calculus-I-please-evaluate-0-1-1-1-x-6-dx-

Question Number 126766 by mnjuly1970 last updated on 24/Dec/20

\dots c a l c u l u s (I) \dots

p l e a s e e v a l u a t e ::

Ψ = \int_{0}^{1} (\frac{1}{1 + x^{6}}) d x = ?

Answered by Ar Brandon last updated on 24/Dec/20

x_{k}^{6} + 1 = 0 \Rightarrow x_{k} = e^{\frac{(2 k + 1) π i}{6}}

\Rightarrow x^{6} + 1 = \underset{k = 0}{\prod^{5}} (x - x_{k})

\Rightarrow Ψ = \int_{0}^{1} \frac{dx}{\underset{k = 0}{\prod^{5}} (x - x_{k})} = \int_{0}^{1} \underset{k = 0}{\sum^{5}} \frac{a_{k}}{x - x_{k}} dx

a_{k} = \frac{1}{{6 x}_{k}^{5}} = - \frac{x_{k}}{6}

Ψ = - \frac{1}{6} \int_{0}^{1} \underset{k = 0}{\sum^{5}} \frac{x_{k}}{x - x_{k}} dx = - \frac{1}{6} {\underset{k = 0}{\sum^{5}} x_{k} \ln ∣ x - x_{k} ∣}_{0}^{1}

= - \frac{1}{6} \underset{k = 0}{\sum^{5}} x_{k} \ln ∣ \frac{x_{k} - 1}{x_{k}} ∣

Commented by mnjuly1970 last updated on 24/Dec/20

t h a n k y o u s o m u c h s i r b r a n d o n . .

Answered by mindispower last updated on 24/Dec/20

= \int_{0}^{1} \frac{1 - x^{6}}{1 - x^{12_{}}}

= \frac{1}{12} \int_{0}^{1} \frac{1 - t^{\frac{1}{2}}}{1 - t} . t^{- \frac{11}{12}}

= \frac{1}{12} (- γ + \int_{0}^{1} \frac{t^{- \frac{11}{12}} - 1}{1 - t} d t) - \frac{1}{12} (- γ + \int_{0}^{1} \frac{1 - t^{\frac{- 5}{12}}}{1 - t} d t)

= \frac{Ψ (\frac{1}{12}) - Ψ (\frac{7}{12})}{12} .

w e h a v e c l o s e f o r Ψ (\frac{p}{q}) = - γ - l n (2 q) - \frac{π}{2} c o t (\frac{p π}{q})

+ 2 \underset{1}{\sum^{[\frac{q - 1}{2}]}} c o s (\frac{2 π n p}{q}) l n (s i n (\frac{π n}{q}))

Commented by mnjuly1970 last updated on 24/Dec/20

g r a t e f u l . . p e a c e b e u p o n y o u

s i r m i n d s . .

Answered by MJS_new last updated on 24/Dec/20

Ψ = \underset{0}{\int^{1}} \frac{d x}{x^{6} + 1} = {[\underset{j = 1}{\sum^{5}} I_{j}]}_{0}^{1}

I_{1} = \frac{1}{3} \underset{0}{\int^{1}} \frac{d x}{x^{2} + 1} = \frac{1}{3} {[\arctan x]}_{0}^{1} = \frac{π}{12}

I_{2} = \frac{1}{12} \underset{0}{\int^{1}} \frac{d x}{x^{2} + \sqrt{3} x + 1} = \frac{1}{6} {[\arctan (2 x + \sqrt{3})]}_{0}^{1} = \frac{π}{72}

I_{3} = \frac{\sqrt{3}}{12} \int \frac{2 x + \sqrt{3}}{x^{2} + \sqrt{3} x + 1} = \frac{\sqrt{3}}{12} {[\ln (x^{2} + \sqrt{3} x + 1)]}_{0}^{1} = \frac{\sqrt{3}}{12} \ln (2 + \sqrt{3})

I_{4} = \frac{1}{12} \int \frac{d x}{x^{2} - \sqrt{3} x + 1} = \frac{1}{6} {[\arctan (2 x - \sqrt{3})]}_{0}^{1} = \frac{5 π}{72}

I_{5} = - \frac{\sqrt{3}}{12} \int \frac{2 x - 3}{x^{2} - \sqrt{3} x + 1} = - \frac{\sqrt{3}}{12} {[\ln (x^{2} - \sqrt{3} x + 1)]}_{0}^{1} = - \frac{\sqrt{3}}{12} \ln (2 - \sqrt{3})

\Rightarrow

Ψ = \frac{π}{6} + \frac{\sqrt{3}}{6} \ln (2 + \sqrt{3})

Commented by mnjuly1970 last updated on 24/Dec/20

t h a n k s a l o t m r m j s . .

Answered by mnjuly1970 last updated on 24/Dec/20

Ψ = \frac{1}{2} \int_{0}^{1 (} \frac{1 + x^{4} - x^{2}) + x^{2} + (1 - x^{4})}{1 + x^{6}} d x

Ψ = \frac{1}{2} [\int_{0}^{1} \frac{1}{1 + x^{2}} d x = Ω_{1}] + \frac{1}{2} [\int_{0}^{1} \frac{x^{2}}{1 + {(x^{3})}^{2}} d x = Ω_{2}]

+ \frac{1}{2} [\int_{0}^{1} \frac{1 - x^{2}}{1 - x^{2} + x^{4}} d x = Ω_{3}]

∴ Ω_{1} = \frac{π}{4}

Ω_{2} \overset{x^{3} = t}{=} \frac{1}{3} \int_{0}^{1} \frac{1}{1 + t^{2}} d t = \frac{π}{12}

Ω_{3} = - \int \frac{1 - x^{- 2}}{x^{2} - 1 + x^{- 2}} d x = - \int_{0}^{1} \frac{1 - x^{- 2}}{{(x + x^{- 1})}^{2} - 3} d x

\overset{x + x^{- 1} = t}{=} - \int_{\infty}^{2} \frac{d t}{t^{2} - 3} = \frac{1}{2 \sqrt{3}} \int_{2}^{\infty} \frac{1}{t - \sqrt{3}} - \frac{1}{t + \sqrt{3}}

= \frac{1}{2 \sqrt{3}} l n {(\frac{t - \sqrt{3}}{t + \sqrt{3}})}_{2}^{\infty} = - \frac{1}{2 \sqrt{3}} l n (\frac{2 - \sqrt{3}}{2 + \sqrt{3}}) ✓

Ψ = \frac{π}{8} + \frac{π}{24} - \frac{1}{4 \sqrt{3}} l n (\frac{2 - \sqrt{3}}{2 + \sqrt{3}}) = \frac{π}{6} + \frac{1}{4 \sqrt{3}} l n (\frac{2 + \sqrt{3}}{2 - \sqrt{3}}) ✓

= \frac{π}{6} + \frac{1}{2 \sqrt{3}} l n (2 + \sqrt{3}) ✓