How to integrate the bump functions,i.e,$\int_a^{b}e^{-\frac{1}{x-a}+\frac{1}{x-b}}dx$,where $aSince $$\lim_{x\to{a}}e^{-\frac{1}{x-a}+\frac{1}{x-b}}=\lim_{x\to{b}}e^{-\frac{1}{x-a}+\frac{1}{x-b}}=0,$$ $e^{-\frac{1}{x-a}+\frac{1}{x-b}}$ is continuous on the interval $[a,b]$ (taking $0$ if  $x=a$ or $b$).
So the integral $\int_a^{b}e^{-\frac{1}{x-a}+\frac{1}{x-b}}dx$ makes sense.But i do not know how to compute this integral?
In particularly, taking $a=0$ and $b=1$,we just need to compute $\int_0^{1}e^{-\frac{1}{x}+\frac{1}{x-1}}dx$. 
My thought:Considering another integration $$I(\epsilon)=\int_0^{1}e^{-\frac{1}{x}+\frac{1}{x-1}}e^{-\epsilon{(\frac{1}{x^2}-\frac{1}{(x-1)^2}})}dx,$$
take the derivative of this with respect to $\epsilon$(assuming it converges uniformly),we get $$I'(\epsilon)=-\int_0^{1}e^{-\frac{1}{x}+\frac{1}{x-1}}e^{-\epsilon{(\frac{1}{x^2}-\frac{1}{(x-1)^2}})}d_{-\frac{1}{x}+\frac{1}{x-1}}.$$
I can not continue,and i do not know whether it work.Can you provide me some methods?
 A: $$\int_a^be^{-\frac{1}{x-a}+\frac{1}{x-b}}~dx$$
$$=\int_{a-\frac{a+b}{2}}^{b-\frac{a+b}{2}}e^{-\frac{1}{x+\frac{a+b}{2}-a}+\frac{1}{x+\frac{a+b}{2}-b}}~d\left(x+\dfrac{a+b}{2}\right)$$
$$=\int_{-\frac{b-a}{2}}^\frac{b-a}{2}e^{-\frac{1}{x+\frac{b-a}{2}}+\frac{1}{x-\frac{b-a}{2}}}~dx$$
$$=\int_{-\frac{b-a}{2}}^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=\int_{-\frac{b-a}{2}}^0e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx+\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=\int_\frac{b-a}{2}^0e^\frac{b-a}{(-x)^2-\frac{(b-a)^2}{4}}~dx+\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx+\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=2\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=2\int_0^\infty e^\frac{b-a}{\left(\frac{b-a}{2}\tanh x\right)^2-\frac{(b-a)^2}{4}}~d\left(\dfrac{b-a}{2}\tanh x\right)$$
$$=(b-a)\int_0^\infty e^{-\frac{b-a}{\frac{(b-a)^2}{4}\text{sech}^2x}}~d(\tanh x)$$
$$=(b-a)\int_0^\infty e^{-\frac{4\cosh^2x}{b-a}}~d(\tanh x)$$
$$=(b-a)\left[e^{-\frac{4\cosh^2x}{b-a}}\tanh x\right]_0^\infty-(b-a)\int_0^\infty\tanh x~d\left(e^{-\frac{4\cosh^2x}{b-a}}\right)$$
$$=8\int_0^\infty e^{-\frac{4\cosh^2x}{b-a}}\sinh x\cosh x\tanh x~dx$$
$$=8\int_0^\infty e^{-\frac{4\cosh^2x}{b-a}}\sinh^2x~dx$$
$$=8\int_0^\infty e^{-\frac{2(\cosh2x+1)}{b-a}}\dfrac{\cosh2x-1}{2}dx$$
$$=4e^{-\frac{2}{b-a}}\int_0^\infty e^{-\frac{2\cosh2x}{b-a}}(\cosh2x-1)~dx$$
$$=2e^{-\frac{2}{b-a}}\int_0^\infty e^{-\frac{2\cosh2x}{b-a}}(\cosh2x-1)~d(2x)$$
$$=2e^{-\frac{2}{b-a}}\int_0^\infty e^{-\frac{2\cosh x}{b-a}}(\cosh x-1)dx$$
$$=2e^{-\frac{2}{b-a}}\left(K_1\left(\dfrac{2}{b-a}\right)-K_0\left(\dfrac{2}{b-a}\right)\right)$$
