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


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

$Q u a t i o n p o s t e d T i m e s a g o$ $i {c a n}^{'} t f i n d i t . a f t e r S o m m e T r y i g o t c l o s e F o f m e$ $\int_{0}^{π} s i n (x^{2}) d x = \frac{π^{3}}{3}_{1} F_{2} (\frac{3}{4}; \frac{3}{2}; \frac{7}{4}, - \frac{π^{4}}{4}) ?$ $s i n (x) = \sum_{k ⩾ 0} \frac{{(- 1)}^{k} x^{2 k + 1}}{(2 k + 1)!}$ $\Rightarrow s i n (x^{2}) = \sum_{k ⩾ 0} \frac{{(- 1)}^{k} x^{4 k + 2}}{(2 k + 1)!}$ $\int_{0}^{π} s i n (x^{2}) d x = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(2 k + 1)!} \int_{0}^{π} x^{4 k + 2} d x$ $= \sum_{k ⩾ 0} \frac{{(- 1)}^{k} π^{4 k + 3}}{(2 k + 1)! (4 k + 3)}$ $= \frac{π^{3}}{3} \sum_{k ⩾ 0} \frac{3 {(- 1)}^{k} π^{4 k}}{(2 k + 1)! (4 k + 3)}$ $(2 k + 1)! = k! . \underset{j = 1}{\prod^{k + 1}} (k + j) = k! . 2^{k} . 3 . . . . (2 k + 1)$ $= \frac{π^{3}}{3} . \sum_{k ⩾ 0} \frac{3 {(- 1)}^{k} π^{4 k}}{k! . 2^{k} . 3 . . . . (2 k + 1) (4 k + 3)}$ $= \frac{π^{3}}{3} \sum_{k ⩾ 0} \frac{2^{k} . 3}{3 . . . (2 k + 1) . (4 k + 3)} . {(\frac{- π^{4}}{4})}^{k} . \frac{1}{k!}$ $= \frac{π^{3}}{3} . \sum_{k ⩾ 0} \frac{2^{k} . 3 . . . . . (4 k - 1)}{3 . . . . . (2 k + 1) . (7 . . . . . (4 k - 1) (4 k + 3)} . {(\frac{- π^{4}}{4})}^{k} . \frac{1}{k!}$ $= \frac{π^{3}}{3} . \sum_{k ⩾ 0} \frac{\frac{1}{4^{k}} . (3) . . . (4 k - 1)}{\frac{1}{2^{k}} . (3) . . . (2 k + 1) . \frac{1}{4^{k}} (7) . . . (4 k + 3)} . {(- \frac{π^{4}}{4})}^{k} . \frac{1}{k!}$ $= \frac{π^{3}}{3} \sum_{k ⩾ 0} \frac{(\frac{3}{4}) . . . . (\frac{3}{4} + k - 1)}{(\frac{3}{2}) . . . . (\frac{3}{2} + k - 1) . (\frac{7}{4}) . . . . (\frac{7}{4} + k - 1)} . {(\frac{- π^{4}}{4})}^{k} . \frac{1}{k!}$ $= \frac{π^{3}}{3}_{1} F_{2} (\frac{3}{4}; \frac{3}{2}; \frac{7}{4}; - \frac{π^{4}}{4})$
Commented by TawaTawa last updated on 12/Feb/20

$God bless you sir . Weldon$
Commented by M±th+et£s last updated on 12/Feb/20

$thank you so much sir . god bless you$
Commented by M±th+et£s last updated on 12/Feb/20

$i will post my solution in another time$
Commented by mind is power last updated on 12/Feb/20

$w i t h e p l e a s u r$
	Answered by M±th+et£s last updated on 12/Feb/20

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