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Question Number 84680 by M±th+et£s last updated on 15/Mar/20

s h o w t h a t

\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{l o g (x y z)}{(1 + x^{2}) (1 + y^{2}) (1 + z^{2})} d x d y d z = \frac{- 3 π^{2} G}{16}

Answered by mind is power last updated on 15/Mar/20

= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{l n (x) + l n (y) + l n (z)}{(1 + x^{2}) (1 + y^{2}) (1 + z^{2})} d x d y d z

b y S y m e t r i e

= 3 \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{l n (x)}{(1 + x^{2}) (1 + y^{2}) (1 + z^{2})} d x d y d z

= 3 \int_{0}^{1} \int_{0}^{1} \frac{d y d z}{(1 + y^{2}) (1 + z^{2})} \int_{0}^{1} \frac{l n (x)}{(1 + x^{2})} d x

- G = \int_{0}^{1} \frac{l n (x)}{1 + x^{2}} \Rightarrow

= - 3 G \int_{0}^{1} \int_{0}^{1} \frac{d y d z}{(1 + y^{2}) (1 + z^{2})} = - 3 G \int_{0}^{1} \frac{d y}{(1 + y^{2})} \int_{0}^{1} \frac{d z}{1 + z^{2}}

= - 3 G . {(\frac{π}{4})}^{2} = \frac{- 3 π^{2} G}{16}

Commented by M±th+et£s last updated on 15/Mar/20

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