${\displaystyle \int _{0}^{1}\int _{0}^{1}(xy)^{xy}\,dx\,dy}$ ist nach Substitution ${\displaystyle x\mapsto {\frac {x}{y}}}$ gleich ${\displaystyle \int _{0}^{1}\int _{0}^{y}x^{x}\,{\frac {dx}{y}}\,dy}$.

Nach dem Vertauschen der Integrationsreihenfolge ist dies ${\displaystyle \int _{0}^{1}\int _{x}^{1}x^{x}\,{\frac {dy}{y}}\,dx=\int _{0}^{1}x^{x}\,(-\log x)\,dx}$.

Wegen ${\displaystyle \int _{0}^{1}x^{x}\,(1+\log x)\,dx=\left[x^{x}\right]_{0}^{1}=0}$ ist ${\displaystyle \int _{0}^{1}x^{x}\,dx=\int _{0}^{1}x^{x}\,(-\log x)\,dx}$.

Teile das Einheitsquadrat ${\displaystyle Q=[0,1]^{2}\,}$ in zwei kongruente Dreiecke, jeweils mit dem Ursprung als Spitze.

${\displaystyle Q_{1}=\{(x,y)\in Q\,:\,x\geq y\}\,,\,Q_{2}=\{(x,y)\in Q\,:\,y\geq x\}}$

${\displaystyle I:=\iint _{Q}{\sqrt {x^{2}+y^{2}}}\,dx\,dy=2\iint _{Q_{1}}{\sqrt {x^{2}+y^{2}}}\,dx\,dy}$

Verwende nun Polarkoordinaten:

${\displaystyle {\begin{bmatrix}x=r\cos \varphi \\y=r\sin \varphi \\\end{bmatrix}}\quad \qquad dx\,dy=r\,dr\,d\varphi \quad \qquad r={\sqrt {x^{2}+y^{2}}}}$

Wegen

${\displaystyle x\geq y\iff r\cos \varphi \geq r\sin \varphi \iff \cos \varphi \geq \sin \varphi \iff 0\leq \varphi \leq {\frac {\pi }{4}}}$
${\displaystyle x\leq 1\iff 0\leq r\leq {\frac {1}{\cos \varphi }}}$

ist ${\displaystyle I=2\int _{0}^{\frac {\pi }{4}}\int _{0}^{\frac {1}{\cos \varphi }}r^{2}\,dr\,d\varphi ={\frac {2}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {d\varphi }{\cos ^{3}\varphi }}}$

${\displaystyle ={\frac {1}{3}}\left[{\frac {\sin \varphi }{\cos ^{2}\varphi }}+\log(\sec \varphi +\tan \varphi )\right]_{0}^{\frac {\pi }{4}}={\frac {1}{3}}\left({\sqrt {2}}+\log \left({\sqrt {2}}+1\right)\right)}$.

Teile den Einheitswürfel ${\displaystyle V=[0,1]^{3}\,}$ in drei kongruente Pyramiden mit dem Ursprung als Spitze.

${\displaystyle V_{1}=\{(x,y,z)\in V\,:\,x\geq y,z\}\,,\,V_{2}=\{(x,y,z)\in V\,:\,y\geq x,z\}\,,\,V_{3}=\{(x,y,z)\in V\,:\,z\geq x,y\}}$

${\displaystyle I:=\iiint _{V}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=3\iiint _{V_{1}}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz}$

Verwende nun Kugelkoordinaten:

${\displaystyle {\begin{bmatrix}x=r\sin \theta \cos \varphi \\y=r\sin \theta \sin \varphi \\z=r\cos \theta \end{bmatrix}}\quad \qquad dx\,dy\,dz=r^{2}\sin \theta \,dr\,d\theta \,d\varphi \quad \qquad r={\sqrt {x^{2}+y^{2}+z^{2}}}}$

