Asymptotics of incomplete Beta function $B_{1/2}(y+1,y)$ when $y\to\infty$ My question concerns the behavior of the incomplete Beta function
$$B_{1/2}(y+1,y)=\int_0^{1/2}x^y (1-x)^{y-1}dx$$
in the large $y$ limit. I have been looking everywhere, but I can't find anything. I was thinking that I can get a good approximation by letting $(1-x)^{y-1}\approx x^y$, thus yielding a solution of the form $\approx 1/(2^{2y+1})$. However, is there any reference that talks about the above function in the large argument limit?
 A: This is straightforward steepest descent (or Laplace's method). Let $f(x):=\ln(x)+\ln(1-x),$  and $g(x)=(1-x)^{-1}$ so that the integral becomes:
$$\int_0^{1/2}g(x)e^{yf(x)}dx.$$
On $[0,1/2]$, $f$ achieves its maximum at $x_0=1/2$. Furthermore, $f''(x)=-1/x^2-1/(1-x)^2$, and $f''(x_0)<0$.
Then Laplace's method gives:
$$\int_0^{1/2}g(x)e^{yf(x)}dx\sim \sqrt{\frac{2\pi}{y\,|f''(x_0)|}}\,g(x_0)\,e^{yf(x_0)}$$
in the sense that the ratio of the LHS and RHS converges to $1$ when $y\to+\infty$.
Can you wrap it up from here? Note the extra $1/\sqrt{y}$ that's going to contribute.
A: I'll play around
and see if anything interesting happens.
$\begin{array}\\
B_{1/2}(y+1,y)
&=\int_0^{1/2}x^y (1-x)^{y-1}dx\\
&\ge\int_0^{1/2}x^y (\frac12)^{y-1}dx\\
&=\dfrac1{2^{y-1}}\int_0^{1/2}x^ydx\\
&=\dfrac1{2^{y-1}}\dfrac{x^{y+1}}{y+1}\big|_0^{1/2}\\
&=\dfrac1{2^{2y}(y+1)}\\
\end{array}
$
and
$\begin{array}\\
B_{1/2}(y+1,y)
&=\int_0^{1/2}x^y (1-x)^{y-1}dx\\
&\le\int_0^{1/2}(\frac12)^y (1-x)^{y-1}dx\\
&=\dfrac1{2^y}\int_0^{1/2} (1-x)^{y-1}dx\\
&=\dfrac1{2^y}\int_{1/2}^1 x^{y-1}dx\\
&=\dfrac1{2^y}\dfrac{x^y}{y}\big|_{1/2}^1\\
&=\dfrac1{2^y}\dfrac{1-(1/2)^y}{y}\\
\end{array}
$
I'll leave it at these bounds.
Don't know which is best.
This has an obvious generalization
to
$B_{z}(y+1,y)
=\int_0^{z}x^y (1-x)^{y-1}dx
$.
A: This is a very empirical approach of the problem since based on numerical simulations.
Computing the values of $B_{1/2}(y+1,y)$ for $0\leq y\leq 5000$, what we can notice is that the logarithm varies as a linear function of $y$. For this range, I obtained $$\log\left(B_{1/2}(y+1,y)\right)\approx -1.38659 y-3.1602$$ which seems to be a very good fit (as shown below)
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & -3.16020 & 6.70 \times 10^{-3} & \{-3.17333,-3.14706\} \\
 b & -1.38659 & 2.32 \times 10^{-6} &
   \{-1.38659,-1.38658\} \\
\end{array}$$ Extrapolated to $n=10000$, this simple model leads, for the logarithm, to a value equal to $-13869.0$ while the exact value should be $\approx -13867.7$ which does not seems to be too bad.
You could notice that $$\log\left(\frac 1 {2^{2y+1}}\right)\approx -1.38629 y-0.693147$$ showing almost the same slope as in the regression (and this is almost $-2\log(2)$).
Finally, considering $$F(k)=\int_0^{\infty } \left(B_{\frac{1}{2}}(y+1,y)-\frac k {2^{2y+1}} \right){}^2 \, dy$$ the minimum of the function is obtained for $k\approx 1.21973$.
