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Question Number 127468 by bramlexs22 last updated on 30/Dec/20

$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^3\sin \left(\lambda \mathrm{x}\right)}{{\mathrm{x}}^4+4}\mathrm{dx}=?$$

Commented by bramlexs22 last updated on 30/Dec/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}\mathrm{sir}$$

Answered by Olaf last updated on 30/Dec/20

$$\mathrm{Let}g(\lambda ,x)=\frac{x^3\sin \left(\lambda x\right)}{x^4+4}$$$$\mathrm{Let}f\left(\lambda \right)={\int }_0^{\mathrm{\infty }}g(\lambda ,x)dx\left(1\right)$$$$f^{\prime }\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{\partial g}{\partial \lambda }(\lambda ,x)dx=\lambda {\int }_0^{\mathrm{\infty }}\frac{x^3\cos \left(\lambda x\right)}{x^4+4}dx$$$$f^{″}\left(\lambda \right)=\int \frac{x^3\cos \left(\lambda x\right)}{x^4+4}dx-{\lambda }^2{\int }_0^{\mathrm{\infty }}\frac{x^3\sin \left(\lambda x\right)}{x^4+4}dx$$$$f^{″}\left(\lambda \right)=\frac{f^{\prime }\left(\lambda \right)}{\lambda }-{\lambda }^{2}f\left(\lambda \right)$$$$\lambda f^{″}\left(\lambda \right)-f^{\prime }\left(\lambda \right)+{\lambda }^{3}f\left(\lambda \right)=0\left(2\right)$$$$\mathrm{Let}t={\lambda }^2,\lambda =t^{1/2}$$$$f^{\prime }\left(\lambda \right)=\frac{df}{d\lambda }=\frac{dt}{d\lambda }.\frac{df}{dt}=2t^{1/2}\frac{df}{dt}\left(3\right)$$$$f^{″}\left(\lambda \right)=\frac{{df}^{\prime }}{d\lambda }=\frac{dt}{d\lambda }.\frac{{df}^{\prime }}{dt}=2t^{1/2}[t^{-1/2}\frac{df}{dt}+2t^{1/2}\frac{d^{2}f}{{dt}^2}]$$$$f^{″}\left(\lambda \right)=2[\frac{df}{dt}+2t\frac{d^{2}f}{{dt}^2}]\left(4\right)$$$$\mathrm{With}\left(2\right),\left(3\right)\mathrm{and}\left(4\right):$$$$2t^{1/2}[\frac{df}{dt}+2t\frac{d^{2}f}{{dt}^2}]-2t^{1/2}\frac{df}{dt}+t^{3/2}f=0$$$$\frac{d^{2}f}{{dt}^2}+\frac{1}{4}f=0$$$$\to f\left(t\right)=\mathrm{Acos}\left(\frac{t}{2}\right)+\mathrm{Bsin}\left(\frac{t}{2}\right)$$$$f\left(\lambda \right)=\mathrm{Acos}\left(\frac{{\lambda }^2}{2}\right)+\mathrm{Bsin}\left(\frac{{\lambda }^2}{2}\right)$$$$\mathrm{A}?\mathrm{B}?$$$$\left(1\right):f\left(0\right)={\int }_0^{\mathrm{\infty }}\frac{x^3\sin \left(0\right)}{x^4+4}dx=0\Rightarrow \mathrm{A}=0$$$$f\left(\lambda \right)=\mathrm{Bsin}\left(\frac{{\lambda }^2}{2}\right)$$$$\left(1\right):f\left(\lambda \right)=-f\left(\lambda \right)\to f\mathrm{is}\mathrm{odd}$$$$\Rightarrow \mathrm{B}=0?????!!!$$$$\mathrm{Sorry}!\mathrm{I}\mathrm{tried}!$$

Answered by bramlexs22 last updated on 30/Dec/20

$$\mathrm{super}\mathrm{nice}\mathrm{integral}$$

Answered by liberty last updated on 30/Dec/20

$$letS\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\lambda x\right)}{x^4+4}dxwhereS\left(0\right)={\int }_0^{\mathrm{\infty }}\frac{1}{x^4+4}dx=\frac{\pi }{8}$$$$differentiatingthefunctionS\left(\lambda \right)over$$$$\lambda once\Rightarrow S^{\prime }\left(\lambda \right)=-{\int }_0^{\mathrm{\infty }}\frac{x\sin \left(\lambda x\right)}{x^4+4}dx$$$$and{S}^{\prime }\left(0\right)=0.Differentiatingthefunction$$$$S\left(\lambda \right)twiceweget{S}^{″}\left(\lambda \right)=-{\int }_0^{\mathrm{\infty }}\frac{x^2\cos \left(\lambda x\right)}{x^4+4}dx$$$$again{S}^{‴}\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{x^3\sin \left(\lambda x\right)}{x^4+4}dx=R\left(\lambda \right)$$$$weknowthat{\int }_0^{\mathrm{\infty }}\frac{\sin \left(\lambda x\right)}{x}dx=\frac{\pi }{2}$$$$thus{S}^{‴}\left(\lambda \right)-\frac{\pi }{2}=R\left(\lambda \right)-{\int }_0^{\mathrm{\infty }}\frac{\sin \left(\lambda x\right)}{x}dx$$$$\frac{x^3\sin \left(\lambda x\right)}{x^4+4}-\frac{\pi }{2}=-4{\int }_0^{\mathrm{\infty }}\frac{\sin \left(\lambda x\right)}{x(x^4+4)}dx$$$$S^{\left(4\right)}\left(\lambda \right)=-4{\int }_0^{\mathrm{\infty }}\frac{\cos \left(\lambda x\right)}{x^4+4}dx$$$$then{S}^{\left(4\right)}\left(\lambda \right)=-4S\left(\lambda \right)\Rightarrow S^{\left(4\right)}\left(\lambda \right)+4S\left(\lambda \right)=0$$$$thecharacteristicequation{\ell }^4+4=0$$$$hasfour{\ell }_0=1+i,{\ell }_1=-1+i,{\ell }_3=-1-i$$$$and{\ell }_4=1-ithenS\left(\lambda \right)=C_1{e}^{-\lambda }\cos \left(\lambda \right)+C_2{e}^{-\lambda }\sin \left(\lambda \right)$$$$+C_3{e}^{\lambda }\cos \left(\lambda \right)+C_4{e}^{\lambda }\sin \left(\lambda \right)$$$$\Rightarrow \mid S\left(\lambda \right)\mid ⩽{\int }_0^{\mathrm{\infty }}\frac{1}{4+x^4}dx=\frac{\pi }{8}$$$$therefore{\int }_0^{\mathrm{\infty }}\frac{x^3\sin \left(\lambda x\right)}{x^4+4}dx=\frac{\pi }{2}{e}^{-\lambda }\cos \left(\lambda \right)$$$$$$$$$$

Commented by Olaf last updated on 30/Dec/20

$$\mathrm{Dear}\mathrm{Mr}\mathrm{Liberty},\mathrm{for}\lambda =0\mathrm{your}$$$$\mathrm{formula}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{work}.$$

Commented by liberty last updated on 30/Dec/20

$$yessir.ithinkthisquestionvalidfor$$$$\lambda >0.$$

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