Wegen

${\displaystyle x\geq y\iff r\sin \theta \cos \varphi \geq r\sin \theta \sin \varphi \iff \cos \varphi \geq \sin \varphi \iff 0\leq \varphi \leq {\frac {\pi }{4}}}$
${\displaystyle x\geq z\iff r\sin \theta \cos \varphi \geq r\cos \theta \iff \tan \theta \geq \sec \varphi \iff \arctan(\sec \varphi )\leq \theta \leq {\frac {\pi }{2}}}$
${\displaystyle x\leq 1\iff 0\leq r\leq {\frac {1}{\sin \theta \,\cos \varphi }}}$

ist ${\displaystyle I=3\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}\int _{0}^{\frac {1}{\sin \theta \,\cos \varphi }}r\cdot r^{2}\,\sin \theta \,dr\,d\theta \,d\varphi }$

${\displaystyle =3\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}\left[{\frac {r^{4}}{4}}\right]_{0}^{\frac {1}{\sin \theta \,\cos \varphi }}\,\sin \theta \,d\theta \,d\varphi ={\frac {3}{4}}\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}\,{\frac {d\varphi }{\cos ^{4}\varphi }}}$.

Substituiere ${\displaystyle \theta ={\text{arccot}}\,s}$, dann ist ${\displaystyle \int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}=\int _{0}^{\cos \varphi }{\sqrt {1+s^{2}}}\,ds=\left[{\frac {s}{2}}{\sqrt {1+s^{2}}}+{\frac {1}{2}}\,{\text{arsinh}}\,s\right]_{0}^{\cos \varphi }}$.

Also ist ${\displaystyle I={\frac {3}{8}}\int _{0}^{\frac {\pi }{4}}{\frac {\sqrt {1+\cos ^{2}\varphi }}{\cos ^{3}\varphi }}\,d\varphi +{\frac {3}{8}}\int _{0}^{\frac {\pi }{4}}{\frac {{\text{arsinh}}(\cos \varphi )}{\cos ^{4}\varphi }}\,d\varphi }$,

wobei ${\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sqrt {1+\cos ^{2}\varphi }}{\cos ^{3}\varphi }}\,d\varphi =\int _{0}^{1}{\frac {\sqrt {1+{\frac {1}{1+x^{2}}}}}{\frac {1}{{\sqrt {1+x^{2}}}^{3}}}}\,{\frac {dx}{1+x^{2}}}=\int _{0}^{1}{\sqrt {2+x^{2}}}\,dx}$

${\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}+{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}+{\text{arsinh}}{\frac {1}{\sqrt {2}}}}$ ist. Und ${\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {{\text{arsinh}}(\cos \varphi )}{\cos ^{4}\varphi }}\,d\varphi }$

${\displaystyle =\left[{\text{arsinh}}(\cos \varphi )\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}{\frac {\sin \varphi }{\sqrt {1+\cos ^{2}\varphi }}}\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)d\varphi }$

${\displaystyle ={\frac {4}{3}}\,{\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {1}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos ^{3}\varphi }}\,d\varphi +{\frac {2}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos \varphi }}\,d\varphi }$,

wobei ${\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos ^{3}\varphi }}\,d\varphi =\int _{0}^{1}{\frac {\frac {x^{2}}{1+x^{2}}}{{\sqrt {1+{\frac {1}{1+x^{2}}}}}\,{\frac {1}{{\sqrt {1+x^{2}}}^{3}}}}}\,{\frac {dx}{1+x^{2}}}=\int _{0}^{1}{\frac {x^{2}}{\sqrt {2+x^{2}}}}\,dx}$

${\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}-{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}-{\text{arsinh}}{\frac {1}{\sqrt {2}}}}$ ist,

und ${\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos \varphi }}\,d\varphi =\int _{0}^{1}{\frac {\frac {x^{2}}{1+x^{2}}}{{\sqrt {1+{\frac {1}{1+x^{2}}}}}\,{\frac {1}{\sqrt {1+x^{2}}}}}}\,{\frac {dx}{1+x^{2}}}}$

${\displaystyle =\int _{0}^{1}{\frac {x^{2}}{{\sqrt {2+x^{2}}}\,(1+x^{2})}}\,dx=\left[{\text{arsinh}}{\frac {x}{\sqrt {2}}}-\arctan {\frac {x}{\sqrt {2+x^{2}}}}\right]_{0}^{1}={\text{arsinh}}{\frac {1}{\sqrt {2}}}-{\frac {\pi }{6}}}$.


