show that ∫xe^(−x^6 ) sin(x^3 ) dx=((Γ((5/6)))/3) 1F1[(5/6);(3/2);((−1)/4)]


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Question Number 82721 by M±th+et£s last updated on 23/Feb/20

$s h o w t h a t$ $\int {x e}^{- x^{6}} s i n (x^{3}) d x = \frac{Γ (\frac{5}{6})}{3} 1 F 1 [\frac{5}{6}; \frac{3}{2}; \frac{- 1}{4}]$
Commented by mind is power last updated on 23/Feb/20

$\int_{R} o r \int_{0}^{+ \infty} ?$
Commented by M±th+et£s last updated on 23/Feb/20

$\int_{- \infty}^{\infty}$ $s o r r y s i r i f o r g a t i t$
	Answered by mind is power last updated on 23/Feb/20

	$s i n (x) = \underset{k = 0}{\sum^{+ \infty}} \frac{{(- 1)}^{k} x^{2 k + 1}}{(2 k + 1)!}$ $s i n (x^{3}) = \underset{k = 0}{\sum^{+ \infty}} \frac{{(- 1)}^{k} x^{6 k + 3}}{(2 k + 1)!}$ $A = \int_{- \infty}^{+ \infty} {x e}^{- x^{6}} s i n (x^{3}) d x = 2 \int_{0}^{+ \infty} {x e}^{- x^{2}} s i n (x^{3}) d x$ $= 2 \int_{0}^{+ \infty} {x e}^{- x^{6}} [\sum_{k ⩾ 0} \frac{{(- 1)}^{k} x^{6 k + 3}}{(2 k + 1)!}]$ $A = 2 \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(2 k + 1)!} \int_{0}^{+ \infty} x^{6 k + 4} e^{- x^{6}} d x$ $u = x^{6} \Rightarrow d x = \frac{u^{- \frac{5}{6}}}{6}$ $\int_{0}^{+ \infty} x^{6 k + 4} e^{- x^{6}} d x = \frac{1}{6} \int_{0}^{+ \infty} u^{k} u^{\frac{4}{6} - \frac{5}{6}} e^{- u} d u$ $= \frac{1}{6} \int_{0}^{+ \infty} u^{k + \frac{5}{6} - 1} e^{- u} d u = \frac{1}{6} Γ (k + \frac{5}{6})$ $= \frac{1}{6 .} (\frac{6 k - 1}{6}) . . . . . . . (\frac{5}{6}) Γ (\frac{5}{6}) =$ $A = \frac{2}{6} \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(2 k + 1)!} \frac{(6 k - 1) . . . . . . 5}{6^{k}} Γ (\frac{5}{6})$ $= \frac{Γ (\frac{5}{6})}{3} . \sum_{k ⩾ 0} \frac{{(- 1)}^{k} (6 k - 1) . . . . . (5)}{(2 k + 1)!! . 2^{k} k! . 6^{k}} . . . . A$ $\frac{(6 k - 1) . . . . . . (5)}{6^{k}} = (k + \frac{5}{6} - 1) . . . . . . . . (\frac{5}{6}) = {(\frac{5}{6})}_{k}$ $\frac{(2 k + 1)!}{2^{k}} = (k + \frac{1}{2}) . . . . . . . . (\frac{3}{2}) = {(\frac{3}{2})}_{k}$ $\Rightarrow (2 k + 1)! = 2^{k} {(\frac{3}{2})}_{k}$ $A \Leftrightarrow \frac{Γ (\frac{5}{6})}{3} . \sum_{k ⩾ 0} \frac{{(- 1)}^{k} {(\frac{5}{6})}_{k}}{2^{k} {(\frac{3}{2})}_{k} . 2^{k} . k!} = \frac{Γ (\frac{5}{6})}{3} . \sum_{k ⩾ 0} \frac{{(- 1)}^{k} {(\frac{5}{6})}_{k}}{{(\frac{3}{2})}_{k} . 4^{k} . k!}$ $= \frac{Γ (\frac{5}{6})}{3} . \sum_{k ⩾ 0} \frac{{(\frac{5}{6})}_{k}}{{(\frac{3}{2})}_{k}} . \frac{{(\frac{- 1}{4})}^{k}}{k!^{}} = \frac{Γ (\frac{5}{6})}{3} ._{1} F_{1} (\frac{5}{6}; \frac{3}{2}; - \frac{1}{4})$
	Commented by M±th+et£s last updated on 23/Feb/20

	$b r i l i a n t s o l u t i o n f r o m a b r i l i a n t p e r s o n$ $t h a n k y o u s o m u c h s i r$
	Commented by mind is power last updated on 23/Feb/20

	$w i t h e p l e a s u r t h a n k y o u$
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