0-x-x-t-1-2-t-5-2-dt-

Question Number 134070 by mohammad17 last updated on 27/Feb/21

\int_{0}^{x} {(x - t)}^{- \frac{1}{2}} {(t)}^{\frac{5}{2}} d t

Commented by mohammad17 last updated on 27/Feb/21

h e l p m e s i r

Answered by Dwaipayan Shikari last updated on 27/Feb/21

\int_{0}^{x} {(x - t)}^{- \frac{1}{2}} t^{\frac{5}{2}} d t t = x u

= x \int_{0}^{1} \frac{1}{\sqrt{x}} {(1 - u)}^{- \frac{1}{2}} x^{\frac{5}{2}} u^{\frac{5}{2}} d u

= x^{3} \int_{0}^{1} {(1 - u)}^{- \frac{1}{2}} u^{\frac{5}{2}} d u = x^{3} \frac{\sqrt{π} Γ (\frac{7}{2})}{Γ (3)} = \frac{5 π}{16} x^{3}

Answered by mathmax by abdo last updated on 27/Feb/21

Φ = \int_{0}^{x} {(x - t)}^{- \frac{1}{2}} t^{\frac{5}{2}} dt \Rightarrow Φ =_{t = xu} \int_{0}^{1} {(x - xu)}^{- \frac{1}{2}} {(xu)}^{\frac{5}{2}} xdu

= x^{\frac{1}{2}} . x^{\frac{5}{2}} \int_{0}^{1} u^{\frac{5}{2} + 1 - 1} {(1 - u)}^{1 - \frac{1}{2} - 1} du = x^{3} \int_{0}^{1} u^{\frac{7}{2} - 1} {(1 - u)}^{\frac{1}{2}} du

= x^{3} B (\frac{7}{2}, \frac{1}{2}) = x^{3} \frac{Γ (\frac{7}{2}) Γ (\frac{1}{2})}{Γ (\frac{7}{2} + \frac{1}{2})} = x^{3} \sqrt{π} \frac{Γ (\frac{7}{2})}{Γ (4)} = \frac{\sqrt{π}}{3!} x^{3} Γ (\frac{7}{2}) but

Γ (\frac{7}{2}) = Γ (\frac{5}{2} + 1) = \frac{5}{2} Γ (\frac{5}{2}) = \frac{5}{2} Γ (\frac{3}{2} + 1) = \frac{5}{2} . \frac{3}{2} Γ (\frac{3}{2})

= \frac{15}{4} Γ (\frac{1}{2} + 1) = \frac{15}{4} \frac{1}{2} Γ (\frac{1}{2}) = \frac{15}{8} \sqrt{π} \Rightarrow

Φ = \frac{\sqrt{π}}{6} x^{3} \frac{15}{8} \sqrt{π} = \frac{5 π}{16} x^{3}