Also ist ${\displaystyle I={\frac {3}{8}}\left({\frac {\sqrt {3}}{2}}+{\text{arsinh}}{\frac {1}{\sqrt {2}}}\right)+{\frac {3}{8}}\left({\frac {4}{3}}\,{\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {1}{3}}\left({\frac {\sqrt {3}}{2}}-{\text{arsinh}}{\frac {1}{\sqrt {2}}}\right)+{\frac {2}{3}}\left({\text{arsinh}}{\frac {1}{\sqrt {2}}}-{\frac {\pi }{6}}\right)\right)}$

${\displaystyle ={\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {\sqrt {3}}{4}}-{\frac {\pi }{24}}}$, wobei ${\displaystyle {\text{arsinh}}{\frac {1}{\sqrt {2}}}=\log({\sqrt {3}}+1)-{\frac {\log 2}{2}}}$ ist.

Es sei ${\displaystyle V=[0,1]^{n}\,}$ der ${\displaystyle n\,}$-dimensionale Einheitswürfel und ${\displaystyle V_{j}=\left\{(x_{1},...,x_{n})\in V\,{\Big |}\,x_{i}^{\alpha _{i}}<x_{j}^{\alpha _{j}}\quad \forall i\neq j\right\}}$.

Für alle Tupel ${\displaystyle (x_{1},...,x_{n})\in V_{j}\,}$ ist dann ${\displaystyle x_{j}^{\alpha _{j}}=\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}}$ und ${\displaystyle x_{i}<x_{j}^{\alpha _{j}/\alpha _{i}}\,\,\forall i\neq j}$.

${\displaystyle V\,}$ ist die abgeschlossene Hülle der disjunkten Vereinigung aller ${\displaystyle V_{j}}$'s. Daher ist ${\displaystyle \int _{V}=\int _{V_{1}}+...+\int _{V_{n}}}$.


${\displaystyle \int _{V_{j}}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}dx_{1}\cdots dx_{n}=\int _{V_{j}}x_{j}^{\alpha _{j}}dx_{1}\cdots dx_{n}}$

${\displaystyle =\int _{0}^{1}\,\int _{0}^{x_{j}^{\alpha _{j}/\alpha _{n}}}\cdots {\widehat {\int _{0}^{x_{j}^{\alpha _{j}/\alpha _{j}}}}}\cdots \int _{0}^{x_{j}^{\alpha _{j}/\alpha _{1}}}x_{j}^{\alpha _{j}}\,dx_{1}\cdots {\widehat {dx_{j}}}\cdots dx_{n}\cdot dx_{j}}$

${\displaystyle =\int _{0}^{1}x_{j}^{\alpha _{j}}\,\,x_{j}^{\alpha _{j}/\alpha _{1}}\cdots {\widehat {x_{j}^{\alpha _{j}/\alpha _{j}}}}\cdots x_{j}^{\alpha _{j}/\alpha _{n}}\,dx_{j}=\int _{0}^{1}x_{j}^{\alpha _{j}\left(1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}\right)}\,dx_{j}}$

${\displaystyle ={\frac {1}{\alpha _{j}\left(1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}\right)+1}}={\frac {\frac {1}{\alpha _{j}}}{1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}+{\frac {1}{\alpha _{j}}}}}={\frac {\frac {1}{\alpha _{j}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}}$.

