Finding $\int^{1}_{0}{p^x(1-p)^{n-x}\exp\left\{-\frac{1}{2} \left(\ln \left(a\frac{p}{1-p}\right)\right)^2\right\}}dp$ I got the following integration and I could not figure it out.I wonder if it has a closed form,
$$\int^{1}_{0}{p^x(1-p)^{n-x}\exp\left\{-\frac{1}{2} \left(\ln \left(a\frac{p}{1-p}\right)\right)^2\right\}}dp$$
where $n>x$ and $n,x,$ and $a$ are some positive constants.
Thanks in advance!
 A: Partial answer:
$$\begin{align}\exp\left\{-\frac12\left(\ln\left(a\frac p{1-p}\right)\right)^2\right\} & =\exp\left\{-\frac12\left((\ln a)^2+2\ln a\ln\left(\frac p{1-p}\right)+\left(\ln\left(\frac p{1-p}\right)\right)^2\right)\right\}\\
 & =\exp\left\{-\frac12(\ln a)^2\right\}\left(\frac p{1-p}\right)^{-\ln a}\exp\left\{-\frac12\left(\ln\left(\frac p{1-p}\right)\right)^2\right\}\end{align}$$
So $$\int_0^1p^x(1-p)^{n-x}\exp\left\{-\frac12\left(\ln\left(a\frac p{1-p}\right)\right)^2\right\}dp=\\
\exp\left\{-\frac12(\ln a)^2\right\}\int_0^1p^{x-\ln a}(1-p)^{n-x+\ln a}\exp\left\{-\frac12\left(\ln\left(\frac p{1-p}\right)\right)^2\right\}dp$$
Then if we let $u=\frac p{1-p}$, we find
$$\int_0^1p^{x-\ln a}(1-p)^{n-x+\ln a}\exp\left\{-\frac12\left(\ln\left(\frac p{1-p}\right)\right)^2\right\}dp= \\
\int_0^{\infty}\frac{u^{x-\ln a}}{(1+u)^{n+2}}\exp\left\{-\frac12\left(\ln u\right)^2\right\}du$$
Then if we let $u=e^y$, we get
$$\int_0^1p^x(1-p)^{n-x}\exp\left\{-\frac12\left(\ln\left(a\frac p{1-p}\right)\right)^2\right\}dp=\\
\exp\left\{-\frac12(\ln a)^2\right\}\int_{-\infty}^{\infty}\frac{e^{(x-\ln a+1)y}}{(1+e^y)^{n+2}}e^{-\frac12y^2}dy$$
Should probably be able to do a contour integral at this point, but I am out of time and have to leave it to another nurse to sew up the patient.
