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Question Number 124461 by Dwaipayan Shikari last updated on 03/Dec/20

$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{7}} } {sin}\left({x}^{\mathrm{7}} \right){dx} \\$$

Commented byDwaipayan Shikari last updated on 03/Dec/20

$${I}\:{have}\:{found}\:\:\frac{\Gamma\left(\frac{\mathrm{8}}{\mathrm{7}}\right)\:{sin}\left(\frac{\pi}{\mathrm{28}}\right)}{\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{14}}} } \\$$

Commented byDwaipayan Shikari last updated on 03/Dec/20

$${My}\:{way} \\$$ $$\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{7}} } {e}^{{ix}^{\mathrm{7}} } −\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{7}} \left(\mathrm{1}+{i}\right)} {dx}\:\:\:\:\:\:{x}^{\mathrm{7}} ={u}\Rightarrow\mathrm{7}{x}^{\mathrm{6}} \Rightarrow\frac{{du}}{{dx}} \\$$ $$=\frac{{e}^{−\frac{\pi}{\mathrm{2}}{i}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {u}^{−\frac{\mathrm{6}}{\mathrm{7}}} {e}^{−{u}\left(\mathrm{1}−{i}\right)} {du}\:−\frac{{e}^{−\frac{\pi}{\mathrm{2}}{i}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {u}^{−\frac{\mathrm{6}}{\mathrm{7}}} {e}^{−{u}\left(\mathrm{1}+{i}\right)} \:\:\:\:\: \\$$ $$=\Gamma\left(\frac{\mathrm{8}}{\mathrm{7}}\right)\left(\frac{{e}^{−\frac{\pi}{\mathrm{2}}{i}} \left(\mathrm{1}−{i}\right)^{−\frac{\mathrm{1}}{\mathrm{7}}} }{\mathrm{2}}−\frac{{e}^{\frac{−\pi}{\mathrm{2}}{i}} \left(\mathrm{1}+{i}\right)^{−\frac{\mathrm{1}}{\mathrm{7}}} }{\mathrm{2}}\right) \\$$ $$=\frac{\Gamma\left(\frac{\mathrm{8}}{\mathrm{7}}\right)}{\:\sqrt[{\mathrm{14}}]{\mathrm{2}^{\mathrm{15}} }}\left({e}^{−\frac{\pi}{\mathrm{2}}{i}+\frac{\pi}{\mathrm{28}}{i}} −{e}^{−\frac{\pi}{\mathrm{2}}{i}−\frac{\pi}{\mathrm{28}}{i}} \right)=\frac{{sin}\left(\frac{\pi}{\mathrm{28}}\right)\Gamma\left(\frac{\mathrm{8}}{\mathrm{7}}\right)}{\:\sqrt[{\mathrm{14}}]{\mathrm{2}}} \\$$

Answered by mnjuly1970 last updated on 03/Dec/20

$${x}^{\mathrm{7}} ={t} \\$$ $$\:\frac{\mathrm{1}}{\mathrm{7}}\int_{\mathrm{0}} ^{\:\infty} {e}^{−{t}} {sin}\left({t}\right)\frac{{dt}}{{t}^{\frac{\mathrm{6}}{\mathrm{7}}} } \\$$ $${I}=\frac{−\mathrm{1}}{\mathrm{7}}{Im}\left(\int_{\mathrm{0}} ^{\:\infty} {e}^{−{t}} {e}^{−{it}} \frac{{dt}}{{t}^{\frac{\mathrm{6}}{\mathrm{7}}} }\right) \\$$ $$\:\:\:=−\frac{\mathrm{1}}{\mathrm{7}}{Im}\left(\int_{\mathrm{0}\:} ^{\:\infty} {e}^{−{t}\left(\mathrm{1}+{i}\right)} \frac{{dt}}{{t}^{\frac{\mathrm{6}}{\mathrm{7}}} }\right) \\$$ $$\:\:\:\overset{{t}\left(\mathrm{1}+{i}\right)={u}} {=}−\frac{\mathrm{1}}{\mathrm{7}}{Im}\left(\frac{\mathrm{1}}{\mathrm{1}+{i}}\int_{\mathrm{0}} ^{\:\infty} {e}^{−{u}} \frac{{du}}{\left(\mathrm{1}+{i}\right)^{−\frac{\mathrm{6}}{\mathrm{7}}} {u}^{\frac{\mathrm{6}}{\mathrm{7}}} }\right) \\$$ $$−\frac{\mathrm{1}}{\mathrm{7}}{Im}\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} }\int_{\mathrm{0}} ^{\:\infty} {e}^{−{u}} {u}^{\frac{−\mathrm{6}}{\mathrm{7}}+\mathrm{1}−\mathrm{1}} \\$$ $$=\frac{−\mathrm{1}}{\mathrm{7}}{Im}\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2}{i}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{14}}} }\Gamma\left(\frac{\mathrm{1}}{\mathrm{7}}\right) \\$$ $$\:=\frac{−\mathrm{1}}{\:\sqrt[{\mathrm{14}}]{\mathrm{2}}\:}\Gamma\left(\frac{\mathrm{8}}{\mathrm{7}}\right){Im}\frac{\mathrm{1}}{\left({e}^{\frac{{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{14}}} } \\$$ $$=\frac{−\mathrm{1}}{\:\sqrt[{\mathrm{14}}]{\mathrm{2}}}\Gamma\left(\frac{\mathrm{8}}{\mathrm{7}}\right){Im}\left\{{cos}\left(\frac{\pi}{\mathrm{28}}\right)−{isin}\left(\frac{\pi}{\mathrm{28}}\right)\right. \\$$ $$\therefore\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{7}} } {sin}\left({x}^{\mathrm{7}} \right){dx}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{14}}]{\mathrm{2}}}\Gamma\left(\frac{\mathrm{8}}{\mathrm{7}}\right){sin}\left(\frac{\pi}{\mathrm{28}}\right)\:\checkmark \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:.....\: \\$$ $$\:\: \\$$ $$\\$$

