1-decompose-F-x-1-x-4-x-3-5-2-calculate-5-dx-x-4-x-3-5-

Question Number 92971 by abdomathmax last updated on 10/May/20

1) d e c o m p o s e F (x) = \frac{1}{x^{4} {(x - 3)}^{5}}

2) c a l c u l a t e \int_{5}^{+ \infty} \frac{d x}{x^{4} {(x - 3)}^{5}}

Commented by mathmax by abdo last updated on 11/May/20

1) F (x) = \sum_{i = 1}^{4} \frac{a_{i}}{x^{i}} + \sum_{i = 1}^{5} \frac{b_{i}}{{(x - 3)}^{i}} a_{i} ? F (x) = \frac{1}{x^{4} {(x - 3)}^{5}}

w e f i n d D_{4} (0) f o r g (x) = {(x - 3)}^{- 5}

g (x) = \sum_{p = 0}^{4} \frac{g^{(p)} (0)}{p!} x^{p} + \frac{x^{5}}{5!} ξ (x) = g (0) + x g^{^{'}} (0) + \frac{x^{2}}{2} g^{(2)} (0) + \frac{x^{3}}{3!} g^{(3} (0)

+ \frac{x^{4}}{4!} g^{(4)} (0) + \frac{x^{5}}{5!} ξ (x)

g (0) = {(- 3)}^{- 5}, g^{^{'}} (x) = - 5 {(x - 3)}^{- 6} \to g^{^{'}} (0) = - 5 {(- 3)}^{- 6}

g^{(2)} (x) = 30 {(x - 3)}^{- 7} \Rightarrow g^{(2)} (0) = 30 {(- 3)}^{- 7}

g^{(3)} (x) = - 210 {(x - 3)}^{- 8} \Rightarrow g^{(3)} (0) = - 210 {(- 3)}^{- 8}

g^{(4)} (x) = 210 \times 8 {(x - 3)}^{- 9} \Rightarrow g^{(4)} (x) = 1680 {(- 3)}^{- 9} \Rightarrow

g (x) = {(- 3)}^{- 5} - 5 {(- 3)}^{- 6} x + 15 {(- 3)}^{- 7} x^{2} - 35 {(- 3)}^{- 8} x^{3}

+ \frac{1680}{4!} {(- 3)}^{- 9} x^{4} + \frac{x^{5}}{5!} ξ (x) \Rightarrow

\frac{g (x)}{x^{4}} = \frac{{(- 3)}^{- 5}}{x^{4}} - \frac{5 {(- 3)}^{- 6}}{x^{3}} + \frac{15 {(- 3)}^{- 7}}{x^{2}} - \frac{35 {(- 3)}^{- 8}}{x} + \frac{1680}{4!} {(- 3)}^{- 9} + x θ (x)

\Rightarrow a_{1} = - \frac{35}{{(- 3)}^{8}}, a_{2} = \frac{15}{{(- 3)}^{7}}, a_{3} = \frac{- 5}{{(- 3)}^{6}}, a_{4} = \frac{1}{{(- 3)}^{5}} l e t f i n d b_{i}

w e f i n d D_{4} (3) f o r h (x) = x^{- 4}

h (x) = \sum_{p = 0}^{4} \frac{h^{(p)} (3)}{p!} {(x - 3)}^{p} + \frac{{(x - 3)}^{5}}{5!} ξ (x - 3)

= h (3) + \frac{x - 3}{1!} h^{^{'}} (3) + \frac{{(x - 3)}^{2}}{2!} h^{(2)} (3) + \frac{{(x - 3)}^{3}}{3!} h^{(3)} (3) + \frac{{(x - 3)}^{4}}{4!} h^{(4)} (3) + \dots

h (3) = 3^{- 4}, h^{^{'}} (x) = - 4 x^{- 5} \to h^{^{'}} (3) = - 4 3^{- 5}

h^{(2)} (x) = 20 x^{- 6} \to h^{(2)} (3) = 20 . 3^{- 6}

h^{(3)} (x) = - 120 x^{- 7} \to h^{(3)} (3) = - 120 . 3^{- 7}

h^{(4)} (x) = 120.7 x^{- 8} = 840 x^{- 8} \Rightarrow

h (x) = 3^{- 4} - 4 . 3^{- 5} (x - 3) + 10 . 3^{- 6} {(x - 3)}^{2} - 20 . 3^{- 7} {(x - 3)}^{3}

