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Question Number 133213 by mathmax by abdo last updated on 20/Feb/21

$$\mathrm{calculate}\mathrm{c}\left(\xi \right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\xi \mathrm{x}\right)}{1+{\mathrm{x}}^4}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 20/Feb/21

$$\mathrm{\Phi }\left(\xi \right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\xi \mathrm{x}\right)}{{\mathrm{x}}^4+1}\mathrm{dx}\Rightarrow 2\mathrm{\Phi }\left(\xi \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(\xi \mathrm{x}\right)}{{\mathrm{x}}^4+1}\mathrm{dx}$$$$=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{i}\xi \mathrm{x}}}{{\mathrm{x}}^4+1}\mathrm{dx}\right)\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{i}\xi \mathrm{z}}}{{\mathrm{z}}^4+1}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{i}\xi \mathrm{z}}}{({\mathrm{z}}^2-\mathrm{i})({\mathrm{z}}^2+\mathrm{i})}$$$$=\frac{{\mathrm{e}}^{\mathrm{i}\xi \mathrm{z}}}{(\mathrm{z}-\sqrt{\mathrm{i}})(\mathrm{z}+\sqrt{\mathrm{i}})(\mathrm{z}-\sqrt{-\mathrm{i}})(\mathrm{z}+\sqrt{-\mathrm{i}})}=\frac{{\mathrm{e}}^{\mathrm{i}\xi \mathrm{z}}}{(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})}$$$$\mathrm{and}{\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})+\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})\}$$$$\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})=\frac{{\mathrm{e}}^{\mathrm{i}\xi {\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}}}{{2\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\left(2\mathrm{i}\right)}=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}.{\mathrm{e}}^{\mathrm{i}\xi (\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}})}$$$$=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}.{\mathrm{e}}^{\frac{\mathrm{i}\xi }{\sqrt{2}}}.{\mathrm{e}}^{-\frac{\xi }{\sqrt{2}}}=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\xi }{\sqrt{2}}}{\mathrm{e}}^{\mathrm{i}(-\frac{\pi }{4}+\frac{\xi }{\sqrt{2}})}$$$$\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})=\frac{{\mathrm{e}}^{-\mathrm{i}\xi {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}}}{-{2\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}(-2\mathrm{i})}=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}.{\mathrm{e}}^{-\mathrm{i}\xi (\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}})}$$$$=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}.{\mathrm{e}}^{-\frac{\mathrm{i}\xi }{\sqrt{2}}}.{\mathrm{e}}^{-\frac{\xi }{\sqrt{2}}}=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\xi }{\sqrt{2}}}{\mathrm{e}}^{\mathrm{i}(\frac{\pi }{4}-\frac{\xi }{\sqrt{2}})}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi .\frac{1}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\xi }{\sqrt{2}}}\{{\mathrm{e}}^{\mathrm{i}(-\frac{\pi }{4}+\frac{\xi }{\sqrt{2}})}+{\mathrm{e}}^{\mathrm{i}(\frac{\pi }{4}-\frac{\xi }{\sqrt{2}})}\}$$$$=\frac{\pi }{2}{\mathrm{e}}^{-\frac{\xi }{\sqrt{2}}}\{{\mathrm{e}}^{\mathrm{i}(\frac{\pi }{4}-\frac{\xi }{\sqrt{2}})}+{\mathrm{e}}^{-\mathrm{i}(\frac{\pi }{4}-\frac{\xi }{\sqrt{2}})}\}$$$$=\frac{\pi }{2}{\mathrm{e}}^{-\frac{\xi }{\sqrt{2}}}\left\{2\cos (\frac{\pi }{4}-\frac{\xi }{\sqrt{2}})\right\}={\mathrm{e}}^{-\frac{\xi }{\sqrt{2}}}\cos (\frac{\pi }{4}-\frac{\xi }{\sqrt{2}})=2\mathrm{\Phi }\left(\xi \right)\Rightarrow$$$$\mathrm{\Phi }\left(\xi \right)=\frac{1}{2}{\mathrm{e}}^{-\frac{\xi }{\sqrt{2}}}\cos (\frac{\pi }{4}-\frac{\xi }{\sqrt{2}})$$

Commented by mnjuly1970 last updated on 21/Feb/21

$$niceverynicesirMax..$$

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