Dirichlet Triple Integral

Evaluate the triple integral I=xα1yβ1zγ1dxdydzI = \iiint { { x }^{ \alpha -1 }{ y }^{ \beta -1 }{ z }^{ \gamma -1 }dxdydz } over the curve (xa)p+(yb)q+(zc)r=1{\left(\frac{x}{a}\right)}^{p} +{\left(\frac{y}{b}\right)}^{q} + {\left(\frac{z}{c}\right)}^{r} = 1.

Solution

Let u=(xa)px=au1pu = {\left(\frac{x}{a}\right)}^{p} \Rightarrow x = a{u}^{\frac{1}{p}} v=(yb)qy=bv1qv = {\left(\frac{y}{b}\right)}^{q} \Rightarrow y = b{v}^{\frac{1}{q}} w=(zc)rz=cw1rw = {\left(\frac{z}{c}\right)}^{r} \Rightarrow z = c{w}^{\frac{1}{r}}

and

dx=apu1p1dudx = \frac{a}{p}{u}^{\frac{1}{p}-1}du dy=bqv1q1dvdy = \frac{b}{q}{v}^{\frac{1}{q}-1}dv dz=crw1r1dwdz = \frac{c}{r}{w}^{\frac{1}{r}-1}dw

After these substitutions, the integral becomes much lighter: I=aαbβcγpqruαp1vβq1wγr1dwdvduI = \frac{{a}^{\alpha}{b}^{\beta}{c}^{\gamma}}{pqr} \iiint { { u }^{ \frac{\alpha}{p} -1 }{ v }^{ \frac{\beta}{q} -1 }{ w }^{ \frac{\gamma}{r} -1 }dwdvdu } evaluated over the curve u+v+w=1u + v + w = 1.

I=aαbβcγpqr0101u01uvuαp1vβq1wγr1dwdvduI=\frac { { a }^{ \alpha }{ b }^{ \beta }{ c }^{ \gamma } }{ pqr } \int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1-u }{ \int _{ 0 }^{ 1-u-v }{ { u }^{ \frac { \alpha }{ p } -1 }{ v }^{ \frac { \beta }{ q } -1 }{ w }^{ \frac { \gamma }{ r } -1 }dwdvdu } } }

I=aαbβcγpqrrγ0101uuαp1vβq1(1uv)γrdvduI=\frac { { a }^{ \alpha }{ b }^{ \beta }{ c }^{ \gamma } }{ pqr } \frac{r}{\gamma} \int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1-u }{ { u }^{ \frac { \alpha }{ p } -1 }{ v }^{ \frac { \beta }{ q } -1 }{ (1-u-v) }^{ \frac { \gamma }{ r } }dvdu } }

Substitute v=(1u)tv = (1-u)t and dv=(1u)dtdv = (1 - u)dt.

I=aαbβcγpqrrγ01uαp1(1u)(βq+γr)01tβq1(1t)γrdt.I=\frac { { a }^{ \alpha }{ b }^{ \beta }{ c }^{ \gamma } }{ pqr } \frac { r }{ \gamma } \int _{ 0 }^{ 1 }{ { u }^{ \frac { \alpha }{ p } -1 }{ (1-u) }^{ \left( \frac { \beta }{ q } +\frac { \gamma }{ r } \right) }\int _{ 0 }^{ 1 }{ { t }^{ \frac { \beta }{ q } -1 }{ (1-t) }^{ \frac { \gamma }{ r } }dt } } .

Here we apply two beta functions

I=aαbβcγpqrrγΓ(αp)Γ(βq+γr+1)Γ(αp+βq+γr+1)Γ(βq)Γ(γr+1)Γ(βq+γr+1).I=\frac { { a }^{ \alpha }{ b }^{ \beta }{ c }^{ \gamma } }{ pqr } \frac { r }{ \gamma } \frac { \Gamma \left( \frac { \alpha }{ p } \right) \Gamma \left( \frac { \beta }{ q } +\frac { \gamma }{ r } +1 \right) }{ \Gamma \left( \frac { \alpha }{ p } +\frac { \beta }{ q } +\frac { \gamma }{ r } +1 \right) } \frac { \Gamma \left( \frac { \beta }{ q } \right) \Gamma \left( \frac { \gamma }{ r } +1 \right) }{ \Gamma \left( \frac { \beta }{ q } +\frac { \gamma }{ r } +1 \right) } .

Applying the gamma function property Γ(n+1)=nΓ(n)\Gamma(n+1) = n\Gamma(n), we arrive at the result: I=aαbβcγpqrΓ(αp)Γ(βq)Γ(γr)Γ(αp+βq+γr+1).I=\frac { { a }^{ \alpha }{ b }^{ \beta }{ c }^{ \gamma } }{ pqr } \frac { \Gamma \left( \frac { \alpha }{ p } \right) \Gamma \left( \frac { \beta }{ q } \right) \Gamma \left( \frac { \gamma }{ r } \right) }{ \Gamma \left( \frac { \alpha }{ p } +\frac { \beta }{ q } +\frac { \gamma }{ r } +1 \right) } .

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
4 years, 11 months ago

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