Transporting goods. Minimize the arc length of $g(x)=\exp(s/\ln(x))$ for $x \in (0,1)$ Say I want to transport goods from my corner of the unit square, $(0,1)$ to a friend's corner, $(1,0)$ who lives a distance of $\sqrt{2}$ units away from me. But route $\sqrt{2}$ is largely closed for construction, and the only option is to travel the scenic route, along the function $\exp(s/\ln(x))$ to reach my friend's corner. Which $s$ should I choose? 

Q: What value of $s$ minimizes the arc length of $g(x)?$

I used Wolfram Alpha as an aid but it said the arc length was infinite when I provided the bounds of $x$ as $(0,1).$ As a work-around I decided to put the bounds as $0.00001$ to $.99999.$ 
 A: A few words on how it can be computed (I don't think there's a closed-form answer). Let $s=a^2$ and choose $x=\exp(-ae^t)$, $y=\exp(-ae^{-t})$ ($t\in\mathbb{R}$) as the parameterization of the curve(s). Then, if $g(a,t)=\exp(t-ae^t)$, the length is equal to $af(a)$, where $$f(a)=\int_{-\infty}^{\infty}\sqrt{g^2(a,t)+g^2(a,-t)}\ dt$$ and, for a minimum, we're solving $f(a)+af'(a)=0$.
The trapezoidal rule is extremely good (!) for these integrals. Here is a PARI/GP script:
myexp(x)=if(x>-default(realbitprecision),exp(x),0);
myint(f,h)=h*(f(0)+2*suminf(n=1,f(n*h)));
myterm(k,a,x)=myexp(k*x-2*a*myexp(x));
mysymm(k,a,x)=myterm(k,a,x)+myterm(k,a,-x);
myfoo(a,x)=sqrt(mysymm(2,a,x));
mygoo(a,x)=mysymm(3,a,x)/myfoo(a,x);
foo(a,h)=myint(x->myfoo(a,x),h);
goo(a,h)=myint(x->mygoo(a,x),h);
answer(h)=solve(a=0.6,0.7,foo(a,h)-a*goo(a,h))^2;

With the default precision of $100$ digits, answer(0.01) and answer(0.02) both produce $$\approx\color{blue}{0.433565595822430815354309488670966602968032246257987930713649607}\cdots$$
(and a few more matching digits -- yes, I couldn't make intnum work out-of-the-box for me).

