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Integration Questions

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

$$\:\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{x}^{\mathrm{3}} \:\mathrm{sin}\:\left(\lambda\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{4}}\:\mathrm{dx}\:=?\: \\$$

Commented bybramlexs22 last updated on 30/Dec/20

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

Answered by Olaf last updated on 30/Dec/20

$$\mathrm{Let}\:{g}\left(\lambda,{x}\right)\:=\:\frac{{x}^{\mathrm{3}} \mathrm{sin}\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}} \\$$ $$\mathrm{Let}\:{f}\left(\lambda\right)\:=\:\int_{\mathrm{0}} ^{\infty} {g}\left(\lambda,{x}\right){dx}\:\left(\mathrm{1}\right) \\$$ $${f}'\left(\lambda\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\partial{g}}{\partial\lambda}\left(\lambda,{x}\right){dx}\:=\:\lambda\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} \mathrm{cos}\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}{dx} \\$$ $${f}''\left(\lambda\right)\:=\:\int\frac{{x}^{\mathrm{3}} \mathrm{cos}\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}{dx}−\lambda^{\mathrm{2}} \int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} \mathrm{sin}\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}{dx} \\$$ $${f}''\left(\lambda\right)\:=\:\frac{{f}'\left(\lambda\right)}{\lambda}−\lambda^{\mathrm{2}} {f}\left(\lambda\right) \\$$ $$\lambda{f}''\left(\lambda\right)\:−{f}'\left(\lambda\right)+\lambda^{\mathrm{3}} {f}\left(\lambda\right)\:=\:\mathrm{0}\:\left(\mathrm{2}\right) \\$$ $$\mathrm{Let}\:{t}\:=\:\lambda^{\mathrm{2}} ,\:\lambda\:=\:{t}^{\mathrm{1}/\mathrm{2}} \\$$ $${f}'\left(\lambda\right)\:=\:\frac{{df}}{{d}\lambda}\:=\:\frac{{dt}}{{d}\lambda}.\frac{{df}}{{dt}}\:=\:\mathrm{2}{t}^{\mathrm{1}/\mathrm{2}} \frac{{df}}{{dt}}\:\left(\mathrm{3}\right) \\$$ $${f}''\left(\lambda\right)\:=\:\frac{{df}'}{{d}\lambda}\:=\:\frac{{dt}}{{d}\lambda}.\frac{{df}'}{{dt}}\:=\:\mathrm{2}{t}^{\mathrm{1}/\mathrm{2}} \left[{t}^{−\mathrm{1}/\mathrm{2}} \frac{{df}}{{dt}}+\mathrm{2}{t}^{\mathrm{1}/\mathrm{2}} \frac{{d}^{\mathrm{2}} {f}}{{dt}^{\mathrm{2}} }\right] \\$$ $${f}''\left(\lambda\right)\:=\:\mathrm{2}\left[\frac{{df}}{{dt}}+\mathrm{2}{t}\frac{{d}^{\mathrm{2}} {f}}{{dt}^{\mathrm{2}} }\right]\:\left(\mathrm{4}\right) \\$$ $$\mathrm{With}\:\left(\mathrm{2}\right),\:\left(\mathrm{3}\right)\:\mathrm{and}\:\left(\mathrm{4}\right)\:\:: \\$$ $$\mathrm{2}\:{t}^{\mathrm{1}/\mathrm{2}} \left[\frac{{df}}{{dt}}+\mathrm{2}{t}\frac{{d}^{\mathrm{2}} {f}}{{dt}^{\mathrm{2}} }\right]−\mathrm{2}{t}^{\mathrm{1}/\mathrm{2}} \frac{{df}}{{dt}}+{t}^{\mathrm{3}/\mathrm{2}} {f}\:=\:\mathrm{0} \\$$ $$\frac{{d}^{\mathrm{2}} {f}}{{dt}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}{f}\:=\:\mathrm{0} \\$$ $$\rightarrow\:{f}\left({t}\right)\:=\:\mathrm{Acos}\left(\frac{{t}}{\mathrm{2}}\right)+\mathrm{Bsin}\left(\frac{{t}}{\mathrm{2}}\right) \\$$ $${f}\left(\lambda\right)\:=\:\mathrm{Acos}\left(\frac{\lambda^{\mathrm{2}} }{\mathrm{2}}\right)+\mathrm{Bsin}\left(\frac{\lambda^{\mathrm{2}} }{\mathrm{2}}\right) \\$$ $$\mathrm{A}\:?\:\mathrm{B}\:? \\$$ $$\left(\mathrm{1}\right)\::\:{f}\left(\mathrm{0}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} \mathrm{sin}\left(\mathrm{0}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}{dx}\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{A}\:=\:\mathrm{0} \\$$ $${f}\left(\lambda\right)\:=\:\mathrm{Bsin}\left(\frac{\lambda^{\mathrm{2}} }{\mathrm{2}}\right) \\$$ $$\left(\mathrm{1}\right)\::\:{f}\left(\lambda\right)\:=\:−{f}\left(\lambda\right)\:\rightarrow\:{f}\:\mathrm{is}\:\mathrm{odd} \\$$ $$\Rightarrow\:\mathrm{B}\:=\:\mathrm{0}\:?????!!! \\$$ $$\mathrm{Sorry}\:!\:\mathrm{I}\:\mathrm{tried}\:! \\$$