Also ist ${\displaystyle \int _{0}^{1}\cdots \int _{0}^{1}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}\,dx_{1}\cdots dx_{n}=\int _{V_{1}}+...+\int _{V_{n}}={\frac {{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}}$.

${\displaystyle I:=\int _{0}^{1}\int _{0}^{1}{\frac {(-\log xy)^{s-2}}{1-xy}}\,(1-x)\,dy\,dx}$

ist nach Substitution ${\displaystyle u=xy\,}$ gleich ${\displaystyle \int _{0}^{1}\int _{0}^{x}{\frac {(-\log u)^{s-2}}{1-u}}\,(1-x)\,{\frac {du}{x}}\,dx}$.

Vertauscht man die Integrationsreihenfolge, so ist ${\displaystyle I=\int _{0}^{1}\int _{u}^{1}\left({\frac {1}{x}}-1\right)dx\,{\frac {(-\log u)^{s-2}}{1-u}}\,du}$

${\displaystyle =\int _{0}^{1}\left(-\log u-(1-u)\right){\frac {(-\log u)^{s-2}}{1-u}}\,du=\int _{0}^{1}\left({\frac {(-\log u)^{s-1}}{1-u}}-(-\log u)^{s-2}\right)du}$.

Substituiert man ${\displaystyle u=e^{-t}\,}$, so ist ${\displaystyle I=\int _{0}^{\infty }\left({\frac {t^{s-1}}{e^{t}-1}}-t^{s-2}\,e^{-t}\right)dt}$.

Für ${\displaystyle {\text{Re}}(s)>1\,}$ ist das ${\displaystyle \Gamma (s)\zeta (s)-\Gamma (s-1)=\Gamma (s)\left(\zeta (s)-{\frac {1}{s-1}}\right)}$.

Wegen der analytischen Fortsetzbarkeit gilt dies auch für ${\displaystyle {\text{Re}}(s)>0\,}$,

wobei der Fall ${\displaystyle s=1\,}$ als Grenzwert ${\displaystyle \gamma \,}$ zu interpretieren ist.

${\displaystyle I:=\int _{0}^{\pi }\int _{0}^{\pi }\int _{0}^{\pi }{\frac {dx\,dy\,dz}{1-\cos x\,\cos y\,\cos z}}}$ ist nach den Substitutionen ${\displaystyle \left[{\begin{matrix}x\mapsto 2\arctan x\\y\mapsto 2\arctan y\\z\mapsto 2\arctan z\end{matrix}}\right]}$ gleich

${\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {{\frac {2dx}{1+x^{2}}}\cdot {\frac {2dy}{1+y^{2}}}\cdot {\frac {2dz}{1+z^{2}}}}{1-{\frac {1-x^{2}}{1+x^{2}}}\cdot {\frac {1-y^{2}}{1+y^{2}}}\cdot {\frac {1-z^{2}}{1+z^{2}}}}}=8\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {dx\,dy\,dz}{(1+x^{2})(1+y^{2})(1+z^{2})-(1-x^{2})(1-y^{2})(1-z^{2})}}}$

${\displaystyle =4\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {dx\,dy\,dz}{x^{2}+y^{2}+z^{2}+x^{2}y^{2}z^{2}}}}$.

Wechselt man von kartesischen Koordinaten zu Kugelkoordinaten ${\displaystyle \left[{\begin{matrix}x=r\,\sin \theta \,\cos \varphi \\y=r\,\sin \theta \,\sin \varphi \\z=r\,\cos \theta \qquad \;\end{matrix}}\right]}$, so ist

${\displaystyle I=4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {r^{2}\,\sin \theta \;dr\,d\theta \,d\varphi }{r^{2}+r^{2}\,\sin ^{2}\theta \,\cos ^{2}\varphi \cdot r^{2}\,\sin ^{2}\theta \,\sin ^{2}\varphi \cdot r^{2}\,\cos ^{2}\theta }}}$

