Proving and deriving a Gamma function I'm having a hard time trying to prove this Gamma function and trying to derive the duplication formula:
a.) Prove that
$$\frac{\Gamma (p)\Gamma (p)}{\Gamma (2p)} = 2\int_0^{1/2}x^{p-1}(1-x)^{p-1}\mathrm dx$$
b.) Make a suitable change of variable (a) and derive the duplication formula for the Gamma function:
$$\Gamma (2p)\Gamma\left(\frac12\right) = 2^{2p-1}\Gamma (p)\Gamma\left(p-\frac12\right)$$
 A: I summarize the proof of the relation between the gamma and the beta
functions as described in section 20.7, p. 680-681 of Angus Taylor's Advanced Calculus, Blaisdell Publishing Company, 1955, and the $B(p,p)$ functional equation in exercise 8, p. 683, from which one can derive the gamma function duplication formula.
An alternative derivation of
$$\Gamma (p)\Gamma (q)=\Gamma (p+q)B(p,q)\qquad (1)$$ 
uses 3 substitutions. The $1^{st}$ one, $t=u/(1+u)$, converts the beta  $B(p,q)$ integral from
$$B(p,q)=\int_{0}^{1}t^{p-1}(1-t)^{q-1}\ \mathrm{d}t\qquad (2)$$
into
$$B(p,q)=\int_{0}^{\infty }\frac{u^{p-1}}{(1+u)^{p+q}}.\ \mathrm{d}u\qquad (3)$$
The $2^{nd}$ one, $t=vu$, transforms the gamma function integral
$$\Gamma (p)=\int_{0}^{\infty }t^{p-1}e^{-t}\ \mathrm{d}t\qquad (4)$$
into
$$\Gamma (p)=\int_{0}^{\infty }v^{p}u^{p-1}e^{-vu}\ \mathrm{d}u.$$
Hence
$$\Gamma (p)\Gamma (q)=\int_{0}^{\infty }u^{p-1}\left( \int_{0}^{\infty
}v^{p+q-1}e^{-v-vu}\ \mathrm{d}v\right) \ \mathrm{d}u.$$
By $(4)$ and making the $3^{rd}$ change of variables, $v=w/(1+u)$, we obtain
$$\int_{0}^{\infty }v^{p+q-1}e^{-v-vu}\ \mathrm{d}v=\frac{\Gamma (p+q)}{%
(1+u)^{p+q}},\qquad (5)$$
and by $(3)$,
$$\Gamma (p)\Gamma (q)=\Gamma (p+q)\int_{0}^{\infty }\frac{u^{p-1}}{%
(1+u)^{p+q}}\ \mathrm{d}u=\Gamma (p+q)B(p,q).\qquad\square$$
The substitution $u=4t(t-1)$ in $$B(p,p)=2\int_{0}^{1/2}\left[ t(t-1)\right]
^{p-1}\ \mathrm{d}t$$ yields the functional equation $$B(p,p)=2^{1-2p}B(p,1/2).$$
A: As the comments (to Chandru1's answer) suggest, your two identities can be derived from the integral formula for the Beta function and its relation to the Gamma function.  I'm not quite sure what Theo meant by "something non-trivial", but here's one way to prove that $ B(\alpha,\beta)=\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}.$
Starting with $ \Gamma(\alpha) = \int_0^{\infty} x^{\alpha} e^{-x} \frac{\mathrm dx}{x} $ and making the substitution $x=y^2$ gives 
$$ \Gamma(\alpha) = 2 \int_0^{\infty} y^{2\alpha-1} e^{-y^2}\mathrm dy. $$
Using this form of $\Gamma$, we get 
$$\Gamma(\alpha) \Gamma(\beta) = 4 \int_0^{\infty} \int_0^{\infty} x^{2\alpha-1} y^{2\beta-1} e^{-(x^2+y^2)}\mathrm dx\mathrm dy. $$
Switching to polar coordinates ($ x=r\cos(\theta), y=r\sin(\theta) $) gives
$$ \Gamma(\alpha) \Gamma(\beta) = 4 \int_0^{\frac{\pi}{2}} \int_0^{\infty} r^{2(\alpha+\beta)-1} \cos(\theta)^{2\alpha-1} \sin(\theta)^{2\beta-1} e^{-r^2}\mathrm dr\mathrm d\theta $$
$$ = \left(2 \int_0^{\frac{\pi}{2}} \cos(\theta)^{2\alpha-1} \sin(\theta)^{2\beta-1} \mathrm d\theta \right) \left( 2\int_0^{\infty} r^{2(\alpha+\beta)-1} e^{-r^2}\mathrm dr \right) $$
$$ = \left(2 \int_0^{\frac{\pi}{2}} \cos(\theta)^{2\alpha-1} \sin(\theta)^{2\beta-1} \mathrm d\theta \right) \Gamma(\alpha+\beta). $$
Therefore,
$$ \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)} = 2 \int_0^{\frac{\pi}{2}} \cos(\theta)^{2\alpha-1} \sin(\theta)^{2\beta-1}\mathrm d\theta $$
and the substitution $z=\sin(\theta)^2 $ yields
$$ \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)} = \int_0^1 (1-z)^{\alpha-1} z^{\beta-1}\mathrm dz.$$
Appropriate selections of $\alpha, \beta$ and observing some symmetry will get the desired identities.
