Integral involving beta function (continuous version of binomial distribution) I'm hoping someone can help me out with a problem that's consumed me for a while.
I'm working with a continuous version of the binomial distribution defined for real $n > 0$ and continuous $0 < \mu < 1$:
$$
P(x) = n \mu^{nx} (1 - \mu)^{n - nx} \frac{\Gamma(n + 1)}{\Gamma(nx + 1) \Gamma(n - nx + 1)}
$$
(The $n$ adjusts for rescaling: the binomial distribution takes in $k$, but we take in $nx$ with $x \in [0, 1]$.)
My goal is to find a way of computing $\int_0^1 P(x)\,dx$ or
$$
\int_0^1 n \mu^{nx} (1 - \mu)^{n - nx} \frac{\Gamma(n + 1)}{\Gamma(nx + 1) \Gamma(n - nx + 1)} \, dx
$$
What I've tried:

*

*It's possible to convert this to the beta function and rewrite in terms of a trigonometric integral found here to get

$$\int_{x=0}^1 \int_{\theta=0}^{\pi/2} \mu^{nx} (1 - \mu)^{n - nx} \frac{2^{n+1}}{\pi} \cos^n \theta \cos ((2nx - n)\theta)\,d\theta\,dx$$
Then, Mathematica eliminates $x$ and produces
$$
\int_{0}^{\pi/2} \frac{2^n \cos^n \theta \left[\theta (\mu^n + (1 - \mu)^n) \sin (n\theta) + ((1 - \mu)^n - \mu^n) \tanh^{-1}(1 - 2\mu) \cos n \theta \right]}{\pi \left(\theta^2 + \left[ \tanh^{-1} (1 - 2u) \right]^2\right)} \, d\theta
$$
which is a pretty interesting integral that's vaguely Bessel-esque (if you graph this for $(n, \mu) = (100, 0.5)$ it's surprising that the answer is almost exactly 1), but not something I've managed to make any progress on.

*

*It seems natural to try and transform the integral so $x, \mu$ are over a different interval than $[0, 1]$, but nothing I've tried there has worked.


*This integral isn't exactly like the incomplete beta function (this is almost exactly the PDF of $\text{Beta}(\alpha = nx + 1, \beta = n(1 - x) + 1)$ applied to $\mu$, instead of switching $\mu$ and $x$ which is what the incomplete beta function can help with.)
Any ideas? I'm beat.
 A: Trying some ideas
Simplifying the expression; I think that we can rewrite
$$P(x)=n \,(1-\mu )^n \ \Gamma (n+1)\,\,\color{red}{\frac {e^{k\, x}}{ \Gamma (n x+1)\,\Gamma (n- nx+1) }}\qquad \qquad k=n \log \left(\frac{\mu }{1-\mu }\right)$$ and the problem could be to approximate
$$f(x)=\frac {1}{ \Gamma (n x+1)\,\Gamma (n- nx+1) }$$
$f(x)$ being symmetric around $x=\frac 12$, what we we could use is
$$f(0)=\frac{1}{\Gamma (n+1)} \qquad f'(0)=\frac{H_n}{\Gamma (n)}$$
$$f\left(\frac{1}{2}\right)=\frac{1}{\Gamma \left(\frac{n}{2}+1\right)^2}\qquad f'\left(\frac{1}{2}\right)=0\qquad f''\left(\frac{1}{2}\right)   =-\frac{8 \psi ^{(1)}\left(\frac{n}{2}+1\right)}{\Gamma \left(\frac{n}{2}\right)^2}$$ and, based of that, build polynomials of even degrees in $(2x-1)$
The simplest one (it only uses $f(0)$ , $f\left(\frac{1}{2}\right)$ and $f'\left(\frac{1}{2}\right)$) is
$$g_1(x)=\frac{1}{\Gamma
   \left(\frac{n}{2}+1\right)^2}+\left(\frac{1}{\Gamma (n+1)}-\frac{1}{\Gamma
   \left(\frac{n}{2}+1\right)^2}\right)(2 x-1)^2 $$
