Question Number 82721 by M±th+et£s last updated on 23/Feb/20
![show that ∫xe^(−x^6 ) sin(x^3 ) dx=((Γ((5/6)))/3) 1F1[(5/6);(3/2);((−1)/4)]](https://www.tinkutara.com/question/Q82721.png)
$${show}\:{that}\: \\ $$$$\int{xe}^{−{x}^{\mathrm{6}} } \:{sin}\left({x}^{\mathrm{3}} \right)\:{dx}=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}{\mathrm{3}}\:\mathrm{1}{F}\mathrm{1}\left[\frac{\mathrm{5}}{\mathrm{6}};\frac{\mathrm{3}}{\mathrm{2}};\frac{−\mathrm{1}}{\mathrm{4}}\right] \\ $$
Commented by mind is power last updated on 23/Feb/20
$$\int_{\mathbb{R}} \:{or}\:\int_{\mathrm{0}} ^{+\infty} ? \\ $$
Commented by M±th+et£s last updated on 23/Feb/20
$$\int_{−\infty} ^{\infty} \\ $$$${sorry}\:{sir}\:{i}\:{forgat}\:{it} \\ $$
Answered by mind is power last updated on 23/Feb/20
![sin(x)=Σ_(k=0) ^(+∞) (((−1)^k x^(2k+1) )/((2k+1)!)) sin(x^3 )=Σ_(k=0) ^(+∞) (((−1)^k x^(6k+3) )/((2k+1)!)) A=∫_(−∞) ^(+∞) xe^(−x^6 ) sin(x^3 )dx=2∫_0 ^(+∞) xe^(−x^2 ) sin(x^3 )dx =2∫_0 ^(+∞) xe^(−x^6 ) [Σ_(k≥0) (((−1)^k x^(6k+3) )/((2k+1)!))] A=2Σ_(k≥0) (((−1)^k )/((2k+1)!))∫_0 ^(+∞) x^(6k+4) e^(−x^6 ) dx u=x^6 ⇒dx=(u^(−(5/6)) /6) ∫_0 ^(+∞) x^(6k+4) e^(−x^6 ) dx=(1/6)∫_0 ^(+∞) u^k u^((4/6)−(5/6)) e^(−u) du =(1/6)∫_0 ^(+∞) u^(k+(5/6)−1) e^(−u) du=(1/6)Γ(k+(5/6)) =(1/(6.))(((6k−1)/6)).......((5/6))Γ((5/6))= A=(2/6)Σ_(k≥0) (((−1)^k )/((2k+1)!))(((6k−1)......5)/6^k )Γ((5/6)) =((Γ((5/6)))/3).Σ_(k≥0) (((−1)^k (6k−1).....(5))/((2k+1)!!.2^k k!.6^k ))....A (((6k−1)......(5))/6^k )=(k+(5/6)−1)........((5/6))=((5/6))_k (((2k+1)!)/2^k )=(k+(1/2))........((3/2))=((3/2))_k ⇒(2k+1)!=2^k ((3/2))_k A⇔((Γ((5/6)))/3).Σ_(k≥0) (((−1)^k ((5/6))_k )/(2^k ((3/2))_k .2^k .k!))=((Γ((5/6)))/3).Σ_(k≥0) (((−1)^k ((5/6))_k )/(((3/2))_k .4^k .k!)) =((Γ((5/6)))/3).Σ_(k≥0) ((((5/6))_k )/(((3/2))_k )).(((((−1)/4))^k )/(k!^ ))=((Γ((5/6)))/3). _1 F_1 ((5/6);(3/2);−(1/4))](https://www.tinkutara.com/question/Q82728.png)
$${sin}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$${sin}\left({x}^{\mathrm{3}} \right)=\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{6}{k}+\mathrm{3}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$${A}=\int_{−\infty} ^{+\infty} {xe}^{−{x}^{\mathrm{6}} } {sin}\left({x}^{\mathrm{3}} \right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {xe}^{−{x}^{\mathrm{2}} } {sin}\left({x}^{\mathrm{3}} \right){dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {xe}^{−{x}^{\mathrm{6}} } \left[\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{6}{k}+\mathrm{3}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}\right] \\ $$$${A}=\mathrm{2}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}\int_{\mathrm{0}} ^{+\infty} {x}^{\mathrm{6}{k}+\mathrm{4}} {e}^{−{x}^{\mathrm{6}} } {dx} \\ $$$${u}={x}^{\mathrm{6}} \Rightarrow{dx}=\frac{{u}^{−\frac{\mathrm{5}}{\mathrm{6}}} }{\mathrm{6}} \\ $$$$\int_{\mathrm{0}} ^{+\infty} {x}^{\mathrm{6}{k}+\mathrm{4}} {e}^{−{x}^{\mathrm{6}} } {dx}=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{+\infty} {u}^{{k}} {u}^{\frac{\mathrm{4}}{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{6}}} {e}^{−{u}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{+\infty} {u}^{{k}+\frac{\mathrm{5}}{\mathrm{6}}−\mathrm{1}} {e}^{−{u}} {du}=\frac{\mathrm{1}}{\mathrm{6}}\Gamma\left({k}+\frac{\mathrm{5}}{\mathrm{6}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}.}\left(\frac{\mathrm{6}{k}−\mathrm{1}}{\mathrm{6}}\right)…….\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)= \\ $$$${A}=\frac{\mathrm{2}}{\mathrm{6}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}\frac{\left(\mathrm{6}{k}−\mathrm{1}\right)……\mathrm{5}}{\mathrm{6}^{{k}} }\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \left(\mathrm{6}{k}−\mathrm{1}\right)…..\left(\mathrm{5}\right)}{\left(\mathrm{2}{k}+\mathrm{1}\right)!!.\mathrm{2}^{{k}} {k}!.\mathrm{6}^{{k}} }….{A} \\ $$$$\frac{\left(\mathrm{6}{k}−\mathrm{1}\right)……\left(\mathrm{5}\right)}{\mathrm{6}^{{k}} }=\left({k}+\frac{\mathrm{5}}{\mathrm{6}}−\mathrm{1}\right)……..\left(\frac{\mathrm{5}}{\mathrm{6}}\right)=\left(\frac{\mathrm{5}}{\mathrm{6}}\right)_{{k}} \\ $$$$\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)!}{\mathrm{2}^{{k}} }=\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)……..\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{k}} \\ $$$$\Rightarrow\left(\mathrm{2}{k}+\mathrm{1}\right)!=\mathrm{2}^{{k}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{k}} \\ $$$${A}\Leftrightarrow\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)_{{k}} }{\mathrm{2}^{{k}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{k}} .\mathrm{2}^{{k}} .{k}!}=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)_{{k}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{k}} .\mathrm{4}^{{k}} .{k}!} \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{5}}{\mathrm{6}}\right)_{{k}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{k}} }.\frac{\left(\frac{−\mathrm{1}}{\mathrm{4}}\right)^{{k}} }{{k}!^{} }=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}{\mathrm{3}}.\:_{\mathrm{1}} {F}_{\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{6}};\frac{\mathrm{3}}{\mathrm{2}};−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 23/Feb/20
$${briliant}\:{solution}\:{from}\:{a}\:{briliant}\:{person} \\ $$$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$
Commented by mind is power last updated on 23/Feb/20
$${withe}\:{pleasur}\:{thank}\:{you} \\ $$