${\displaystyle =4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {dr}{1+{\Big (}r\,\sin \theta \,{\sqrt {\cos \theta }}\,{\sqrt {\cos \varphi \,\sin \varphi }}{\Big )}^{4}}}\,\sin \theta \;d\theta \,d\varphi }$

Das ist nach Substitution ${\displaystyle t=r\,\sin \theta \,{\sqrt {\cos \theta }}\,{\sqrt {\sin \varphi \,\cos \varphi }}}$ gleich ${\displaystyle 4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {dt\,d\theta \,d\varphi }{(1+t^{4})\,{\sqrt {\cos \theta }}\,{\sqrt {\sin \varphi \,\cos \varphi }}}}}$

${\displaystyle =4\int _{0}^{\infty }{\frac {dt}{1+t^{4}}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\sqrt {\cos \theta }}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\varphi }{\sqrt {\sin \varphi \,\cos \varphi }}}=4\cdot {\frac {{\sqrt {2}}\,\pi }{4}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2\,{\sqrt {2\pi }}}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2{\sqrt {\pi }}}}={\frac {1}{4}}\,\left[\Gamma \left({\frac {1}{4}}\right)\right]^{4}}$.

Setze ${\displaystyle R(\lambda )=\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy}$.

Differenziere nach ${\displaystyle \lambda \,}$ :

${\displaystyle R'(\lambda )=\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot {\frac {3\lambda ^{2}}{xy}}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy}$

Substituiere ${\displaystyle z={\frac {\lambda ^{3}}{xy}}\,\Rightarrow \,x={\frac {\lambda ^{3}}{yz}}\,\Rightarrow \,dx=-{\frac {\lambda ^{3}}{yz^{2}}}\,dz}$ :

${\displaystyle R'(\lambda )=\int _{0}^{\infty }\int _{\infty }^{0}e^{-\left({\frac {\lambda ^{3}}{yz}}+y+z\right)}\cdot {\frac {3z}{\lambda }}\cdot {\frac {\lambda ^{-2}}{y^{{\frac {1}{3}}-1}\cdot z^{{\frac {1}{3}}-1}}}\cdot y^{{\frac {2}{3}}-1}\cdot {\frac {-\lambda ^{3}}{yz^{2}}}\,dz\,dy}$

${\displaystyle =3\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(y+z+{\frac {\lambda ^{3}}{yz}}\right)}\cdot y^{{\frac {1}{3}}-1}\cdot z^{{\frac {2}{3}}-1}\,dz\,dy=3\cdot R(\lambda )}$

${\displaystyle \Rightarrow \,R(\lambda )=R(0)\cdot e^{-3\lambda }}$ mit ${\displaystyle R(0)=\int _{0}^{\infty }\int _{0}^{\infty }e^{-(x+y)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy}$

${\displaystyle =\int _{0}^{\infty }x^{{\frac {1}{3}}-1}\,e^{-x}\,dx\cdot \int _{0}^{\infty }y^{{\frac {2}{3}}-1}\,e^{-y}\,dy=\Gamma \left({\frac {1}{3}}\right)\cdot \Gamma \left({\frac {2}{3}}\right)={\frac {\pi }{\sin {\frac {\pi }{3}}}}={\frac {2\pi }{\sqrt {3}}}}$

Setze ${\displaystyle F(z)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(x_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}$.

