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Since $$\lim_{x\to{a}}e^{-\frac{1}{x-a}+\frac{1}{x-b}}=\lim_{x\to{b}}e^{-\frac{1}{x-a}+\frac{1}{x-b}}=0,$$ $e^{-\frac{1}{x-a}+\frac{1}{x-b}}$ 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?

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  • $\begingroup$ Very interesting question - and a nice bump function. The area seems to be around .007 so I suggest the name "Bond function". $\endgroup$ – Bennett Gardiner Nov 29 '13 at 15:30
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    $\begingroup$ I have an idea of using the fact that the Laplace transform of $e^{-1/t}$ is $2 K_1(2 \sqrt{s})/\sqrt{s}$. The convolution theorem would then state that $$\int_0^t dt' \, e^{-1/t'} e^{-1/(t-t')} = \frac{4}{i 2 \pi} \int_{c-i\infty}^{c+i \infty} ds \, e^{s t} \, \frac{\left [ K_1(2 \sqrt{s})\right ]^2}{s}$$ but I have not yet been able to crack the integral this way either yet. $\endgroup$ – Ron Gordon Nov 29 '13 at 17:27
  • $\begingroup$ Nice idea Ron, will the result just be given by the residue at $s=0$? It seems impossible to calculate the residue via a Laurent series. $\endgroup$ – Bennett Gardiner Nov 30 '13 at 0:29
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    $\begingroup$ This make me to think about math.stackexchange.com/questions/145015. $\endgroup$ – Harry Peter Dec 9 '13 at 13:58
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$\int_a^be^{-\frac{1}{x-a}+\frac{1}{x-b}}~dx$

$=\int_{a-\frac{a+b}{2}}^{b-\frac{a+b}{2}}e^{-\frac{1}{x+\frac{a+b}{2}-a}+\frac{1}{x+\frac{a+b}{2}-b}}~d\left(x+\dfrac{a+b}{2}\right)$

$=\int_{-\frac{b-a}{2}}^\frac{b-a}{2}e^{-\frac{1}{x+\frac{b-a}{2}}+\frac{1}{x-\frac{b-a}{2}}}~dx$

$=\int_{-\frac{b-a}{2}}^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$

$=\int_{-\frac{b-a}{2}}^0e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx+\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$

$=\int_\frac{b-a}{2}^0e^\frac{b-a}{(-x)^2-\frac{(b-a)^2}{4}}~dx+\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$

$=\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx+\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$

$=2\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$

$=2\int_0^\infty e^\frac{b-a}{\left(\frac{b-a}{2}\tanh x\right)^2-\frac{(b-a)^2}{4}}~d\left(\dfrac{b-a}{2}\tanh x\right)$

$=(b-a)\int_0^\infty e^{-\frac{b-a}{\frac{(b-a)^2}{4}\text{sech}^2x}}~d(\tanh x)$

$=(b-a)\int_0^\infty e^{-\frac{4\cosh^2x}{b-a}}~d(\tanh x)$

$=(b-a)\left[e^{-\frac{4\cosh^2x}{b-a}}\tanh x\right]_0^\infty-(b-a)\int_0^\infty\tanh x~d\left(e^{-\frac{4\cosh^2x}{b-a}}\right)$

$=8\int_0^\infty e^{-\frac{4\cosh^2x}{b-a}}\sinh x\cosh x\tanh x~dx$

$=8\int_0^\infty e^{-\frac{4\cosh^2x}{b-a}}\sinh^2x~dx$

$=8\int_0^\infty e^{-\frac{2(\cosh2x+1)}{b-a}}\dfrac{\cosh2x-1}{2}dx$

$=4e^{-\frac{2}{b-a}}\int_0^\infty e^{-\frac{2\cosh2x}{b-a}}(\cosh2x-1)~dx$

$=2e^{-\frac{2}{b-a}}\int_0^\infty e^{-\frac{2\cosh2x}{b-a}}(\cosh2x-1)~d(2x)$

$=2e^{-\frac{2}{b-a}}\int_0^\infty e^{-\frac{2\cosh x}{b-a}}(\cosh x-1)~dx$

$=2e^{-\frac{2}{b-a}}\left(K_1\left(\dfrac{2}{b-a}\right)-K_0\left(\dfrac{2}{b-a}\right)\right)$

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