# 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.

• Looks like:$$\frac{1}{2} \sqrt{\pi } G_{1,2}^{2,0}\left(\frac{4}{c}| \begin{array}{c} \frac{3}{2} \\ 0,1 \\ \end{array} \right)$$ or in Mathematica code: 1/2 Sqrt[\[Pi]] MeijerG[{{}, {3/2}}, {{0, 1}, {}}, 4/c] Commented Jan 11, 2021 at 17:35
• Many thanks! Any clue on how such Meijer G-function behaves for $c \to \infty$? I re-edited my question in this sense. Commented Jan 11, 2021 at 17:51
• There is a "closed" expression in terms of modified Bessel functions, if that helps you, $\displaystyle\frac2c\,e^{-2/c}(K_1(2/c)-K_0(2/c)$ (if I'm not mistaken).
– user436658
Commented Jan 11, 2021 at 19:02
• @ProfessorVector How did you get that? Commented Jan 11, 2021 at 19:05
• @Semiclassical I wrote up my derivation (see below), without Mathematica, but using Wolfram Alpha for numerical checks.
– user436658
Commented Jan 12, 2021 at 7:53

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.
• It looks like the root of the issue with Mathematica's Integrate is that it's bad at recognizing integral representations like this (c.f. mathematica.stackexchange.com/questions/4728/…). Using the suggested internal function gives results which agree with you. Moreover, I think my numerical disagreement was due to an error on my part. So I withdraw my doubt and concur with your conclusion. Commented Jan 12, 2021 at 8:23
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.
• Its probably a silly question, but: where does the term including $\gamma$ come from? Using the asymptotic relations provided at dlmf.nist.gov/10.30 I was concluding that $K_1(2/c)\sim 2^{-1}\Gamma(1) c$ while $K_0(z)\sim -\log(2/c)$, thus $$\frac{2}{c} e^{-2/c} \left(K_1\left(\frac{2}{c}\right)-K_0\left(\frac{2}{c}\right)\right)\sim e^{-2/c}(1-2\log(c/2)/c)\sim (1-2/c)(1-2\log(c/2)/c)$$ which is essentially of the same asymptotic order of the expression you obtained - I just wantes to check whether I'm missing anything stupid. Commented Jan 13, 2021 at 16:55
• @JackLondon $\gamma$ comes from the Taylor expansion of the Bessel K function wolframalpha.com/input/?i=series+BesselK%5B0%2C+z%5D Commented Jan 13, 2021 at 19:01