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It is well-known that $10! = 6! 7! $, I want to prove it via algebraic manipulations of Gamma function, i.e show that: $$ \int_{0}^{+\infty} x^7 e^{-x} \, dx \int_{0}^{+\infty} y^6 e^{-y} \, dy = \int_{0}^{+\infty} z^{10} e^{-z} \, dz \qquad (1) $$ I tried to write the LHS as:

$$ \int_{0}^{+\infty} \int_{0}^{+\infty} x (xy)^6 e^{-(x+y)} \, dx dy \qquad (2)$$

and go for the substitution $u = x + y $, $v=xy$, but that leads to complicated calculations. Is there a reasonable way/substitution to turn the double integral in $(2)$ into the RHS in $(1)$. Thanks in advance for any contribution.

GaC
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  • I think the easiest way is to integrate by part several times and use the identity $10!=6!7!$... – Antoine Apr 11 '17 at 11:42
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    Use the identity you want to prove, well that sounds promising. xD – Dirk Apr 11 '17 at 11:42
  • I would try use this identity that seems easier to manipulate $$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{k=1}^\infty\left(1+\frac{z}k\right)e^{-z/k}$$ for $z\in\Bbb C{\setminus}({-}\Bbb N)$. – Masacroso Apr 11 '17 at 12:21
  • Nice suggestion, I'll give it a thought and maybe edit the post, thank you :) – GaC Apr 11 '17 at 12:30
  • Isn't it possible to use B (beta) integrals ? – Jean Marie Apr 11 '17 at 12:49
  • I tried to give a look at Beta function's most famous identities, but was not able to do much... – GaC Apr 11 '17 at 12:53
  • have you tried polar coordinates? – tired Apr 12 '17 at 07:19
  • Normally, conversion of two $\Gamma$-function integrals into a single integral is the way to derive the Beta-function identity: $\Gamma(m)\Gamma(n) = \Gamma(m+n) B(m,n)$. You can set your $x=u^2$, $y=v^2$, then change from $(u,v)$ to polar coordinates $(r,\theta)$. The $r$-integral will become $\Gamma(m+n)$ and the $\theta$ integral the Beta function. – jcandy Oct 20 '17 at 08:09

1 Answers1

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The goal as I understand it is to use some nifty change of variables to turn this double integral $$ LHS=\int_{0}^{+\infty} \int_{0}^{+\infty} x (xy)^6 e^{-(x+y)} \, dx dy=6!\times 7!, $$ into $$ RHS=\int_0^{+\infty} z^{10} e^{-z} \, dz=10!. $$ This appears to me to be an exercise in manipulating the product of Gamma functions or as it is otherwise known a Beta function. A common substitution made when dealing with such a double integral turns the exponential weighting into a Gaussian $$ x=s^2 , \ y=t^2 , \ \frac{\partial (x,y)}{\partial (s,t)}=4st, \\LHS=\int_{0}^{+\infty} \int_{0}^{+\infty} x (xy)^6 e^{-(x+y)} \, dx dy=4\int_{0}^{+\infty} \int_{0}^{+\infty} s^2 (st)^{13} e^{-(s^2+t^2)} \, ds dt. $$ This particular substitution is valuable because it makes it simpler to use the polar parametrization of the integration region which is desirable because it reduces the Gaussian weight function to single variable dependence $$ s=r\cos{(\theta)} ,\ t=r\sin{(\theta)} , \ \frac{\partial (s,t)}{\partial (r,\theta)}=r, \\LHS=4\int_{0}^{+\infty} e^{-r^2} \, r^{29}dr\int_{0}^{\pi/2} \cos^{15}{(\theta)} \sin^{13}{(\theta)} d\theta. $$ The $\theta$-integral is easily evaluated using the trigonometric substitution we all learn in high school $$ \int_{0}^{\pi/2} \cos^{15}{(\theta)} \sin^{13}{(\theta)} \ d\theta \\=\int_{0}^{\pi/2} \cos^{14}{(\theta)} (1-\cos^{2}{(\theta)})^6\cos{(\theta)}\sin{(\theta)} \ du, \\u=\cos^{2}{(\theta)} , \ \frac{du}{d\theta}=-2\cos{(\theta)}\sin{(\theta)}, \\=\frac{1}{2}\int_{0}^{1} u^7 (1-u)^6 \ du=1/48048. $$ Finally, the radial integral may be shown to be proportional to the RHS by using the first substitution I introduced and integrating by parts $$ r=\sqrt{z} , \ \frac{dr}{dz}=\frac{1}{2}z^{-1/2} , \\\int_{0}^{+\infty} e^{-r^2} \, r^{29}dr=\frac{1}{2}\int_{0}^{+\infty} e^{-z} \, z^{14} dz=\frac{14\times13\times12\times11}{2}\int_{0}^{+\infty} e^{-z} \, z^{10} dz, $$ hence $$ LHS=\frac{4\times14\times13\times12\times11}{2\times48048}\int_{0}^{+\infty} e^{-z} \, z^{10} dz=\int_{0}^{+\infty} e^{-z} \, z^{10} dz=RHS. $$ As required, however it is worth noting that the quickest way to prove this identity is to simply use the definition of factorial and cancel down $$ LHS=6!7!=\frac{6\times5\times4\times3\times2}{8\times9\times10}10!=10!=RHS. $$ Unsurprisingly this is mirrored by using integration by parts on the integral representation of the Gamma function.

Joseph Johnson
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