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Question Number 200061 by universe last updated on 13/Nov/23

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x\sin x}{(x^2+1)(x^2+4)}dx=??$$

Answered by witcher3 last updated on 13/Nov/23

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\mathrm{xsin}\left(\mathrm{x}\right)}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+4)}=\mathrm{Im}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{xe}}^{\mathrm{ix}}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+4)}\mathrm{dx}=\mathrm{A}$$$$\mathrm{Residue}\mathrm{Theorem}:$$$$\mathrm{over}{\mathrm{C}}_{\mathrm{R}},{\mathrm{Re}}^{\mathrm{ia}};\mathrm{a}\in [0,\pi ]\mathrm{and}\mathrm{R}\to \mathrm{\infty }$$$${\mathrm{e}}^{\mathrm{iRcos}\left(\mathrm{a}\right)-\mathrm{Rsin}\left(\mathrm{a}\right)}\to 0$$$$\mathrm{A}=2\mathrm{i}\pi .[\mathrm{Res}(\mathrm{f},\mathrm{i})+\mathrm{Res}(\mathrm{f},2\mathrm{i})]$$$$2\mathrm{i}\pi .[\frac{{\mathrm{ie}}^{-1}}{2\mathrm{i}\left(3\right)}+\frac{{2\mathrm{ie}}^{-2}}{(1-4)\left(4\mathrm{i}\right)}]$$$$=\mathrm{i}\pi [\frac{1}{3\mathrm{e}}-\frac{1}{{3\mathrm{e}}^2}]$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\mathrm{xsin}\left(\mathrm{x}\right)}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+4)}=\frac{\pi }{3\mathrm{e}}(1-{\mathrm{e}}^{-1})$$

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