+ \frac{840}{4!} {(x - 3)}^{4} + \dots \Rightarrow

\frac{h (x)}{{(x - 3)}^{5}} = \frac{1}{3^{4} {(x - 3)}^{5}} - \frac{4}{. 3^{5} {(x - 3)}^{4}} + \frac{10}{3^{6} {(x - 3)}^{3}} - \frac{20}{3^{7} {(x - 3)}^{2}}

+ \frac{840}{4! (x - 3)} + \dots \Rightarrow b_{1} = \frac{840}{4!}, b_{2} = \frac{20}{3^{7}}, b_{3} = \frac{10}{3^{6}}, b_{4} = \frac{4}{3^{5}}, b_{5} = \frac{1}{3^{4}}

t h e c o m p o s i t i o n o f F (x) i s k n o w n \dots

r e m a r k f o r a_{i} w e c a n u s e D_{3} (0) f o r g (x) = {(x - 3)}^{- 5} o n l y \dots

Commented by abdomathmax last updated on 11/May/20

2) w e c a n f i n d t h i s i n t e g r a l w i t h o u t u s i n g

d e c o m p o s i t i o n! l e t I = \int_{5}^{+ \infty} \frac{d x}{x^{4} {(x - 3)}^{5}} \Rightarrow

I = \int_{5}^{+ \infty} \frac{d x}{{(\frac{x}{x - 3})}^{4} {(x - 3)}^{9}} c h a n g e m e n t \frac{x}{x - 3} = t g i v e

x = t x - 3 t \Rightarrow (1 - t) x = - 3 t \Rightarrow x = \frac{3 t}{t - 1} \Rightarrow

\frac{d x}{d t} = \frac{3 (t - 1) - 3 t}{{(t - 1)}^{2}} = \frac{- 3}{{(t - 1)}^{2}} a n d x - 3 = \frac{3 t}{t - 1} - 3

= \frac{3 t - 3 t + 3}{t - 1} = \frac{3}{(t - 1)} \Rightarrow

I = \int_{\frac{5}{2}}^{1} \frac{- 3 d t}{{(t - 1)}^{2} t^{4} {(\frac{3}{t - 1})}^{9}} = \frac{1}{3^{8}} \int_{1}^{\frac{5}{2}} \frac{{(t - 1)}^{9}}{{(t - 1)}^{2} . t^{4}} d t

= \frac{1}{3^{8}} \int_{1}^{\frac{5}{2}} \frac{{(t - 1)}^{7}}{t^{4}} d t \Rightarrow

3^{8} \times I = \int_{1}^{\frac{5}{2}} \frac{\sum_{k = 0}^{7} C_{7}^{k} t^{k} {(- 1)}^{7 - k}}{t^{4}} d t

= - \int_{1}^{\frac{5}{2}} \sum_{k = 0}^{7} {(- 1)}^{k} C_{7}^{k} t^{k - 4} d t

= - \sum_{k = 0 a n d k \neq 3}^{7} {(- 1)}^{k} C_{7}^{k} {[\frac{1}{k - 3} t^{k - 3}]}_{1}^{\frac{5}{2}}

+ C_{7}^{3} \int_{1}^{\frac{5}{2}} \frac{d t}{t}

= - \sum_{k = 0 a n d k \neq 3}^{7} \frac{{(- 1)}^{k}}{k - 3} C_{7}^{k} {{(\frac{5}{2})}^{k - 3} - 1}

+ C_{7}^{3} l n (\frac{5}{2})

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

+ C_{7}^{3} l n (\frac{5}{2}) .

Commented by abdomathmax last updated on 11/May/20

k \neq 3