0-x-7-1-x-10-dx-

Question Number 122862 by Dwaipayan Shikari last updated on 20/Nov/20

\int_{0}^{\infty} \frac{x^{7}}{{(1 + x)}^{10}} d x

Answered by MJS_new last updated on 20/Nov/20

= \underset{0}{\int^{\infty}} \underset{j = 3}{\sum^{10}} \frac{{(- 1)}^{j + 1} (\begin{matrix} 7 \\ j - 3 \end{matrix})}{{(x + 1)}^{j}} = {[\underset{j = 3}{\sum^{10}} \frac{{(- 1)}^{j + 1} (\begin{matrix} 7 \\ j - 3 \end{matrix})}{(j - 1) {(x + 1)}^{j - 1}}]}_{0}^{\infty} =

= - \underset{j = 3}{\sum^{10}} \frac{{(- 1)}^{j + 1} (\begin{matrix} 7 \\ j - 3 \end{matrix})}{(j - 1)} =

= \frac{1}{2} - \frac{7}{3} + \frac{21}{4} - 7 + \frac{35}{6} - 3 + \frac{7}{8} - \frac{1}{9} = \frac{1}{72}

Commented by Dwaipayan Shikari last updated on 20/Nov/20

G r e a t w a y s i r!

Commented by Dwaipayan Shikari last updated on 20/Nov/20

\int_{0}^{\infty} \frac{x^{7}}{{(1 + x)}^{10}} d x \frac{x}{1 + x} = t \Rightarrow \frac{1}{{(1 + x)}^{2}} = \frac{d t}{d x} a n d 1 + x = \frac{1}{1 - t}

= \int_{0}^{1} \frac{x^{7}}{{(1 + x)}^{8}} d t = \int_{0}^{1} t^{7} (1 - t) d t

= β (8, 2) = \frac{Γ (8) Γ (2)}{Γ (10)} = \frac{7!}{9!} = \frac{1}{72}