\left.\mathrm{2}\right){deduce}\:{the}\:{value}\:{of}\:\int_{−\infty} ^{\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\$$$$\left.\mathrm{3}\right){calculate}\:\int_{−\infty} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{ix}^{\mathrm{2}} }\:\:{and}\:\int_{−\infty} ^{\infty} \:\frac{{dx}}{\mathrm{1}−{ix}^{\mathrm{2}} } \\$$$$\left.\mathrm{4}\right){deduce}\:{the}\:{value}\:{of}\:\int_{−\infty} ^{\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} } \\$$
Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19
$$\:\:\:{let}\:\:{A}=\int_{−\infty} ^{\infty} \:\frac{{dx}}{\mathrm{1}+{ix}}=\mathrm{2}{i}\pi{lim}_{{x}−>{i}} \:\left(\frac{{x}−{i}}{\mathrm{1}+{ix}}\right)\:=\mathrm{2}\pi \\$$$${B}=\int_{−\infty} ^{\infty} \frac{{dx}}{\mathrm{1}−{ix}}=−\mathrm{2}{i}\pi\:{lim}_{{z}−>\:\:\:\:} \:\left(\frac{{z}+{i}}{\mathrm{1}−{iz}}\right)=\mathrm{2}\pi \\$$$${A}+{B}=\int_{−\infty} ^{\infty} \:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow\:{C}=\int_{−\infty} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{{A}+{B}}{\mathrm{2}}={A}={B}=\mathrm{2}\pi \\$$$${D}=\int_{−\infty} ^{\infty} \frac{{dx}}{\mathrm{1}+{ix}^{\mathrm{2}} }\:=\:\mathrm{2}{i}\pi\:{lim}_{{z}−>{e}^{{i}\frac{\pi}{\mathrm{4}}} } \:\left(\frac{\mathrm{1}}{{z}+{e}^{{i}\frac{\pi}{\mathrm{4}}} }\right)=\pi{e}^{{i}\frac{\pi}{\mathrm{4}}} \\$$$${E}=\int_{−\infty} ^{\infty} \:\frac{{dx}}{\mathrm{1}−{ix}^{\mathrm{2}} }=\:−\mathrm{2}{i}\pi\:{lim}_{{z}−>\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} } \:\left(\frac{\mathrm{1}}{{z}+{e}^{−{i}\frac{\pi}{\mathrm{4}}} }\right)=\pi{e}^{−{i}\frac{\pi}{\mathrm{4}}} \: \\$$$${D}+{E}=\int_{−\infty} ^{\infty} \frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:\Rightarrow\:{F}=\int_{−\infty} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:=\:\frac{{D}+{E}}{\mathrm{2}}=\frac{\pi\left({e}^{{i}\frac{\pi}{\mathrm{4}}} +{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)}{\mathrm{2}}=\pi{cos}\left(\frac{\pi}{\mathrm{4}}\right)=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\$$
Commented by mathmax by abdo last updated on 05/Aug/19
$$\left.\mathrm{3}\right)\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\mathrm{1}+{ix}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{i}}\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{i}}}\:=\frac{\mathrm{1}}{{i}}\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} −{i}} \\$$$$=\frac{\mathrm{1}}{{i}}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} −\left(\sqrt{{i}}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{i}}\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −\left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{i}}\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)} \\$$$$=\frac{−{i}}{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\int_{−\infty} ^{+\infty} \left\{\:\frac{\mathrm{1}}{{x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}+{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right\}{dx} \\$$$$=−\frac{{i}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \left\{\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:−\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}+{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right\} \\$$$$=−\frac{{i}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\left\{{i}\pi−\left(−{i}\pi\right)\right\}\:=−\frac{{i}}{\mathrm{2}}\left(\mathrm{2}{i}\pi\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \:=\pi\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\$$$$\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\mathrm{1}−{ix}^{\mathrm{2}} }\:={conj}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\mathrm{1}+{ix}^{\mathrm{2}} }\right)\:=\pi\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \\$$$$\left.\mathrm{4}\right)\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{ix}^{\mathrm{2}} \right)\left(\mathrm{1}−{ix}^{\mathrm{2}} \right)} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\:\left\{\frac{\mathrm{1}}{\mathrm{1}−{ix}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{ix}^{\mathrm{2}} }\right\}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\pi\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:+\pi\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right\} \\$$$$=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)\:=\pi\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:.\right. \\$$$$\\$$
Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19
$${Please}\:{check}\:{if}\:{there}\:{are}\:{some}\:{mistake}\::\:{cause}\:{normally}\:{C}=\mathrm{2}\left[{arctanx}\right]_{\mathrm{0}} ^{\infty} =\pi \\$$
Commented by mathmax by abdo last updated on 05/Aug/19
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{proved}\:{thst}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{a}}\:={i}\pi\:{if}\:{im}\left({a}\right)>\mathrm{0}\:{and}\:−{i}\pi\:{if} \\$$$${im}\left({a}\right)<\mathrm{0}\:\:{so}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\mathrm{1}+{ix}}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{i}\left({x}+\frac{\mathrm{1}}{{i}}\right)}\:=\frac{\mathrm{1}}{{i}}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}−{i}}\:=\frac{\mathrm{1}}{{i}}\left({i}\pi\right) \\$$$$=\pi\:\:{also}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\mathrm{1}−{ix}}\:=\frac{\mathrm{1}}{{i}}\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\frac{\mathrm{1}}{{i}}−{x}}\:=−\frac{\mathrm{1}}{{i}}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−\frac{\mathrm{1}}{{i}}} \\$$$$=−\frac{\mathrm{1}}{{i}}\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}+{i}}\:=−\frac{\mathrm{1}}{{i}}\left(−{i}\pi\right)\:=\pi \\$$$$\left.\mathrm{2}\right)\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}−{i}\right)\left({x}+{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{−\infty} ^{+\infty} \left\{\frac{\mathrm{1}}{{x}−{i}}−\frac{\mathrm{1}}{{x}+{i}}\right\}{dx} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{i}}\:−\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}+{i}}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{{i}\pi\:−\left(−{i}\pi\right)\right\}\:=\pi \\$$