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Integration Questions

Question Number 130979 by mnjuly1970 last updated on 31/Jan/21

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\$$$$\:\:\:\:\:\:\:\:\:\:\phi\:=\int_{\mathrm{0}} ^{\:\infty} {log}^{\mathrm{2}} \left({x}\right){sin}\left({x}^{\mathrm{2}} \right){dx}=? \\$$$$\:\:\:\:\:{i}\:{had}\:{solved}\:{that}\:{already}\:\:{and}: \\$$$$\:\:\:{answ}\:\::\:=\:−\frac{\pi^{\mathrm{2}} \sqrt{\mathrm{2}\pi}\:}{\mathrm{32}} \\$$$$\:\:\:\:\: \\$$$$\:\:\: \\$$$$\:\:\:\:\: \\$$

Commented by Dwaipayan Shikari last updated on 31/Jan/21

$$\int_{\mathrm{0}} ^{\infty} \frac{{tanx}}{{x}}{dx}\:\:\Rightarrow{Diverges} \\$$

Commented by Dwaipayan Shikari last updated on 31/Jan/21

$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} {sinx}\:{dtdx} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}−{i}}−\frac{\mathrm{1}}{{t}+{i}}{dt} \\$$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\frac{\pi}{\mathrm{2}} \\$$

Commented by mnjuly1970 last updated on 31/Jan/21

$${yes}\:{it}\:{is}\:{convergent} \\$$$$\:{i}\:{will}\:{prepare} \\$$$$\:{other}\:{question}... \\$$

Answered by mnjuly1970 last updated on 31/Jan/21

Answered by mnjuly1970 last updated on 31/Jan/21

Answered by Dwaipayan Shikari last updated on 31/Jan/21

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {sin}\left({x}^{\mathrm{2}} \right){dx} \\$$$${I}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}^{\mathrm{2}} } −{x}^{{a}} {e}^{−{ix}^{\mathrm{2}} } {dx}\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={iu}\:\:\:\:{x}^{\mathrm{2}} =−{ij} \\$$$${I}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} {x}^{{a}−\mathrm{1}} {e}^{−{u}} {du}\:−{x}^{{a}−\mathrm{1}} {e}^{−{j}} {dj} \\$$$${I}\left({a}\right)=\frac{\left({i}\right)^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} }{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} {u}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} {e}^{−{u}} +\left(−{i}\right)^{{a}−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} {j}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} {e}^{−{j}} {dj} \\$$$$=\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right) \\$$$${I}'\left({a}\right)=\frac{\Gamma'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{4}}{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi}{\mathrm{4}}\:\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}{sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right) \\$$$${I}''\left({a}\right)=\frac{\Gamma''\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{8}}{cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi}{\mathrm{4}}.\frac{\Gamma'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{4}}{sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi}{\mathrm{16}}\Gamma'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right){sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{32}}.\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right){cos}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right) \\$$$${I}''\left(\mathrm{0}\right)=\frac{\Gamma''\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\pi\Gamma'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\:\mathrm{8}\sqrt{\mathrm{2}}}−\frac{\pi^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{32}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\$$$$=\frac{\pi\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\pi\sqrt{\pi}}{\mathrm{8}\sqrt{\mathrm{2}}}\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\pi^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{32}}\:\:\:\: \\$$$$=\frac{\pi\sqrt{\pi}}{\mathrm{8}\sqrt{\mathrm{2}}}\left(\gamma+{log}\left(\mathrm{4}\right)\right)−\frac{\pi^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{32}}−\frac{\pi\left(\gamma+{log}\left(\mathrm{4}\right)\right.}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}} \\$$