Commented byDwaipayan Shikari last updated on 03/Dec/20

$${Thanking}\:{you} \\$$

Commented bymnjuly1970 last updated on 03/Dec/20

$${grateful} \\$$ $${your}\:{questions}\:{are}\:{very}\:{stunning}.. \\$$

Answered by mathmax by abdo last updated on 03/Dec/20

$$\mathrm{let}\:\mathrm{A}\:=\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}^{\mathrm{7}} } \mathrm{sin}\left(\mathrm{x}^{\mathrm{7}} \right)\mathrm{dx}\:\Rightarrow\mathrm{A}=_{\mathrm{x}^{\mathrm{7}} =\mathrm{t}} \:\:\frac{\mathrm{1}}{\mathrm{7}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}} \mathrm{sin}\left(\mathrm{t}\right)\:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{7}}−\mathrm{1}} \mathrm{dt} \\$$ $$=\frac{\mathrm{1}}{\mathrm{7}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{−\frac{\mathrm{6}}{\mathrm{7}}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{sint}\:\mathrm{dt}\:\:=\frac{\mathrm{1}}{\mathrm{7}}\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{−\frac{\mathrm{6}}{\mathrm{7}}} \:\mathrm{e}^{−\mathrm{t}+\mathrm{it}} \mathrm{dt}\right)\mathrm{but} \\$$ $$\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{−\frac{\mathrm{6}}{\mathrm{7}}} \:\mathrm{e}^{−\left(\mathrm{1}−\mathrm{i}\right)\mathrm{t}} \mathrm{dt}\:=_{\left(\mathrm{1}−\mathrm{i}\right)\mathrm{t}=\mathrm{u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{u}^{−\frac{\mathrm{6}}{\mathrm{7}}} }{\left(\mathrm{1}−\mathrm{i}\right)^{−\frac{\mathrm{6}}{\mathrm{7}}} }\:\mathrm{e}^{−\mathrm{u}} \:\:\frac{\mathrm{du}}{\mathrm{1}−\mathrm{i}} \\$$ $$=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{1}−\frac{\mathrm{6}}{\mathrm{7}}} }\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{u}^{−\frac{\mathrm{6}}{\mathrm{7}}} \:\mathrm{e}^{−\mathrm{u}} \:\mathrm{du}\:\:=\left(\mathrm{1}−\mathrm{i}\right)^{−\frac{\mathrm{1}}{\mathrm{7}}} \:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{7}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{u}} \mathrm{du} \\$$ $$=\left(\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{−\frac{\mathrm{1}}{\mathrm{7}}} \:×\Gamma\left(\frac{\mathrm{1}}{\mathrm{7}}\right)\:=\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} }\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{28}}} \:×\Gamma\left(\frac{\mathrm{1}}{\mathrm{7}}\right) \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{14}}} }×\Gamma\left(\frac{\mathrm{1}}{\mathrm{7}}\right)\left\{\mathrm{cos}\left(\frac{\pi}{\mathrm{28}}\right)+\mathrm{isin}\left(\frac{\pi}{\mathrm{28}}\right)\right\}\:\Rightarrow \\$$ $$\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{x}^{\mathrm{7}} \:} \:\mathrm{sin}\left(\mathrm{x}^{\mathrm{7}} \right)\mathrm{dx}\:=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{7}}\right)}{\mathrm{7}}×\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{28}}\right)}{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{14}}} } \\$$

Commented bymathmax by abdo last updated on 03/Dec/20

$$\frac{\mathrm{1}}{\mathrm{7}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{7}}\right)=\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}\right)=\Gamma\left(\frac{\mathrm{8}}{\mathrm{7}}\right) \\$$

Commented byDwaipayan Shikari last updated on 03/Dec/20

$${Thanking}\:{you} \\$$

Commented bymathmax by abdo last updated on 03/Dec/20

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\$$