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Question Number 114135 by mnjuly1970 last updated on 17/Sep/20

\dots . A d v a n c e d m a t h e m a t i c s \dots

i :: p r o v e t h a t : Ω = \frac{1}{π} \int_{0}^{\infty} \frac{1}{{(x^{2} - x + 1)}^{2} \sqrt{x}} d x = 1

i i :: e v a l u a t e :: Φ = \int_{0}^{1} x^{2} l n (x) l n (1 - x) d x = ? ? ?

\dots . m . n . j u l y . 1970 \dots .

Answered by MJS_new last updated on 17/Sep/20

\int \frac{d x}{{(x^{2} - x + 1)}^{2} \sqrt{x}} =

[t = \sqrt{x} \to d x = 2 \sqrt{x} d t]

= 2 \int \frac{d t}{{(t^{4} - t^{2} + 1)}^{2}} =

[{Ostrogradski}^{'} s Method]

= \frac{t (t^{2} + 1)}{3 (t^{4} - t^{2} + 1)} + \frac{1}{3} \int \frac{t^{2} + 5}{t^{4} - t^{2} + 1} d t

\frac{1}{3} \int \frac{t^{2} + 5}{t^{4} - t^{2} + 1} d t =

= \frac{1}{18} \int (\frac{4 \sqrt{3} t + 15}{t^{2} + \sqrt{3} t + 1} - \frac{4 \sqrt{3} t - 15}{t^{2} - \sqrt{3} t + 1}) d t =

[using formula]

= \frac{\sqrt{3}}{9} \ln \frac{t^{2} + \sqrt{3} t + 1}{t^{2} - \sqrt{3} t + 1} + \arctan (2 t - \sqrt{3}) + \arctan (2 t + \sqrt{3})

\Rightarrow

\int \frac{d x}{{(x^{2} - x + 1)}^{2} \sqrt{x}} =

= \frac{(x + 1) \sqrt{x}}{3 (x^{2} - x + 1)} + \ln \frac{x + 1 + \sqrt{3 x}}{x + 1 - \sqrt{3 x}} + \arctan (2 \sqrt{x} - \sqrt{3}) + \arctan (2 \sqrt{x} + \sqrt{3}) + C

\Rightarrow \frac{1}{π} \underset{0}{\int^{\infty}} \frac{d x}{{(x^{2} - x + 1)}^{2} \sqrt{x}} = 1

Commented by mnjuly1970 last updated on 17/Sep/20

t h a n k y o u s o m u c h m r

Answered by mathmax by abdo last updated on 18/Sep/20

let prove thst \int_{0}^{\infty} \frac{dx}{{(x^{2} - x + 1)}^{2} \sqrt{x}} = π

changement \sqrt{x} = t give I = \int_{0}^{\infty} \frac{2 tdt}{{(t^{4} - t^{2} + 1)}^{2} t}

= 2 \int_{0}^{\infty} \frac{dt}{{(t^{4} - t^{2} + 1)}^{2}} = \int_{- \infty}^{+ \infty} \frac{dt}{{(t^{4} - t^{2} + 1)}^{2}} let φ (z) = \frac{1}{{(z^{4} - z^{2} + 1)}^{2}}

poles of φ ? z^{4} - z^{2} + 1 = 0 \Rightarrow u^{2} - u + 1 = 0 (u = z^{2})

Δ = - 3 \Rightarrow u_{1} = \frac{1 + 3 i}{2} = e^{\frac{i π}{3}} and u_{2} = \frac{1 - 3 i}{2} = e^{- \frac{i π}{3}}

\Rightarrow z^{4} - z^{2} + 1 = (z^{2} - e^{\frac{i π}{3}}) (z^{2} - e^{- \frac{i π}{3}}) = (z - e^{\frac{i π}{6}}) (z + e^{\frac{i π}{6}}) (z - e^{- \frac{i π}{6}}) (z + e^{- i \frac{π}{6}})

\Rightarrow φ (z) = \frac{1}{{(z - e^{\frac{i π}{6}})}^{2} {(z + e^{\frac{i π}{6}})}^{2} {(z - e^{- \frac{i π}{6}})}^{2} {(z + e^{- \frac{i π}{6}})}^{2}}

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{6}}) + Res (φ, - e^{- \frac{i π}{6}})}

Res (φ, e^{\frac{i π}{6}}) = \lim_{z \to e^{\frac{i π}{6}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{6}})}^{2} φ (z)}^{(1)}

= \lim_{z \to e^{\frac{i π}{6}}} {\frac{1}{{(z + e^{\frac{i π}{6}})}^{2} {(z^{2} - e^{- \frac{i π}{3}})}^{2}}}^{(1)}

