Difficulty in solving the following integral How to solve the following definite integral:
$$I= \int_{\pi/4}^{3\pi/4} e^{-e^{\frac{1}{x}}}dx$$
I have tried solving this using Feynman's trick but to me, it turns out there is no obvious way to parametrize this function. I am very weak in contour integration and hence couldn't apply it here. I couldn't think of any obvious substitution as well.
The only thing I can see is that $1/x$ is the derivative of $log$ $x$ which is the inverse of exponential. So probably this integral requires exploitation of that but I can't see how.
Also, I prefer answers that do not go beyond undergraduate knowledge in mathematics (if possible) since I encountered it in an undergraduate exam.
 A: Try this: Let $t= e^{\frac1x}$
$$\int_{\pi/4}^{3\pi/4} e^{-e^{\frac{1}{x}}}dx
= - \int_{e^{\frac4\pi}}^{e^{\frac4{3\pi}}}\frac{e^{-t}}{t\ln^2t}dt= J(0)
$$
where
$J(a)= - \int_{e^{\frac4\pi}}^{e^{\frac4{3\pi}}}\frac{t^{a-1} e^{-t}}{\ln^2t}dt
$
$$J’’(a) =  - \int_{e^{\frac4\pi}}^{e^{\frac4{3\pi}}}{t^{a-1} e^{-t}}dt = \Gamma(a, e^{\frac4{3\pi}}) -\Gamma(a, e^{\frac4\pi}) 
$$
A: Just for the fun.
It would be amuzing to approximate the integral using the $[2,2]$ Padé approximant built around $\frac \pi 2$ to make
$$ e^{-e^{\frac{1}{x}}}=\frac {a_0+a_1 \left(x-\frac{\pi }{2}\right)+a_2 \left(x-\frac{\pi
   }{2}\right)^2} {1+b_1 \left(x-\frac{\pi }{2}\right)+b_2 \left(x-\frac{\pi
   }{2}\right)^2 }$$ The problem is that the constants are quite ugly but perfectly usable.
As usual, integrating will give two arctangents and two logarithms.
I shall not write any formula but, numerically, the result is $0.221184$ to be compared with the "exact" $0.221115$.
The inverse symbolic calculator provides, as an approximation,
$$\frac 1{10}\left(\log (\pi ) \,\,\Gamma \left(\frac{13}{24}\right)\right)^{\sqrt[3]{2}}=0.221115422\cdots$$to be compared with the "exact" $      0.221115434\cdots$
