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

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Question Number 65837 by mathmax by abdo last updated on 04/Aug/19

$$1)calculate{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+ix}and{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1-ix}$$$$2)deducethevalueof{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+x^2}$$$$3)calculate{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+{ix}^2}and{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1-{ix}^2}$$$$4)deducethevalueof{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+x^4}$$

Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19

$$letA={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+ix}=2i\pi {lim}_{x->i}\left(\frac{x-i}{1+ix}\right)=2\pi$$$$B={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1-ix}=-2i\pi {lim}_{z->}\left(\frac{z+i}{1-iz}\right)=2\pi$$$$A+B={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{2dx}{1+x^2}\Rightarrow C={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+x^2}=\frac{A+B}{2}=A=B=2\pi$$$$D={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+{ix}^2}=2i\pi {lim}_{z->e^{i\frac{\pi }{4}}}\left(\frac{1}{z+e^{i\frac{\pi }{4}}}\right)=\pi e^{i\frac{\pi }{4}}$$$$E={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1-{ix}^2}=-2i\pi {lim}_{z->e^{-i\frac{\pi }{4}}}\left(\frac{1}{z+e^{-i\frac{\pi }{4}}}\right)=\pi e^{-i\frac{\pi }{4}}$$$$D+E={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{2dx}{1+x^4}\Rightarrow F={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+x^4}=\frac{D+E}{2}=\frac{\pi (e^{i\frac{\pi }{4}}+e^{-i\frac{\pi }{4}})}{2}=\pi cos\left(\frac{\pi }{4}\right)=\frac{\pi }{\sqrt{2}}$$

Commented by mathmax by abdo last updated on 05/Aug/19

$$3){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1+{ix}^2}=\frac{1}{i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+\frac{1}{i}}=\frac{1}{i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2-i}$$$$=\frac{1}{i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2-{\left(\sqrt{i}\right)}^2}=\frac{1}{i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2-{\left(e^{\frac{i\pi }{4}}\right)}^2}=\frac{1}{i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(x-e^{\frac{i\pi }{4}})(x+e^{\frac{i\pi }{4}})}$$$$=\frac{-i}{2e^{\frac{i\pi }{4}}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\{\frac{1}{x-e^{\frac{i\pi }{4}}}-\frac{1}{x+e^{\frac{i\pi }{4}}}\}dx$$$$=-\frac{i}{2}{e}^{-\frac{i\pi }{4}}\{{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x-e^{\frac{i\pi }{4}}}-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x+e^{\frac{i\pi }{4}}}\}$$$$=-\frac{i}{2}{e}^{-\frac{i\pi }{4}}\{i\pi -(-i\pi )\}=-\frac{i}{2}\left(2i\pi \right)e^{-\frac{i\pi }{4}}=\pi e^{-\frac{i\pi }{4}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1-{ix}^2}=conj\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1+{ix}^2}\right)=\pi e^{\frac{i\pi }{4}}$$$$4){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1+x^4}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(1+{ix}^2)(1-{ix}^2)}$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\{\frac{1}{1-{ix}^2}+\frac{1}{1+{ix}^2}\}dx=\frac{1}{2}\{\pi e^{\frac{i\pi }{4}}+\pi e^{-\frac{i\pi }{4}}\}$$$$=\frac{\pi }{2}(2cos\left(\frac{\pi }{4}\right)=\pi \frac{\sqrt{2}}{2}=\frac{\pi }{\sqrt{2}}.$$$$$$

Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19

$$Pleasecheckiftherearesomemistake:causenormallyC=2{\left[arctanx\right]}_0^{\mathrm{\infty }}=\pi$$

Commented by mathmax by abdo last updated on 05/Aug/19

$$1)wehaveprovedthst{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x-a}=i\pi ifim\left(a\right)>0and-i\pi if$$$$im\left(a\right)<0so{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1+ix}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{i(x+\frac{1}{i})}=\frac{1}{i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x-i}=\frac{1}{i}\left(i\pi \right)$$$$=\pi also{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1-ix}=\frac{1}{i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{\frac{1}{i}-x}=-\frac{1}{i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x-\frac{1}{i}}$$$$=-\frac{1}{i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x+i}=-\frac{1}{i}(-i\pi )=\pi$$$$2){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1+x^2}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(x-i)(x+i)}=\frac{1}{2i}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\{\frac{1}{x-i}-\frac{1}{x+i}\}dx$$$$=\frac{1}{2i}\{{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x-i}-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x+i}\}=\frac{1}{2i}\{i\pi -(-i\pi )\}=\pi$$

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