${\displaystyle F'(z)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(x_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,{\frac {-n\,z^{n-1}}{x_{1}\,x_{2}\cdots x_{n-1}}}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}$

${\displaystyle =n\,\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left({\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}+x_{2}+...+x_{n-1}+x_{1}\right)}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\cdot {\frac {x_{1}^{\frac {1}{n}}}{z}}\cdot {\frac {-z^{n}}{x_{1}^{2}\,x_{2}\cdots x_{n-1}}}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}$

Nach Substitution ${\displaystyle u_{1}={\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}}$ ist ${\displaystyle du_{1}={\frac {-z^{n}}{x_{1}^{2}\,x_{2}\cdots x_{n-1}}}\,dx_{1}}$ und wegen ${\displaystyle x_{1}={\frac {z^{n}}{u_{1}\cdot x_{2}\cdots x_{n-1}}}}$ ist ${\displaystyle x_{1}^{\frac {1}{n}}={\frac {z}{u_{1}^{\frac {1}{n}}\,x_{2}^{\frac {1}{n}}\cdots x_{n-1}^{\frac {1}{n}}}}}$.

${\displaystyle F'(z)=-n\,\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(u_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{u_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,x_{2}^{{\frac {1}{n}}-1}\cdots x_{n-1}^{{\frac {n-2}{n}}-1}\,u_{1}^{{\frac {n-1}{n}}-1}\,du_{1}\,dx_{2}\cdots dx_{n-1}}$

Also ist ${\displaystyle F'(z)=-n\,F(z)}$, woraus ${\displaystyle F(z)=C\cdot e^{-nz}}$ folgt.

Und ${\displaystyle C=F(0)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-(x_{1}+x_{2}+...+x_{n-1})}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}$

${\displaystyle =\int _{0}^{\infty }x_{1}^{{\frac {1}{n}}-1}\,e^{-x_{1}}\,dx_{1}\cdot \int _{0}^{\infty }x_{2}^{{\frac {2}{n}}-1}\,e^{-x_{2}}\,dx_{2}\cdots \int _{0}^{\infty }x_{n-1}^{{\frac {n-1}{n}}-1}\,e^{-x_{n-1}}\,dx_{n-1}}$

${\displaystyle =\Gamma \left({\frac {1}{n}}\right)\cdot \Gamma \left({\frac {2}{n}}\right)\cdots \Gamma \left({\frac {n-1}{n}}\right)={\frac {1}{\sqrt {n}}}\,{\sqrt {2\pi }}^{\,n-1}}$.

Substituiere ${\displaystyle u=xy}$:

${\displaystyle I=\int _{0}^{1}\int _{0}^{y}{\frac {1}{(1-u)\,{\sqrt {{\frac {u}{y}}\,(1-y)}}}}\,{\frac {du}{y}}\,dy=\int _{0}^{1}\int _{0}^{y}{\frac {du}{(1-u)\,{\sqrt {u}}}}\,{\frac {dy}{{\sqrt {y}}\,{\sqrt {1-y}}}}}$

Vertausche die Integrationsreihenfolge:

${\displaystyle I=\int _{0}^{1}\int _{u}^{1}{\frac {dy}{{\sqrt {y}}{\sqrt {1-y}}}}\,{\frac {du}{(1-u)\,{\sqrt {u}}}}=\int _{0}^{1}2\arccos({\sqrt {u}}\,)\,{\frac {du}{(1-u)\,{\sqrt {u}}}}}$

Substituiere ${\displaystyle u=w^{2}}$:

${\displaystyle I=\int _{0}^{1}2\arccos w\,{\frac {2w\,dw}{(1-w^{2})\,w}}=4\int _{0}^{1}{\frac {\arccos w}{1-w^{2}}}\,dw}$

Substituiere ${\displaystyle w=\cos t}$:

${\displaystyle I=4\int _{\frac {\pi }{2}}^{0}{\frac {t}{1-\cos ^{2}t}}\,(-\sin t)\,dt=4\int _{0}^{\frac {\pi }{2}}{\frac {t}{\sin t}}\,dt}$

Substituiere ${\displaystyle t=2\arctan s}$:

${\displaystyle I=8\int _{0}^{1}{\frac {\arctan s}{\frac {2s}{1+s^{2}}}}\,{\frac {2ds}{1+s^{2}}}=8\int _{0}^{1}{\frac {\arctan s}{s}}\,ds=8G}$