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Question Number 117221 by bemath last updated on 10/Oct/20

$$\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{\cos \left(3x\right)}{x^2+4}dx=?$$

Commented by bemath last updated on 10/Oct/20

$$thankyousirs$$

Answered by AbduraufKodiriy last updated on 10/Oct/20

Answered by bobhans last updated on 10/Oct/20

$$\mathrm{Note}\int \underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{}}\left(\frac{1}{{\mathrm{x}}^2+4}\right)\mathrm{dx}=\frac{\pi }{2}$$$$\int \underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{}}\frac{\sin \mathrm{x}}{\mathrm{x}}\mathrm{dx}=\frac{\pi }{2}$$$$\mathrm{I}\left(\mathrm{t}\right)=2\int \underset{0}{\stackrel{\mathrm{\infty }}{}}\frac{\cos \left(\mathrm{tx}\right)}{{\mathrm{x}}^2+4}\mathrm{dx};\mathrm{I}\left(0\right)=\frac{\pi }{2}$$$${\mathrm{I}}^{\prime }\left(\mathrm{t}\right)=-2\int \underset{0}{\stackrel{\mathrm{\infty }}{}}\frac{\mathrm{xsin}\left(\mathrm{tx}\right)}{{\mathrm{x}}^2+4}\mathrm{dx}=-2\int \underset{0}{\stackrel{\mathrm{\infty }}{}}\frac{{\mathrm{x}}^2\sin \left(\mathrm{tx}\right)}{\mathrm{x}({\mathrm{x}}^2+4)}$$$${\mathrm{I}}^{\prime }\left(\mathrm{t}\right)=-2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{({\mathrm{x}}^2+4)\sin \left(\mathrm{tx}\right)}{\mathrm{x}({\mathrm{x}}^2+4)}\mathrm{dx}+8\int \underset{0}{\stackrel{\mathrm{\infty }}{}}\frac{\sin \left(\mathrm{tx}\right)}{\mathrm{x}({\mathrm{x}}^2+4)}\mathrm{dx}$$$$=-2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\sin \mathrm{u}}{\mathrm{u}}\mathrm{du}+8\int \underset{0}{\stackrel{\mathrm{\infty }}{}}\frac{\sin \left(\mathrm{tx}\right)}{\mathrm{x}({\mathrm{x}}^2+4)}\mathrm{dx}$$$$=-\pi +8\int \underset{0}{\stackrel{\mathrm{\infty }}{}}\frac{\sin \left(\mathrm{tx}\right)}{\mathrm{x}({\mathrm{x}}^2+4)}\mathrm{dx}$$$${\mathrm{I}}^{″}\left(\mathrm{t}\right)=8\int \underset{0}{\stackrel{\mathrm{\infty }}{}}\frac{\cos \left(\mathrm{tx}\right)}{{\mathrm{x}}^2+4}\mathrm{dx}=4\mathrm{I}\left(\mathrm{t}\right)$$$$\mathrm{I}\left(\mathrm{t}\right)={\mathrm{C}}_1{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{-2\mathrm{t}}$$$$\Rightarrow {\mathrm{C}}_1+{\mathrm{C}}_2=\frac{\pi }{2};{2\mathrm{C}}_1-{2\mathrm{C}}_2=0$$$${\mathrm{C}}_1=0\wedge {\mathrm{C}}_2=\frac{\pi }{2}$$$$\mathrm{I}\left(\mathrm{t}\right)=\frac{\pi }{2}{\mathrm{e}}^{-2\mathrm{t}}\iff \mathrm{I}\left(3\right)=\frac{\pi }{2{\mathrm{e}}^6}$$$$$$

Answered by Bird last updated on 11/Oct/20

$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(3x\right)}{x^2+4}dx\Rightarrow$$$$I{=}_{x=2t}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(6t\right)}{4(t^2+1)}\left(2dt\right)$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(6t\right)}{t^2+1}dt$$$$=\frac{1}{2}Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{6it}}{t^2+1}dt\right)but$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{6iz}}{z^2+1}dz?=2i\pi Res(f,i)$$$$=2i\pi .\frac{e^{6i\left(i\right)}}{2i}=\pi e^{-6}=\frac{\pi }{e^6}\Rightarrow$$$$I=\frac{\pi }{2e^6}$$

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