Answered by bramlexs22 last updated on 30/Dec/20

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

Answered by liberty last updated on 30/Dec/20

$${let}\:{S}\left(\lambda\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{cos}\:\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}\:{dx}\:{where}\:{S}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{4}}{dx}=\frac{\pi}{\mathrm{8}} \\$$ $${differentiating}\:{the}\:{function}\:{S}\left(\lambda\right)\:{over} \\$$ $$\lambda\:{once}\:\Rightarrow\:{S}'\left(\lambda\right)=−\int_{\mathrm{0}} ^{\infty} \:\frac{{x}\:\mathrm{sin}\:\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}\:{dx} \\$$ $${and}\:{S}'\left(\mathrm{0}\right)\:=\:\mathrm{0}.\:{Differentiating}\:{the}\:{function} \\$$ $${S}\left(\lambda\right)\:{twice}\:{we}\:{get}\:{S}''\left(\lambda\right)=−\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}\:{dx} \\$$ $${again}\:{S}'''\left(\lambda\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\mathrm{3}} \:\mathrm{sin}\:\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}\:{dx}\:=\:{R}\left(\lambda\right) \\$$ $${we}\:{know}\:{that}\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sin}\:\left(\lambda{x}\right)}{{x}}\:{dx}\:=\:\frac{\pi}{\mathrm{2}} \\$$ $${thus}\:{S}'''\left(\lambda\right)−\frac{\pi}{\mathrm{2}}={R}\left(\lambda\right)−\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}\:\left(\lambda{x}\right)}{{x}}{dx} \\$$ $$\:\frac{{x}^{\mathrm{3}} \:\mathrm{sin}\:\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}−\frac{\pi}{\mathrm{2}}\:=\:−\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:\left(\lambda{x}\right)}{{x}\left({x}^{\mathrm{4}} +\mathrm{4}\right)}{dx} \\$$ $${S}^{\left(\mathrm{4}\right)} \left(\lambda\right)\:=\:−\mathrm{4}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{cos}\:\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}{dx}\: \\$$ $${then}\:{S}^{\left(\mathrm{4}\right)} \left(\lambda\right)=−\mathrm{4}{S}\left(\lambda\right)\Rightarrow{S}^{\left(\mathrm{4}\right)} \left(\lambda\right)+\mathrm{4}{S}\left(\lambda\right)=\mathrm{0} \\$$ $${the}\:{characteristic}\:{equation}\:\ell^{\mathrm{4}} +\mathrm{4}=\mathrm{0} \\$$ $${has}\:{four}\:\ell_{\mathrm{0}} =\mathrm{1}+{i}\:,\ell_{\mathrm{1}} =−\mathrm{1}+{i}\:,\ell_{\mathrm{3}} =−\mathrm{1}−{i} \\$$ $${and}\:\ell_{\mathrm{4}} =\mathrm{1}−{i}\:{then}\:{S}\left(\lambda\right)={C}_{\mathrm{1}} {e}^{−\lambda} \mathrm{cos}\:\left(\lambda\right)+{C}_{\mathrm{2}} {e}^{−\lambda} \mathrm{sin}\:\left(\lambda\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{C}_{\mathrm{3}} {e}^{\lambda} \mathrm{cos}\:\left(\lambda\right)+{C}_{\mathrm{4}} {e}^{\lambda} \mathrm{sin}\:\left(\lambda\right) \\$$ $$\Rightarrow\:\mid{S}\left(\lambda\right)\mid\:\leqslant\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\mathrm{4}+{x}^{\mathrm{4}} }\:{dx}\:=\frac{\pi}{\mathrm{8}} \\$$ $${therefore}\:\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\mathrm{3}} \:\mathrm{sin}\:\left(\lambda{x}\right)}{{x}^{\mathrm{4}} +\mathrm{4}}\:{dx}\:=\:\frac{\pi}{\mathrm{2}}{e}^{−\lambda} \:\mathrm{cos}\:\left(\lambda\right)\: \\$$ $$\\$$ $$\\$$

Commented byOlaf last updated on 30/Dec/20

$$\mathrm{Dear}\:\mathrm{Mr}\:\mathrm{Liberty},\:\mathrm{for}\:\lambda\:=\:\mathrm{0}\:\mathrm{your}\: \\$$ $$\mathrm{formula}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{work}. \\$$

Commented byliberty last updated on 30/Dec/20

$${yes}\:{sir}.\:{i}\:{think}\:{this}\:{question}\:{valid}\:{for} \\$$ $$\lambda\:>\:\mathrm{0}\:. \\$$