Answered by mathmax by abdo last updated on 31/Jan/21

$$\Phi=_{\mathrm{x}=\sqrt{\mathrm{t}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{t}}\right)\mathrm{sin}\left(\mathrm{t}\right)\left(\mathrm{2t}\right)\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)\:\mathrm{tsint}\:\mathrm{dt} \\$$$$\varphi\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{a}+\mathrm{1}} \:\mathrm{sint}\:\mathrm{dt}\:\:=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\left(\mathrm{a}+\mathrm{1}\right)\mathrm{lnt}} \mathrm{sint}\:\mathrm{dt}\:\Rightarrow\varphi^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \mathrm{lnt}\:.\mathrm{e}^{\left(\mathrm{a}+\mathrm{1}\right)\mathrm{lnt}} \mathrm{sint}\:\mathrm{dt} \\$$$$\Rightarrow\varphi^{\left(\mathrm{2}\right)} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \mathrm{ln}^{\mathrm{2}} \mathrm{t}\:.\mathrm{t}^{\mathrm{a}+\mathrm{1}} \mathrm{sint}\:\mathrm{dt}\:\Rightarrow\varphi^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \mathrm{ln}^{\mathrm{2}} \mathrm{t}\:.\mathrm{tsint}\:\mathrm{dt}\:=\mathrm{2}\Phi \\$$$$\varphi\left(\mathrm{a}\right)\:=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{a}+\mathrm{1}} \:\mathrm{e}^{−\mathrm{it}} \mathrm{dt}\right)\:\mathrm{and}\: \\$$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{a}+\mathrm{1}} \:\mathrm{e}^{−\mathrm{it}} \mathrm{dt}\:=_{\mathrm{it}=\mathrm{u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{u}}{\mathrm{i}}\right)^{\mathrm{a}+\mathrm{1}} \:\mathrm{e}^{−\mathrm{u}} \:\frac{\mathrm{du}}{\mathrm{i}} \\$$$$=\frac{\mathrm{1}}{\mathrm{i}^{\mathrm{a}+\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\mathrm{u}^{\mathrm{a}+\mathrm{2}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{u}} \:\mathrm{du}\:=\frac{\mathrm{1}}{\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\mathrm{a}+\mathrm{2}} }×\Gamma\left(\mathrm{a}+\mathrm{2}\right) \\$$$$=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}\left(\mathrm{a}+\mathrm{2}\right)} .\Gamma\left(\mathrm{a}+\mathrm{2}\right)\:=\Gamma\left(\mathrm{a}+\mathrm{2}\right)\left\{\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{a}+\mathrm{2}\right)\right)−\mathrm{isin}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{a}+\mathrm{2}\right)\right)\right\}\:\Rightarrow \\$$$$\varphi\left(\mathrm{a}\right)=\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}+\pi\right).\Gamma\left(\mathrm{a}+\mathrm{2}\right)\:=−\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right).\Gamma\left(\mathrm{a}+\mathrm{2}\right)\:\:\:\left(\mathrm{witha}>−\mathrm{2}\right) \\$$$$\Gamma\left(\mathrm{a}+\mathrm{2}\right)\:=\Gamma\left(\mathrm{a}+\mathrm{1}\:+\mathrm{1}\right)\:=\left(\mathrm{a}+\mathrm{1}\right)\Gamma\left(\mathrm{a}+\mathrm{1}\right)\:=\mathrm{a}\left(\mathrm{a}+\mathrm{1}\right)\Gamma\left(\mathrm{a}\right)\:\Rightarrow \\$$$$\varphi\left(\mathrm{a}\right)\:=−\mathrm{a}\left(\mathrm{a}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right).\Gamma\left(\mathrm{a}\right)=\mathrm{w}\left(\mathrm{a}\right).\Gamma\left(\mathrm{a}\right)\:\Rightarrow \\$$$$\varphi^{'} \left(\mathrm{a}\right)\:=\mathrm{W}^{,} \left(\mathrm{a}\right).\Gamma\left(\mathrm{a}\right)+\mathrm{w}\left(\mathrm{a}\right).\Gamma^{'} \left(\mathrm{a}\right) \\$$$$\mathrm{w}^{'} \left(\mathrm{a}\right)=−\left\{\left(\mathrm{2a}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)+\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{a}\right)\frac{\pi}{\mathrm{2}}\mathrm{cos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)\right\}\:\Rightarrow \\$$$$\varphi^{,} \left(\mathrm{a}\right)=−\left\{\left(\mathrm{2a}+\mathrm{1}\right)\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{a}\right)\mathrm{cos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)\right\}.\Gamma\left(\mathrm{a}\right) \\$$$$−\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{a}\right)\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right).\Gamma^{'} \left(\mathrm{a}\right).....\mathrm{be}\:\mathrm{continued}... \\$$$$\\$$