- Show that Wallis’s product formula can be written as
\[ \sqrt {\frac {\pi}{2}} = \lim _ {n \rightarrow \infty} \frac {2 ^ {2 n} (n !) ^ {2}}{(2 n + 1) !} (2 n + 1) ^ {1 / 2}. \]
As a result, prove the following identity:
\[ \Gamma (s) \Gamma (s + 1 / 2) = \sqrt {\pi} 2 ^ {1 - 2 s} \Gamma (2 s). \]
Solution
1. Wallis’s product is \(\frac{\pi}{2} = \prod_{n=1}^{\infty} \frac{4n^2}{4n^2 - 1} = \lim_{n \to \infty} \frac{2^{4n}(n!)^4}{((2n)!)^2 (2n+1)}\).
standard form of Wallis’s product.
2. Taking square roots and rearranging gives \(\sqrt{\frac{\pi}{2}} = \lim_{n \to \infty} \frac{2^{2n}(n!)^2}{(2n+1)!}(2n+1)^{1/2}\).
1, using \((2n+1)! = (2n)!(2n+1)\).
3. For the duplication formula, use the integral representation \(\Gamma(s) = \int_0^\infty t^{s-1} e^{-t}\,dt\).
definition of the Gamma function.
4. \(\Gamma(s)\Gamma(s + 1/2) = \int_0^\infty \int_0^\infty x^{s-1} y^{s-1/2} e^{-(x+y)}\,dx\,dy\).
3.
5. Substitute \(x = u^2\), \(y = v^2\): \(\Gamma(s)\Gamma(s+1/2) = 4\int_0^\infty \int_0^\infty u^{2s-1} v^{2s} e^{-(u^2 + v^2)}\,du\,dv\).
4, change of variables.
6. Switch to polar coordinates \(u = r\cos\theta\), \(v = r\sin\theta\): the integral becomes \(4\int_0^{\pi/2}\int_0^\infty r^{4s-1}(\cos\theta)^{2s-1}(\sin\theta)^{2s} e^{-r^2}\,dr\,d\theta\).
5.
7. The \(r\)-integral is \(\int_0^\infty r^{4s-1} e^{-r^2}\,dr = \frac{1}{2}\Gamma(2s)\) (substituting \(r^2 = t\)).
6.
8. The \(\theta\)-integral is \(\int_0^{\pi/2} (\cos\theta)^{2s-1}(\sin\theta)^{2s}\,d\theta = \frac{1}{2}B(s, s + 1/2) = \frac{\Gamma(s)\Gamma(s+1/2)}{2\Gamma(2s + 1/2)}\).
beta function.
9. Combining: \(\Gamma(s)\Gamma(s+1/2) = 4 \cdot \frac{1}{2}\Gamma(2s) \cdot \frac{\Gamma(s)\Gamma(s+1/2)}{2\Gamma(2s+1/2)}\), which simplifies (using \(\Gamma(2s+1/2) = \frac{\Gamma(2s)\Gamma(1/2)}{2^{2s-1}\Gamma(s)}\)) to the duplication formula.
7 and 8, and the standard identity \(\Gamma(2s) = \frac{2^{2s-1}}{\sqrt{\pi}}\Gamma(s)\Gamma(s+1/2)\).
10. Hence \(\Gamma(s)\Gamma(s+1/2) = \sqrt{\pi}\,2^{1-2s}\,\Gamma(2s)\).
9.
11. Q.E.D.
2 and 10.