= - \lim_{z \to e^{\frac{i π}{6}}} \frac{2 (z + e^{\frac{i π}{6}}) {(z^{2} - e^{- \frac{i π}{3}})}^{2} + 4 z (z^{2} - e^{- \frac{i π}{3}}) {(z + e^{\frac{i π}{6}})}^{2}}{{(z + e^{\frac{i π}{6}})}^{4} {(z^{2} - e^{- \frac{i π}{3}})}^{4}}

= - {2 \lim}_{z \to e^{\frac{i π}{6}}} \frac{(z^{2} - e^{- \frac{i π}{3}}) + 2 z (z + e^{\frac{i π}{6}})}{{(z + e^{\frac{i π}{6}})}^{3} {(z^{2} - e^{- \frac{i π}{3}})}^{3}} \dots . be continued \dots .

Commented by mathmax by abdo last updated on 18/Sep/20

let I = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{4} - x^{2} + 1)}^{2}} let try parametric method

f (a) = \int_{- \infty}^{+ \infty} \frac{dx}{x^{4} - x^{2} + a} with a > \frac{1}{4} we have f^{^{'}} (a) = - \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{4} - x^{2} + a)}^{2}}

I = - f^{^{'}} (1) let explicit f (a)

φ (z) = \frac{1}{z^{4} - z^{2} + a} poles of φ ?

u^{2} - u + a = 0 (u = z^{2}) \to Δ = 1 - 4 a < 0 \Rightarrow u_{1} = \frac{1 + i \sqrt{4 a - 1}}{2}

u_{2} = \frac{1 - i \sqrt{4 a - 1}}{2} we have ∣ u_{1} ∣ = \frac{1}{2} \sqrt{1 + 4 a - 1} = \sqrt{a} \Rightarrow

u_{1} = \sqrt{a} e^{iar 4 \tan \sqrt{4 a - 1}} and u_{2} = \sqrt{a} e^{- iarctan (\sqrt{4 a - 1})} \Rightarrow

z^{4} - z^{2} + a = (z^{2} - \sqrt{a} e^{iarctan (\sqrt{4 a} - 1)}) (z^{2} - \sqrt{a} e^{- iarftan (\sqrt{4 a - 1})}) \Rightarrow

φ (z) = \frac{1}{(z - \sqrt{\sqrt{a}} e^{\frac{i}{2} \arctan (\sqrt{4 a - 1})}) (z + \sqrt{\sqrt{a}} e^{\frac{i}{2} \arctan (\sqrt{4 a - 1})}) (z - \sqrt{\sqrt{a}} e^{- \frac{i}{2} \arctan (\sqrt{4 a - 1})}) (z + \sqrt{\sqrt{a}} e^{- \frac{i}{2} \arctan (\sqrt{{a - 1})})}

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, \sqrt{\sqrt{a}} e^{\frac{i}{2} \arctan (\sqrt{4 a} - 1}) + Res (φ, - \sqrt{\sqrt{a}} e^{- \frac{i}{2} \arctan (\sqrt{4 a - 1})}}

Res (φ, (^{4} \sqrt{a}) e^{\frac{i}{2} \arctan (\sqrt{4 a - 1})})

= \frac{1}{2 (^{4} \sqrt{a}) e^{\frac{i}{2} ar 4 \tan (\sqrt{4 a - 1})} \times \sqrt{a} (2 i \sin (\arctan (\sqrt{4 a - 1})}

= \frac{e^{- \frac{i}{2} \arctan (\sqrt{4 a - 1})}}{4 i a^{\frac{3}{4}} \sin (\arctan (\sqrt{4 a - 1})}

Res (φ, - (^{4} \sqrt{a}) e^{- \frac{i}{2} \arctan (\sqrt{4 a - 1})})

= \frac{1}{- 2 (^{4} \sqrt{a}) e^{- \frac{i}{2} \arctan (\sqrt{4 a - 1})} (\sqrt{a}) (- 2 i \sin (\arctan (\sqrt{4 a - 1})}

= \frac{e^{\frac{i}{2} \arctan (\sqrt{4 a - 1})}}{4 i a^{\frac{3}{4}} \sin (\arctan (\sqrt{4 a - 1})} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {\frac{a^{- \frac{3}{4}}}{4 i} \times \frac{e^{- \frac{i}{2} \arctan (\sqrt{4 a - 1})}}{\sin (\arctan (\sqrt{4 a - 1})} + \frac{a^{- \frac{3}{4}}}{4 i} \times \frac{e^{\frac{i}{2} \arctan (\sqrt{4 a - 1})}}{\sin (\arctan (\sqrt{4 a - 1})}}

= \frac{π}{2 a^{\frac{3}{4}} \sin (\arctan (\sqrt{4 a - 1})} \times 2 \cos (\arctan (\sqrt{4 a - 1})

\dots . be continued \dots .