calculate-by-complex-method-0-dx-x-2-x-1-

Question Number 86375 by mathmax by abdo last updated on 28/Mar/20

c a l c u l a t e b y c o m p l e x m e t h o d \int_{0}^{\infty} \frac{d x}{x^{2} - x + 1}

Commented by mathmax by abdo last updated on 30/Mar/20

f i r s t l e t f i n d t h e v a l u e o f t h i s i n t e g r a l

w e h a v e \int \frac{d x}{x^{2} - x + 1} = \int \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} =_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int \frac{1}{u^{2} + 1} \times \frac{\sqrt{3}}{2} d u

= \frac{2}{\sqrt{3}} a r c t a n (u) + c = \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + C \Rightarrow

\int_{0}^{\infty} \frac{d x}{x^{2} - x + 1} = {[\frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}})]}_{0}^{+ \infty} = \frac{2}{\sqrt{3}} {\frac{π}{2} + a r c t a n (\frac{1}{\sqrt{3}})}

= \frac{2}{\sqrt{3}} {\frac{π}{2} + \frac{π}{6}} = \frac{2}{\sqrt{3}} {\frac{4 π}{6}} = \frac{2}{\sqrt{3}} \times \frac{2 π}{3} = \frac{4 π}{3 \sqrt{3}}

n o w l e t u s e c o m p l e x m e t h o d t h e p r o b l e m c o m e f r o m 0!

x^{2} - x + 1 = 0 \to Δ = - 3 \Rightarrow x_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} a n d x_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}}

\Rightarrow \int \frac{d x}{x^{2} - x + 1} = \int \frac{d x}{(x - e^{\frac{i π}{3}}) (x - e^{- \frac{i π}{3}})} = \frac{1}{2 i s i n (\frac{π}{3})} \int (\frac{1}{x - e^{\frac{i π}{3}}} - \frac{1}{x - e^{- \frac{i π}{3}}}) d x

= \frac{1}{i \sqrt{3}} l n (\frac{x - e^{\frac{i π}{3}}}{x - e^{- \frac{i π}{3}}}) + C \Rightarrow

\int_{0}^{\infty} \frac{d x}{x^{2} - x + 1} = \frac{1}{i \sqrt{3}} {[l n (\frac{x - e^{\frac{i π}{3}}}{x - e^{- \frac{i π}{3}}})]}_{0}^{+ \infty} = \frac{1}{i \sqrt{3}} {- l n (e^{\frac{2 i π}{3}})}

= - \frac{1}{i \sqrt{3}} (\frac{2 i π}{3}) = \frac{2 π}{3 \sqrt{3}} n o t c o r r e c t t h e p r o b l e m c o m e s f r o m 0

l e t c h a n g e 0 b y 1

\int_{1}^{+ \infty} \frac{d x}{x^{2} - x + 1} = \frac{2}{\sqrt{3}} {[a r c t a n (\frac{2 x - 1}{\sqrt{3}})]}_{1}^{+ \infty}

= \frac{2}{\sqrt{3}} {\frac{π}{2} - a r c t a n (\frac{1}{\sqrt{3}})} = \frac{2}{\sqrt{3}} {\frac{π}{2} - \frac{π}{6}} = \frac{2}{\sqrt{3}} {\frac{π}{3}} = \frac{2 π}{3 \sqrt{3}}

c o m p l e x m e t h o d \to \int_{1}^{+ \infty} \frac{d x}{x^{2} - x + 1} = \frac{1}{i \sqrt{3}} {[l n (\frac{x - e^{\frac{i π}{3}}}{x - e^{- \frac{i π}{3}}})]}_{1}^{+ \infty}

= \frac{1}{i \sqrt{3}} (- l n (\frac{1 - e^{\frac{i π}{3}}}{1 - e^{- \frac{i π}{3}}})) w e h a v e

\frac{1 - e^{\frac{i π}{3}}}{1 - e^{- \frac{i π}{3}}} = \frac{1 - \frac{1}{2} - i \frac{\sqrt{3}}{2}}{1 - \frac{1}{2} + i \frac{\sqrt{3}}{2}} = \frac{\frac{1}{2} - i \frac{\sqrt{3}}{2}}{\frac{1}{2} + \frac{i \sqrt{3}}{2}} = \frac{e^{- \frac{i π}{3}}}{e^{\frac{i π}{3}}} = e^{- \frac{2 i π}{3}} \Rightarrow

l n (\dots) = - \frac{2 i π}{3} \Rightarrow \int_{1}^{+ \infty} \frac{d x}{x^{2} - x + 1} = \frac{1}{i \sqrt{3}} (\frac{- 2 i π}{3}) = \frac{- 2 π}{3 \sqrt{3}} (p r o b l e m e o f -)

i h a v e 2 r e m a r k h e r e

d o n t u s e c o m p l e x m e t h o d i f y o u h a v e 0 i n t h e i n t e g r a l

t h e f u n c t i o n x \to \frac{1}{x^{2} - x + 1} i s ⩾ 0 \Rightarrow \int_{1}^{+ \infty} \frac{d x}{x^{2} - x} ⩾ 0 s o

y o u m u s t e r a d i c a t e - i n t h e v a l u e \dots .

Commented by redmiiuser last updated on 31/Mar/20

s i r c a n y o u p l s c o m m e n t

o n m y p r o c e s s .

Answered by redmiiuser last updated on 30/Mar/20

l e t x^{2} - x = z

{(z + 1)}^{- 1}

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . z^{n}

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . {(x^{2} - x)}^{n}

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . \underset{k = 0}{\sum^{n}} \underset{k}{\overset{n}{C}} . {(- 1)}^{n - k} . {(x^{2})}^{k} . {(x)}^{n - k}

= \underset{n = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{n}} {(- 1)}^{2 n - k} . \underset{k}{\overset{n}{C}} . x^{2 k + n - k}

{\int_{0}}^{\infty} \underset{n = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{n}} i^{2 n - k} . \underset{k}{\overset{n}{C}} . x^{k + n} . d x

= \frac{\underset{n = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{n}} i^{2 n - k} . \underset{k}{\overset{n}{C}} . {(\infty)}^{k + n + 1}}{k + n + 1}

Commented by redmiiuser last updated on 30/Mar/20

s i r c a n y o u p l s c h e c k

t h e a n s w e r .

Commented by mathmax by abdo last updated on 01/Apr/20

n o t c o r r e c t s i r b e c s u s e \frac{1}{1 + z} = \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{n} i f w e h a v e ∣ z ∣< 1

y o u c a n t h i s i f y o u h a v e ∣ x^{2} - x ∣< 1 \dots

Commented by redmiiuser last updated on 01/Apr/20

s i r c a n y o u p l s c h e c k

\frac{1}{z + 1} e x p a n s i o n b y

{T a l o y r}^{'} s f o r m u l a .

Commented by redmiiuser last updated on 03/Apr/20

f o r z < 1 l i k e z = - l

∴ {(1 + z)}^{- 1} = {(1 - l)}^{- 1}

= \underset{n = 0}{\sum^{\infty}} l^{n}