Question-109863

Question Number 109863 by mathdave last updated on 26/Aug/20

Answered by mathdave last updated on 26/Aug/20

s o l u t i o n

r e c a l l t h a t

_{a} D_{x}^{- α} f (t) =_{a} I_{x}^{α} f (t) = \frac{1}{Γ (α)} \int_{a}^{x} {(x - t)}^{α - 1} f (t) d t f o r

x ≻ a

l e t α = \frac{1}{3} a = 0 x = 1 f (t) = t^{4}

_{0} I_{1}^{\frac{1}{3}} f (t^{4}) = \frac{1}{Γ (\frac{1}{3})} \int_{0}^{1} {(1 - t)}^{\frac{1}{3} - 1} t^{4} d t

= \frac{1}{Γ (\frac{1}{3})} \int_{0}^{1} t^{5 - 1} {(1 - t)}^{\frac{1}{3} - 1} d t = \frac{1}{Γ (\frac{1}{3})} β (5, \frac{1}{3})

= \frac{1}{Γ (\frac{1}{3})} ∙ \frac{Γ (5) . Γ (\frac{1}{3})}{Γ (5 + \frac{1}{3})} = \frac{Γ (5)}{Γ (\frac{16}{3})} = \frac{4!}{\frac{13}{3}!} = \frac{24}{\frac{13}{3} \times \frac{10}{3} \times \frac{7}{3} \times \frac{4}{3} \times \frac{1}{3} \times 0!}

∵_{0} I_{1}^{\frac{1}{3}} f (t^{3}) = \frac{729}{455}

b y m a t h d a v e

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