I'd like to derive an asymptotic expression for the following function (as $m \rightarrow \infty$) involving the incomplete Beta function:
$$ F(m) = \frac{1}{2} B_{1/2}(m, m) - B_{1/2}(m+1, m) $$
where $m$ is a non-negative integer.
I've attempted to use Laplace's method, but the contributions I get from each of the terms cancel out.
Could anyone suggest an alternative approach? Is there perhaps a way of getting higher order asymptotic terms from Laplace's method?
The definition I am working from for the incomplete Beta function is:
$$ B_z(a,b) := \int_0^z \alpha^{a-1} (1-\alpha)^{b-1} d\alpha.$$
We have
$$F(m)=\frac{1}{2} B_{1/2}(m, m) - B_{1/2}(m+1, m)\qquad\qquad\qquad\qquad\qquad\qquad\ \ $$
$$=\frac{1}{2}\left[\int_0^{1/2} x^{m-1} (1-x)^{m-1} dx-2\int_0^{1/2} x^{m} (1-x)^{m-1} dx\right]$$
$$=\frac{1}{2}\int_0^{1/2}x^{m-1}(1-x)^{m-1}(1-2x)dx\qquad\qquad\qquad\qquad\quad$$
$$=\frac{1}{2}\int_0^{1/2}[x^{m-1}(1-x)^{m}-x^{m}(1-x)^{m-1}]dx\qquad\qquad\qquad$$
$$=\frac{1}{2}\left[\frac{z^m(1-z)^m}{m}\right]_0^{1/2}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ \ $$
$$=\frac{1}{m{2}^{1+2m}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ $$
which has limit zero as $m\to\infty$.
The working shows that for any $z\in[0,1]$:
$$\frac{1}{2}B_z(m,m)-B_z(m+1,m)=\frac{z^m(1-z)^m}{2m},$$
and for any $z\in[0,1]$ the limit as $m\to\infty$ is zero.