Question Number 200061 by universe last updated on 13/Nov/23
$$\:\:\:\:\int_{−\infty} ^{+\infty} \frac{{x}\mathrm{sin}{x}\:}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}\:\:=\:\:\:?? \\ $$
Answered by witcher3 last updated on 13/Nov/23
![∫_(−∞) ^∞ ((xsin(x))/((x^2 +1)(x^2 +4)))=Im∫_(−∞) ^∞ ((xe^(ix) )/((x^2 +1)(x^2 +4)))dx=A Residue Theorem: over C_R ,Re^(ia) ;a∈[0,π] and R→∞ e^(iRcos(a)−Rsin(a)) →0 A=2iπ.[Res(f,i)+Res(f,2i)] 2iπ.[((ie^(−1) )/(2i(3)))+((2ie^(−2) )/((1−4)(4i)))] =iπ[(1/(3e))−(1/(3e^2 ))] ∫_(−∞) ^∞ ((xsin(x))/((x^2 +1)(x^2 +4)))=(π/(3e))(1−e^(−1) )](https://www.tinkutara.com/question/Q200072.png)
$$\int_{−\infty} ^{\infty} \frac{\mathrm{xsin}\left(\mathrm{x}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}=\mathrm{Im}\int_{−\infty} ^{\infty} \frac{\mathrm{xe}^{\mathrm{ix}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}\mathrm{dx}=\mathrm{A} \\ $$$$\mathrm{Residue}\:\mathrm{Theorem}: \\ $$$$\mathrm{over}\:\mathrm{C}_{\mathrm{R}} ,\mathrm{Re}^{\mathrm{ia}} ;\mathrm{a}\in\left[\mathrm{0},\pi\right]\:\mathrm{and}\:\mathrm{R}\rightarrow\infty \\ $$$$\mathrm{e}^{\mathrm{iRcos}\left(\mathrm{a}\right)−\mathrm{Rsin}\left(\mathrm{a}\right)} \rightarrow\mathrm{0} \\ $$$$\mathrm{A}=\mathrm{2i}\pi.\left[\mathrm{Res}\left(\mathrm{f},\mathrm{i}\right)+\mathrm{Res}\left(\mathrm{f},\mathrm{2i}\right)\right] \\ $$$$\mathrm{2i}\pi.\left[\frac{\mathrm{ie}^{−\mathrm{1}} }{\mathrm{2i}\left(\mathrm{3}\right)}+\frac{\mathrm{2ie}^{−\mathrm{2}} }{\left(\mathrm{1}−\mathrm{4}\right)\left(\mathrm{4i}\right)}\right] \\ $$$$=\mathrm{i}\pi\left[\frac{\mathrm{1}}{\mathrm{3e}}−\frac{\mathrm{1}}{\mathrm{3e}^{\mathrm{2}} }\right] \\ $$$$\int_{−\infty} ^{\infty} \frac{\mathrm{xsin}\left(\mathrm{x}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}=\frac{\pi}{\mathrm{3e}}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{1}} \right) \\ $$