calculate-1-dx-1-x-3-

Question Number 201266 by Mathspace last updated on 02/Dec/23

c a l c u l a t e \int_{1}^{\infty} \frac{d x}{\sqrt{1 + x^{3}}}

Answered by witcher3 last updated on 03/Dec/23

\frac{1}{x^{3}} = y

\Leftrightarrow \frac{1}{3} \int_{0}^{1} \frac{y^{- \frac{5}{6}}}{\sqrt{1 + y}} dy

{(1 + y)}^{- \frac{1}{2}} = 1 + \sum_{n ⩾ 1} (- \frac{1}{2}) \dots . (- \frac{1}{2} - n + 1) . \frac{x^{n}}{n!}

\frac{y^{- \frac{1}{6}}}{\sqrt{1 + y}} = \sum_{n ⩾ 0} \frac{Γ (\frac{1}{2} + n) {(- y)}^{n} y^{- \frac{5}{6}}}{Γ (\frac{1}{2}) n!}

\Leftrightarrow \frac{1}{3} \int_{0}^{1} \sum_{n ⩾ 0} \frac{Γ (\frac{1}{2} + n)}{Γ (\frac{1}{2}) n!} {(- 1)}^{n} . y^{- \frac{5}{6} + n} dy = \frac{1}{3} \sum_{n ⩾ 0} {(- 1)}^{n} \frac{Γ (\frac{1}{2} + n)}{Γ (\frac{1}{2}) n!} . \frac{1}{n + \frac{1}{6}}

= 2 \sum_{n ⩾ 0} \frac{\frac{Γ (n + \frac{1}{2})}{Γ (\frac{1}{2})} . \frac{Γ (n + \frac{1}{6})}{Γ (\frac{1}{6})}}{\frac{Γ (\frac{7}{6} + n)}{Γ (\frac{7}{6})}} \frac{{(- 1)}^{n}}{n!}

= 2_{2} F_{1} (\frac{1}{2}, \frac{1}{6}, \frac{7}{6}, - 1) \approx 1, 89

Answered by Calculusboy last updated on 03/Dec/23

S o l u t i o n : \int_{1}^{\infty} {(1 + x^{3})}^{- \frac{1}{2}} d x

l e t x^{- 3} + 1 = t^{2} x = {(t^{2} - 1)}^{- \frac{1}{3}} d x = - \frac{1}{3} {(t^{2} - 1)}^{- \frac{4}{3}} \cdot 2 t d t = - \frac{2}{3} {(t^{2} - 1)}^{- \frac{4}{3}} t d t

w h e n x = \infty t = 1 a n d w h e n x = 1 t = \infty

I = \int_{\infty}^{1} \frac{- \frac{2}{3} {(t^{2} + 1)}^{- \frac{4}{3}} t d t}{{[1 + {(t^{2} - 1)}^{- 1}]}^{\frac{1}{2}}} \Leftrightarrow I = \frac{2}{3} \int_{1}^{\infty} \frac{{(t^{2} - 1)}^{- \frac{4}{3}} t d t}{\frac{{(t^{2})}^{\frac{1}{2}}}{{(t^{2} - 1)}^{\frac{1}{2}}}}

I = \frac{2}{3} \int_{1}^{\infty} \frac{d t}{{(t^{2} - 1)}^{\frac{5}{6}}} = 1.8947