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Question Number 197376 by megrex last updated on 15/Sep/23

D o e s a n y o n e k n o w h o w t o p r o v e t h i s ?

\int \int \int_{V} \frac{d x d y d z}{1 + x^{4} + y^{4} + z^{4}} = \frac{Γ^{4} (\frac{1}{4})}{4^{4}}

w h e r e V i s t h e u n i t c u b e {[0, 1]}^{3}

T h a n k y o u .

Answered by witcher3 last updated on 15/Sep/23

\frac{1}{1 + x^{4} + y^{4} + z^{4}} = \int_{0}^{\infty} e^{- t (1 + x^{4} + y^{4} + z^{4})} dt

I = \int \int \int_{V} \frac{dxdydz}{1 + x^{4} + y^{4} + z^{4}} = \int \int \int_{V} \int_{0}^{\infty} e^{- t (1 + x^{4} + y^{4} + z^{4})} dt

= \int_{0}^{\infty} e^{- t} {\int_{0}^{1} e^{- {tx}^{4}} dx \int_{0}^{1} e^{- {ty}^{4}} dy \int_{0}^{1} e^{- {tz}^{4}} dz} dt

\int_{0}^{1} e^{- {ts}^{4}} ds, t \in [0, \infty [, \int_{0}^{\infty} e^{- y}

= \frac{1}{4} \int_{0}^{\infty} \frac{y^{\frac{1}{4} - 1}}{\sqrt[4]{t}} e^{- y} dy

= \frac{1}{4 \sqrt[4]{t}} . Γ (\frac{1}{4})

I = \int_{0}^{\infty} e^{- t} {\frac{Γ (\frac{1}{4})}{{4 t}^{\frac{1}{4}}}}^{3} = \frac{Γ^{3} (\frac{1}{4})}{4^{3}} \int_{0}^{\infty} t^{- \frac{3}{4}} e^{- t} dt

= \frac{Γ^{3} (\frac{1}{4})}{4^{3}} Γ (\frac{1}{4}) = \frac{Γ^{4} (\frac{1}{4})}{4^{3}} \dots may bee error of calculus

I didint find 4^{4}

Commented by megrex last updated on 16/Sep/23

V e r y n i c e l y d o n e - t h a n k y o u f o r y o u r t i m e

(I w i l l d o u b l e c h e c k m y e x t r a f a c t o r) .

T h i s c o u l d e a s i l y b e g e n e r a l i s e d t o

a d - d i m e n s i o n a l c u b e w i t h

a r b i t r a r y p o w e r s o f x^{p}

- w h i c h i s r e a l l y n i c e!

Commented by witcher3 last updated on 16/Sep/23

withe Pleasur