find-0-dx-1-x-3-

Question Number 29003 by abdo imad last updated on 03/Feb/18

f i n d \int_{0}^{\infty} \frac{d x}{1 + x^{3}} .

Commented by abdo imad last updated on 03/Feb/18

l e t p u t I = \int_{0}^{\infty} \frac{d x}{1 + x^{3}} l e t p u t x^{3} = t \Leftrightarrow x = t^{\frac{1}{3}} a n d

I = \int_{0}^{\infty} \frac{1}{1 + t} \frac{1}{3} t^{\frac{1}{3} - 1} d t = \frac{1}{3} \int_{0}^{\infty} \frac{t^{\frac{1}{3} - 1}}{1 + t} d t b u t w e k n o w t h a t

\int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} w i t h 0 < a < 1 d u e t o t h a t

\int_{0}^{\infty} \frac{t^{\frac{1}{3} - 1}}{1 + t} d t = \frac{π}{s i n (\frac{π}{3})} = \frac{π}{\frac{\sqrt{3}}{2}} = \frac{2 π}{\sqrt{3}} \Rightarrow I = \frac{2 π}{3 \sqrt{3}} .

Answered by sma3l2996 last updated on 03/Feb/18

w e h a v e \frac{1}{1 + x^{3}} = \frac{1}{(x + 1) (x^{2} - x + 1)} = \frac{a}{1 + x} + \frac{b x + c}{x^{2} - x + 1}

a = \frac{1}{3}, c = \frac{2}{3}, b = - \frac{1}{3}

\frac{1}{1 + x^{3}} = \frac{1}{3} (\frac{1}{1 + x} - \frac{x - 2}{x^{2} - x + 1})

s o \int_{0}^{\infty} \frac{d x}{1 + x^{3}} = \frac{1}{3} \int_{0}^{\infty} (\frac{1}{x + 1} - \frac{2 x - 4}{2 (x^{2} - x + 1)}) d x

= \frac{1}{3} (\underset{x \to \infty}{l i m l n} (x + 1) - \frac{1}{2} \int_{0}^{\infty} \frac{2 x - 1 - 3}{x^{2} - x + 1} d x)

= \frac{1}{3} \underset{x \to \infty}{l i m} (l n (x + 1) - \frac{1}{2} l n (x^{2} - x + 1)) - \frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{2} - x + 1}

\int_{0}^{\infty} \frac{d x}{x^{2} - x + 1} = \int_{0}^{\infty} \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} = \frac{4}{3} \int_{0}^{\infty} \frac{d x}{{(\frac{2 x - 1}{\sqrt{3}})}^{2} + 1}

t = \frac{2 x - 1}{\sqrt{3}} \Rightarrow d x = \frac{\sqrt{3}}{2} d t

\int_{0}^{\infty} \frac{d x}{x^{2} - x + 1} = \frac{2 \sqrt{3}}{3} \int_{0}^{\infty} \frac{d t}{t^{2} + 1} = \frac{2 \sqrt{3}}{3} (\underset{x \to \infty}{l i m a r c t a n t} - 0)

= \frac{\sqrt{3} π}{3}

A = \underset{x \to \infty}{l i m} (l n (x + 1) - \frac{1}{2} l n (x^{2} - x + 1))

= \underset{x \to \infty}{l i m} (l n (x (1 + \frac{1}{x})) - \frac{1}{2} l n (x^{2} (1 - \frac{1}{x} + \frac{1}{x^{2}}))

= \underset{x \to \infty}{l i m} (l n x + l n (1 + 1 / x) - l n (x) - \frac{1}{2} l n (1 - 1 / x + 1 / x^{2}))

= \underset{x \to \infty}{l i m} (l n (1 + \frac{1}{x}) - \frac{1}{2} l n (1 - \frac{1}{x} + \frac{1}{x^{2}})) = 0

s o \int_{0}^{\infty} \frac{d x}{1 + x^{3}} = \frac{1}{3} \times (0) - \frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{2} - x + 1} = - \frac{1}{2} \times \frac{\sqrt{3} π}{\sqrt{3}}

= - \frac{2 \sqrt{3} π}{3}

Commented by abdo imad last updated on 03/Feb/18

w e h a v e \frac{1}{1 + x^{3}} > 0 \forall x ⩾ 0 s o \int_{0}^{\infty} \frac{d x}{1 + x^{3}} > 0 s o h o w d o y o u

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