We use the beta function $\text B(\alpha,\beta)=\int_0^1 t^{\alpha-1}(1-t)^{\beta-1}\text dt$ to prove the duplication formula for the gamma function:

$$\Gamma(2\alpha) = \frac {2^{2\alpha -1}\Gamma\left(\alpha+\frac 1 2\right)\Gamma(\alpha)}{\sqrt \pi}$$

Is there a similar formula for the duplication of the beta function? I.e., can I find $\text B(2\alpha,\beta)$ and/or $\text B(\alpha,2\beta)$ from $\text B(\alpha,\beta)$ and perhaps other functions? Can something like that be found for the incomplete beta function $\text B_x(\alpha,\beta)=\int_0^x t^{\alpha-1}(1-t)^{\beta-1}\text dt$?

And more specifically, can I find $\text B_x(2\alpha,\alpha)$ from $\text B_x(\alpha,\alpha)$?


Interesting question. We may start from the definition of the Beta function:

$$B(m, n) = \int_0^1 t^{m-1}(1-t)^{n-1}\ \text{d}t$$

and rewrite it when $m\to 2m$:

$$B(2m, n) = \int_0^1 t^{2m-1}(1-t)^{n-1}\ \text{d}t$$

$$B(2m, n) = \int_0^1 t^{2m-1} \frac{(1-t)^n}{1-t}\ \text{d}t$$

Now, since the range of integration is $[0, 1]$, we are allowed to make use of the geometric series

$$\frac{1}{1-t} = \sum_{k = 0}^{+\infty} t^k$$


$$B(2m, n) = \int_0^1 t^{2m-1} (1-t)^n \sum_{k = 0}^{+\infty} t^k\ \text{d}t = \sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1} (1-t)^n t^k\ \text{d}t$$

And easily write:

$$\sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1+k} (1-t)^n\ \text{d}t$$

Calling now

$$2m-1+k = a ~~~~~~~~~~~ n = b$$

We notice that the integral is well known:

$$ \int_0^1 t^a (1-t)^b\ \text{d}t \equiv B(a+1, b+1)$$

Then we end up with the partial result (re-expanding $a$ and $b$):

$$B(2m, n) = \sum_{k = 0}^{+\infty} B(2m+k, n+1)$$

That series does exist and it does converge to a known result:

$$\sum_{k = 0}^{+\infty} B(2m+k, n+1) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$

What you end up with is a sort of recursive relation for the Beta function:

$$B(2m, n) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$


The above expression can be simplified!

$$\frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)} \equiv \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$

What we obtained is actually nothing than what we would have obtained by simply substituting at the beginning $m\to 2m$ in the Gamma function / Beta function definition.

$$B(2m, n) = \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$

This really suggest that such a particular duplication formula for the Beta function may not exist at all, since all you need is the Gamma function and ITS duplication formula, through which you can evaluate $\Gamma(2m)$.

Seems like that this is the only "duplication formula" for the beta function.

(Also, I found nothing on reviews or literature).


Note that $$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},$$ so $$B(2a,b) = \frac{\Gamma(2a)\Gamma(b)}{\Gamma(2a+b)} = 2^{2a-1} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} \cdot \frac{\Gamma(a+1/2)\Gamma(a+b)}{\Gamma(1/2)\Gamma(2a+b)},$$ or $$B(2a,b) = 2^{2a-1} \frac{B(a,b) B(a,a+b)}{B(a,1/2)}.$$ This is about as good as I could get.


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