Let $v,w\in \Bbb R^3$ two orthogonal vectors of length $1$. Calculate:
$$\int_{\Bbb R^3}\langle x,v \rangle^{20} \;\langle x,w \rangle^{16}\;e^{-|x|}\;dx_1dx_2dx_3$$
My attempt: Using the coarea formula:
$$\int_{\Bbb R^3}\langle x,v \rangle^{20} \;\langle x,w \rangle^{16}\;e^{-|x|}\;dx_1dx_2dx_3=\lim_{r \to \infty }\int_{B_r}\langle x,v \rangle^{20} \;\langle x,w \rangle^{16}\;e^{-|x|}\;dx_1dx_2dx_3 $$
$$=\lim_{r \to \infty}\int_0^r\int_{\rho S^2}\langle x,v \rangle^{20} \;\langle x,w \rangle^{16}\;e^{-|x|}dS(x)d\rho$$$$=\lim_{r \to \infty}\int_0^r\rho ^2\int_{ S^2}\langle \rho y,v \rangle^{20} \;\langle \rho y,w \rangle^{16}\;e^{-|\rho y|}dS(y)d\rho=\biggr(\lim_{r \to \infty}\int_0^r\rho ^{38}e^{-\rho}d\rho\biggr)\biggr(\int_{ S^2}\langle y,v \rangle^{20} \;\langle y,w \rangle^{16}dS(y)\biggr)=\Gamma(39)\int_{ S^2}\langle y,v \rangle^{20} \;\langle y,w \rangle^{16}dS$$
Now I need to calculate $\int_{ S^2}\langle y,v \rangle^{20} \;\langle y,w \rangle^{16}dS$, which is the part I'm having trouble with. Since $v,w$ are orthogonal vectors of length $1$, I can assume without loss of generality (since the surface integral over a sphere doesn't change under orthogonal transformation) that $v=(0,0,1), w=(0,1,0)$. Then I'm left with $\int_{ S^2}y_3^{20}y_2^{16}dS$, which I'm pretty stuck with. Using the parametrization $(\varphi, \theta) \mapsto (\cos\theta \sin \varphi,\sin \theta \sin \varphi, \cos \varphi)$, $0\leq \varphi \leq \pi$, $0 \leq \theta \leq 2\pi$, we get:
$$\int_0^{2\pi}\int_0^{\pi}\cos ^{20}\varphi \sin^{17}\varphi \sin^{16} \theta d\varphi d\theta$$
Any ideas how to continue from here?
Best Answer
This is a pedestrian answer, i am just continuing the calculus of the two (extracted) integrals from the last lines of the OP.
(1)
First of all, recall the relation for the Beta-Function $B$, that is relevant for us: $$ \begin{aligned} &\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} =B(x,y) =2\int_0^{\pi/2} \sin^{2x-1} t\; \cos^{2y-1} t\; d t\ . \\[3mm] &\text{This gives immediately} \\ &\int_0^\pi \cos^{20}s\; \sin^{17} s\; ds = 2 \int_0^{\pi/2} \cos^{20}s\; \sin^{17} s\; ds \\ &\qquad= B\left(\frac{20+1}2,\; \frac{17+1}2\right) = \frac { \Gamma\left(\frac{21}2\right) \Gamma(9) } { \Gamma\left(\frac{39}2\right) } \\ & \qquad = \frac { \Gamma\left(\frac{21}2\right) \cdot 8!} { \frac{37}2\cdot \frac{35}2\cdot \frac{33}2\cdot \frac{31}2\cdot \frac{29}2\cdot \frac{27}2\cdot \frac{25}2\cdot \frac{23}2\cdot \frac{21}2\cdot \Gamma\left(\frac{21}2\right) } \\ &\qquad= \frac{8!\cdot 2^9} { 37\cdot 35\cdot 33\cdot 31\cdot 29\cdot 27\cdot 25\cdot 23\cdot 21}\ . \\[3mm] &\text{In the same spirit} \\ &\int_0^{2\pi} \sin^{16} t\; dt = 2B\left(\frac{16+1}2,\; \frac{0+1}2\right) = 2\cdot \frac { \Gamma\left(\frac{17}2\right) \Gamma\left(\frac{1}2\right) } { \Gamma(9) } \\ & \qquad = 2\cdot \frac { \frac{15}2\cdot \frac{13}2\cdot \frac{11}2\cdot \frac{9}2\cdot \frac{7}2\cdot \frac{5}2\cdot \frac{3}2\cdot \frac{1}2\cdot \Gamma\left(\frac12\right)^2 } { 8!}\ . \end{aligned} $$
(2)
Following the OP and the above computational path we have for the initial integral $$ J = \iiint_{\Bbb R^3}x_1^{20}\;x_2^{16}\; e^{-\sqrt{x_1^2+x_2^2+x_3^2}}\; dx_1\; dx_2\; dx_3 $$ by using a passage to spherical coordinates, $(x_1,x_2,x_3)$ to $(r,s,t)$, with $x_1=r\sin s\cos t$, $x_2=r\sin s\sin t$, $x_3=r\cos s$, $dx_1\; dx_2\; dx_3=r^{\color{red}2}\; \sin s\; dr\, ds\, dt$ the formula $$ \begin{aligned} J &= 2\cdot \Gamma(\color{red}{39}) \cdot \frac{\Gamma(21/2)\;\Gamma(9)}{\Gamma(37/2)} \cdot \frac{\Gamma(17/2)\;\Gamma(1/2)}{\Gamma(9)} \\ &= 2\cdot\frac{\Gamma(39)}{\Gamma(39/2)} \cdot \Gamma\left(\frac{20+1}2\right) \Gamma\left(\frac{16+1}2\right) \Gamma\left(\frac{ 0+1}2\right) \\ &=2\cdot 2^{19}\; 38!!\cdot 19!!\cdot 15!!\cdot \pi \\ &=2\cdot 2^{19}\cdot 2^{19}\cdot 19!!\cdot 19!!\cdot 15!!\cdot \pi \ . \end{aligned} $$ But we could also go the following way, first use a change of variables $x=r\cos t$, $y=r\sin t$, $z=z$, thus $$ \begin{aligned} J &= \int_{r\in[0,\infty)} \int_{t\in[0,2\pi]} r^{20+16}\; \cos^{20} t\; \sin^{16}t\; r\; dr\; dt \int_{z\in\Bbb R} e^{-\sqrt{r^2+t^2}} \; dz \\ &= 2B(21/2, 17/2) \int_{r\in[0,\infty)} \int_{z\in\Bbb R} r^{37}\; e^{-\sqrt{r^2+t^2}} \; dz \; dr \qquad\text{ now }r=R\cos s\dots \\ &= 2B(21/2, 17/2) \int_{R\in[0,\infty)} \int_{s\in[0,\pi]} R^{37}\; \cos s^{37} \; e^{-R} \; R\; dR\; ds \\ &= 2B(21/2, 17/2)\cdot B(19,1/2)\cdot \Gamma(39)\ . \end{aligned} $$ We introduce instead of the beta functions the gamma functions, obtain the same result.
Or we could have taken instead of $x_1^{20}x_2^{16}$ any other of the "same shape", $x_j^{20}x_k^{16}$, with $1\le j,k\le 3$, $j\ne k$, then perform the same substitutions, transform to obtain beta functions and gamma functions, then finally only gamma functions. Same result.