calculate-1-dx-x-4-x-1-

Question Number 38119 by maxmathsup by imad last updated on 22/Jun/18

c a l c u l a t e \int_{1}^{+ \infty} \frac{d x}{x^{4} \sqrt{x - 1}}

Commented by math khazana by abdo last updated on 22/Jun/18

c h a n g e m e n t \sqrt{x - 1} = t g i v e x - 1 = t^{2} \Rightarrow

I = \int_{0}^{+ \infty} \frac{2 t d t}{{(1 + t^{2})}^{4} t} = 2 \int_{0}^{+ \infty} \frac{d t}{{(1 + t^{2})}^{4}}

= \int_{- \infty}^{+ \infty} \frac{d t}{{(1 + t^{2})}^{4}} l e t c o n s i d e r t h e c o m p l e x

f u n c t i o n f (z) = \frac{1}{{(z^{2} + 1)}^{4}}

f (z) = \frac{1}{{(z - i)}^{4} {(z + i)}^{4}} s o t h e p o l e s o f f a r e i a n d - i

(w i t h m u l t i p l i c i t y 4)

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π R e s (f, i)

R e s (f, i) = {l i m}_{z \to i} \frac{1}{(4 - 1)!} {{(z - i)}^{4} f (z)}^{(3)}

= {l i m}_{z \to i} \frac{1}{3!} {{(z + i)}^{- 4}}^{(3)} b u t

{(z + i)}^{- 4}}^{(1)} = - 4 {(z + i)}^{- 5}

{{(z + i)}^{- 4}}^{(2)} = 20 {(z + i)}^{- 6}

{{(z + i)}^{- 4}}^{(3)} = - 120 {(z + i)}^{- 7}

R e s (f, i) = {l i m}_{z \to i} \frac{1}{6} (- 120) {(z + i)}^{- 7}

= - 20 {(2 i)}^{- 7} = \frac{- 20}{2^{7} i^{7}} = - \frac{2^{2} . 5}{2^{2} . 2^{5} (- i)} = \frac{5}{32 i}

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π \frac{5}{32 i} = \frac{5 π}{16} \Rightarrow

I = \frac{5 π}{16} .