1-decompose-F-x-1-x-1-5-2x-1-4-2-find-1-F-x-dx-

Question Number 133957 by mathmax by abdo last updated on 25/Feb/21

1) decompose F (x) = \frac{1}{{(x + 1)}^{5} {(2 x - 1)}^{4}}

2) find \int_{1}^{\infty} F (x) dx

Answered by Olaf last updated on 25/Feb/21

F (x) = \frac{1}{{(x + 1)}^{5} {(2 x - 1)}^{4}}

F (x) = \frac{1}{81 {(x + 1)}^{5}} + \frac{8}{243 {(x + 1)}^{4}} + \frac{40}{729 {(x + 1)}^{3}}

+ \frac{160}{2187 {(x + 1)}^{2}} + \frac{570}{6561 (x + 1)} + \frac{2}{243 {(x - \frac{1}{2})}^{4}}

- \frac{20}{729 {(x - \frac{1}{2})}^{3}} + \frac{40}{729 {(x - \frac{1}{2})}^{2}} - \frac{560}{6561 (x - \frac{1}{2})}

\int_{1}^{\infty} F (x) d x = [- \frac{1}{324 {(x + 1)}^{4}} - \frac{8}{729 {(x + 1)}^{3}}

- \frac{20}{729 {(x + 1)}^{2}} - \frac{160}{2187 (x + 1)} + \frac{570}{6561} \ln (x + 1)

- \frac{2}{729 {(x - \frac{1}{2})}^{3}} + \frac{10}{729 {(x - \frac{1}{2})}^{2}} - \frac{40}{729 (x - \frac{1}{2})}

{- \frac{560}{6561} \ln (x - \frac{1}{2})]}_{1}^{\infty}

= \frac{27051}{139968} - \frac{1120}{6561} \ln 2

Commented by mathmax by abdo last updated on 26/Feb/21

thank you sir

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

2) Φ = \int_{1}^{\infty} \frac{dx}{{(x + 1)}^{5} {(2 x - 1)}^{4}} \Rightarrow Φ = \int_{1}^{\infty} \frac{dx}{{(\frac{x + 1}{2 x - 1})}^{5} {(2 x - 1)}^{9}}

we do the changement \frac{x + 1}{2 x - 1} = t \Rightarrow x + 1 = 2 tx - t \Rightarrow (1 - 2 t) x = - 1 - t \Rightarrow

x = \frac{- 1 - t}{1 - 2 t} = \frac{t + 1}{2 t - 1} \Rightarrow \frac{dx}{dt} = \frac{2 t - 1 - 2 (t + 1)}{{(2 t - 1)}^{2}} = \frac{- 3}{{(2 t - 1)}^{2}} and

2 x - 1 = \frac{2 t + 2}{2 t - 1} - 1 = \frac{2 t + 2 - 2 t + 1}{2 t - 1} = \frac{3}{2 t - 1} \Rightarrow

Φ = \int_{2}^{\frac{1}{2}} \frac{1}{t^{5} {(\frac{3}{2 t - 1})}^{9}} \frac{- 3 dt}{{(2 t - 1)}^{2}} = \frac{1}{3^{8}} \int_{\frac{1}{2}}^{2} \frac{{(2 t - 1)}^{9}}{{(2 t - 1)}^{2} t^{5}} dt

= \frac{1}{3^{8}} \int_{\frac{1}{2}}^{2} \frac{{(2 t - 1)}^{7}}{t^{5}} dt \Rightarrow 3^{8} Φ = \int_{\frac{1}{2}}^{2} \frac{1}{t^{5}} \sum_{k = 0}^{7} C_{7}^{k} {(2 t)}^{k} {(- 1)}^{7 - k}

= - \sum_{k = 0}^{7} 2^{k} {(- 1)}^{k} C_{7}^{k} \int_{\frac{1}{2}}^{2} t^{k - 5} dt

= - \sum_{k = 0 and k \neq 4}^{7} {(- 2)}^{k} C_{7}^{k} {[\frac{1}{k - 4} t^{k - 4}]}_{\frac{1}{2}}^{2} - {(- 2)}^{4} C_{7}^{4} {[\ln ∣ t ∣]}_{\frac{1}{2}}^{2}

= - \sum_{k = 0 and k \neq 4}^{7} {(- 2)}^{k} \frac{C_{7}^{k}}{k - 4} (2^{k - 4} - \frac{1}{2^{k - 4}}) - 2^{4} C_{7}^{4} (2 \ln 2) \Rightarrow

Φ = - \frac{1}{3^{8}} {\sum_{k = 0 and k \neq 4}^{7} \frac{{(- 2)}^{k} C_{7}^{k}}{k - 4} (2^{k - 4} - \frac{1}{2^{k - 4}}) + 2^{5} \ln (2) C_{7}^{4}}