Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 81340 by mind is power last updated on 11/Feb/20

Quation posted Times ago i can′t find it .after Somme Try i got close Fofme ∫_0 ^π sin(x^2 )dx=(π^3 /3) _1 F_2 ((3/4);(3/2);(7/4),−(π^4 /4))? sin(x)=Σ_(k≥0) (((−1)^k x^(2k+1) )/((2k+1)!)) ⇒sin(x^2 )=Σ_(k≥0) (((−1)^k x^(4k+2) )/((2k+1)!)) ∫_0 ^π sin(x^2 )dx=Σ_(k≥0) (((−1)^k )/((2k+1)!))∫_0 ^π x^(4k+2) dx =Σ_(k≥0) (((−1)^k π^(4k+3) )/((2k+1)!(4k+3))) =(π^3 /3)Σ_(k≥0) ((3(−1)^k π^(4k) )/((2k+1)!(4k+3))) (2k+1)!=k!.Π_(j=1) ^(k+1) (k+j)=k!.2^k .3....(2k+1) =(π^3 /3).Σ_(k≥0) ((3(−1)^k π^(4k) )/(k!.2^k .3....(2k+1)(4k+3))) =(π^3 /3)Σ_(k≥0) ((2^k .3)/(3...(2k+1).(4k+3))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3).Σ_(k≥0) ((2^k .3.....(4k−1))/(3.....(2k+1).(7.....(4k−1)(4k+3))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3).Σ_(k≥0) (((1/4^k ).(3)...(4k−1))/((1/2^k ).(3)...(2k+1).(1/4^k )(7)...(4k+3))).(−(π^4 /4))^k .(1/(k!)) =(π^3 /3)Σ_(k≥0) ((((3/4))....((3/4)+k−1))/(((3/2))....((3/2)+k−1).((7/4))....((7/4)+k−1))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3) _1 F_2 ((3/4);(3/2);(7/4);−(π^4 /4))

$${Quation}\:\:{posted}\:{Times}\:{ago} \\ $$$${i}\:{can}'{t}\:{find}\:{it}\:.{after}\:{Somme}\:{Try}\:{i}\:{got}\:{close}\:{Fofme} \\ $$$$\int_{\mathrm{0}} ^{\pi} {sin}\left({x}^{\mathrm{2}} \right){dx}=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\:\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{3}}{\mathrm{2}};\frac{\mathrm{7}}{\mathrm{4}},−\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\right)? \\ $$$${sin}\left({x}\right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$\Rightarrow{sin}\left({x}^{\mathrm{2}} \right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{4}{k}+\mathrm{2}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$\int_{\mathrm{0}} ^{\pi} {sin}\left({x}^{\mathrm{2}} \right){dx}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\pi} {x}^{\mathrm{4}{k}+\mathrm{2}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{4}{k}+\mathrm{3}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!\left(\mathrm{4}{k}+\mathrm{3}\right)} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{4}{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!\left(\mathrm{4}{k}+\mathrm{3}\right)} \\ $$$$\left(\mathrm{2}{k}+\mathrm{1}\right)!={k}!.\underset{{j}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\prod}}\left({k}+{j}\right)={k}!.\mathrm{2}^{{k}} .\mathrm{3}....\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{4}{k}} }{{k}!.\mathrm{2}^{{k}} .\mathrm{3}....\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{4}{k}+\mathrm{3}\right)} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{{k}} .\mathrm{3}}{\mathrm{3}...\left(\mathrm{2}{k}+\mathrm{1}\right).\left(\mathrm{4}{k}+\mathrm{3}\right)}.\left(\frac{−\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{{k}} .\mathrm{3}.....\left(\mathrm{4}{k}−\mathrm{1}\right)}{\mathrm{3}.....\left(\mathrm{2}{k}+\mathrm{1}\right).\left(\mathrm{7}.....\left(\mathrm{4}{k}−\mathrm{1}\right)\left(\mathrm{4}{k}+\mathrm{3}\right)\right.}.\left(\frac{−\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\frac{\mathrm{1}}{\mathrm{4}^{{k}} }.\left(\mathrm{3}\right)...\left(\mathrm{4}{k}−\mathrm{1}\right)}{\frac{\mathrm{1}}{\mathrm{2}^{{k}} }.\left(\mathrm{3}\right)...\left(\mathrm{2}{k}+\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{4}^{{k}} }\left(\mathrm{7}\right)...\left(\mathrm{4}{k}+\mathrm{3}\right)}.\left(−\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)....\left(\frac{\mathrm{3}}{\mathrm{4}}+{k}−\mathrm{1}\right)}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)....\left(\frac{\mathrm{3}}{\mathrm{2}}+{k}−\mathrm{1}\right).\left(\frac{\mathrm{7}}{\mathrm{4}}\right)....\left(\frac{\mathrm{7}}{\mathrm{4}}+{k}−\mathrm{1}\right)}.\left(\frac{−\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\:\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{3}}{\mathrm{2}};\frac{\mathrm{7}}{\mathrm{4}};−\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\right) \\ $$$$ \\ $$

Commented by TawaTawa last updated on 12/Feb/20

God bless you sir. Weldon

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Weldon} \\ $$

Commented by M±th+et£s last updated on 12/Feb/20

thank you so much sir . god bless you

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir}\:.\:\mathrm{god}\:\mathrm{bless}\:\mathrm{you} \\ $$

Commented by M±th+et£s last updated on 12/Feb/20

i will post my solution in another time

$$\mathrm{i}\:\mathrm{will}\:\mathrm{post}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{in}\:\mathrm{another}\:\mathrm{time} \\ $$

Commented by mind is power last updated on 12/Feb/20

withe pleasur

$${withe}\:{pleasur} \\ $$

Answered by M±th+et£s last updated on 12/Feb/20

Terms of Service

Privacy Policy

Contact: info@tinkutara.com