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Question Number 148064 by tabata last updated on 25/Jul/21

Commented by tabata last updated on 25/Jul/21

$$helpmesirincomplexnumber$$

Answered by Olaf_Thorendsen last updated on 25/Jul/21

$$\mathrm{Q}6.$$$$\mathrm{I}={\int }_0^{\pi }\frac{d\theta }{2-\cos \theta }$$$$\mathrm{Let}t=\tan \frac{\theta }{2}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{1}{2-\frac{1-t^2}{1+t^2}}.\frac{2dt}{1+t^2}$$$$\mathrm{I}=2{\int }_0^{\mathrm{\infty }}\frac{dt}{2(1+t^2)-(1-t^2)}$$$$\mathrm{I}=2{\int }_0^{\mathrm{\infty }}\frac{dt}{3t^2+1}=\frac{2}{\sqrt{3}}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{3}dt}{3t^2+1}$$$$\mathrm{I}=\frac{2}{\sqrt{3}}{\left[\arctan \left(\sqrt{3}t\right)\right]}_0^{\mathrm{\infty }}=\frac{\pi }{\sqrt{3}}$$

Answered by Olaf_Thorendsen last updated on 25/Jul/21

$$\mathrm{J}={\int }_0^{\mathrm{\infty }}\frac{\cos \left(2x\right)}{x^2+1}dx=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(2x\right)}{x^2+1}dx$$$$\mathrm{J}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{2ix}}{x^2+1}dx$$$$\mathrm{J}=\frac{1}{2}.(2i\pi .\mathrm{Res}(\frac{e^{2ix}}{x^2+1},+i))$$$$\mathrm{J}=i\pi .\underset{x\to +i}{\lim }\left(\frac{e^{2ix}}{x+i}\right)=i\pi .\frac{e^{-2}}{2i}=\frac{\pi }{2e^2}$$

Answered by qaz last updated on 25/Jul/21

$${\int }_0^{\mathrm{\infty }}\frac{\mathcal{L}\left\{\cos \left(\mathrm{ax}\right)\right\}}{{\mathrm{x}}^2+1}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{s}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+{\mathrm{s}}^2)}\mathrm{dx}=\frac{\pi }{2(\mathrm{s}+1)}$$$${\int }_0^{\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{\mathrm{x}}^2+1}\mathrm{dx}={\mathcal{L}}^{-1}\left\{\frac{\pi }{2(\mathrm{s}+1)}\right\}(\mathrm{a}=2)=\frac{\pi }{{2\mathrm{e}}^2}$$

Answered by mathmax by abdo last updated on 25/Jul/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{\mathrm{x}}^2+1}\mathrm{dx}\Rightarrow 2\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{2\mathrm{ix}}}{{\mathrm{x}}^2+1}\mathrm{dx}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{2\mathrm{iz}}}{{\mathrm{z}}^2+1}$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{2\mathrm{iz}}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})$$$$=2\mathrm{i}\pi \times \frac{{\mathrm{e}}^{2\mathrm{i}\left(\mathrm{i}\right)}}{2\mathrm{i}}=\pi {\mathrm{e}}^{-2}\Rightarrow \mathrm{I}=\frac{\pi }{{2\mathrm{e}}^2}★$$

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