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Evaluate: $ I = \int_{0}^{1} \sqrt[3]{x log\left(\frac{1}{x}\right)} dx $

(10 marks)
asked Aug 21, 2017 by randomisation (1,920 points)  

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Let $ x = e^{-t} \implies dx = -e^{-t} dt \\ x = 0 \implies t = \infty \ and\  x = 1 \implies t = 0 $

Thus, $ I = \int_{0}^{\infty} \sqrt[3]{-e^{-t} t}\: e^{-t} \ dt \\  = -\int_{0}^{\infty} t^{1/3} e^{\frac{-4t}{3}} \ dt \\ $

Now let,  $ 4t/3 = u \implies dt = \frac{3}{4} du, \\ \implies I = - \left( \frac{3}{4} \right)^{4/3} \int_{0}^{\infty} u^{1/3} e^{-u} \ dt = - \left( \frac{3}{4} \right)^{4/3} \Gamma(4/3) $
answered Aug 21, 2017 by randomisation (1,920 points)  
selected Aug 21, 2017 by randomisation
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