How to solve this integral in terms of $c$? Let $c>0$, is there a "closed form" solution of the integral
$$
\int_0^1 \exp\left(-\frac{1}{cx(1-x)}\right) dx
$$
or a special function (of $c$) giving the solution of the integral above? So far, I've only managed to obtain some upper bounds, for example by
$$
\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1}{cx}\right) dx = c (1-e^{-1/c}),
$$
or lower bounds, for example by
$$
\int_0^{1/2}\exp\left( -\frac{1}{cx^2}\right)dx +\int_{1/2}^1 \exp\left( -\frac{1}{c(1-x)^2}\right)dx=2a\left(
\frac{\text{erf}(2/\sqrt{c})-1}{\sqrt{c}}
\right)+e^{-4/c}
$$
where $a=1.77245$ and $\text{erf}(\cdot)$ is the Gauss error function, which might be not so sharp. Any ideas/known results?
EDIT: my primary interest is to determine the rate at which the integral converges to $1$ as $c\to \infty$, thus also approximate solutions/sharp bounds would be appreciated.
 A: With the substitution $x=\frac12(1+\tanh t)$, we obtain $$\int^1_0 e^{-\frac1{c\,x\,(1-x)}}\,dx=\frac12\,\int^\infty_{-\infty}\frac1{\cosh^2 t}\,e^{-\frac4c\,\cosh^2 t}\,dt=\int^\infty_0\frac1{\cosh^2 t}\,e^{-\frac4c\,\cosh^2 t}\,dt,$$ so let's investigate
$$I(x)=\int^\infty_0\frac1{\cosh^2 t}\,e^{-x\,\cosh^2 t}\,dt.$$
Since $$\frac{d}{dt}\frac1{\cosh t}e^{-x\,\cosh^2t}=\left(-\frac{\sinh t}{\cosh^2t}-2x\sinh t\right)e^{-x\,\cosh^2t},$$ we have
$$\int^\infty_0\left(-\frac{\sinh^2 t}{\cosh^2t}-2x\sinh^2 t\right)e^{-x\,\cosh^2t}\,dt=\int^\infty_0\sinh t\,\frac{d}{dt}\frac1{\cosh t}e^{-x\,\cosh^2t}\,dt=-\int^\infty_0e^{-x\,\cosh^2t}\,dt$$ by partial integration. Now we know that
$$1-\frac{\sinh^2 t}{\cosh^2t}=\frac1{\cosh^2t},$$ this means
$$\int^\infty_0\frac1{\cosh^2 t}\,e^{-x\,\cosh^2 t}\,dt=2x\int^\infty_0\sinh^2t\,e^{-x\,\cosh^2 t}\,dt.$$ We make use of the well-known identities
$$\sinh^2t=\frac12(\cosh2t-1)$$ and $$\cosh^2t=\frac12(\cosh2t+1)$$ to get
$$I(x)=x\,e^{-x/2}\int^\infty_0(\cosh2t-1)\,e^{-x/2\,\cosh2t}\,dt.$$ Substituting $t\to t/2$ and taking into account the integral representation $$K_\nu(z)=\int^\infty_0e^{-z\cosh t}\,\cosh(\nu t)\,dt$$ (this is formula 10.32.9 in https://dlmf.nist.gov/10.32), we finally arrive at
$$I(x)=\frac{x}2\,e^{-x/2}(K_1(x/2)-K_0(x/2)).$$
As this almost looked too good to be true, I checked it numerically, with $x=4/c$ and various values of $c$ between $0.1$ and $10$, and found perfect agreement.
A: As OP wanted the behaviour of the solution at infinity, starting from amazing Professor Vector solution, I expanded it at $c=\infty$
and got
$$\frac{2}{c} e^{-2/c} \left(K_1\left(\frac{2}{c}\right)-K_0\left(\frac{2}{c}\right)\right)\sim \frac{2 \gamma }{c}-\frac{2}{c}-\frac{2 \log c}{c}+1$$
Where $\gamma$ in Euler-Mascheroni constant.
Hope this helps
