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This is a bit of recreational integration. The following, rather attractive integral is quite straightforward via residues:

$$\int_0^1 x^{-x}(1-x)^{x-1}\sin \pi x\,\mathrm{d}x=\frac{\pi}{e}$$ Motivated mainly by curiosty, I have painstakingly spent hours trying to prove this nifty result without the advances of complex analysis - to no avail. I have also extensively searched the net for such a solution without success. Now of course the integrand is naturally underpinned by a complex expression, so such a solution would probably be a bit outlandish - but it would be interesting to see whether it is feasible.

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This is not really an answer, but it might be the first step in getting one. Let us represent the integral as a double one, but with the integrand containing no expressions of the form $f(x)^{g(x)}$, with the base and exponent both variable.

Indeed, letting $\ell(u):=u-\ln u$ for $u>0$, note that for $x\in(0,1)$ $$(1)\qquad \Gamma(x)x^{-x}=\int_0^\infty e^{-x\ell(u)}\,du=\int_0^\infty e^{-x\ell(u)}\,\frac{du}u $$ and hence $$(2)\qquad \Gamma(1-x)(1-x)^{x-1}=\int_0^\infty e^{(x-1)\ell(v)}\,dv=\int_0^\infty e^{(x-1)\ell(v)}\,\frac{dv}v. $$ Multiplying $(1)$ and $(2)$ and using Euler's reflection formula $\Gamma(x)\Gamma(1-x)=\pi/\sin\pi x$, we see that the integral in question equals $$\frac1\pi\,\int_0^1dx\,\sin^2\pi x\,\int_0^\infty\int_0^\infty du\,dv\,e^{-x\ell(u)+(x-1)\ell(v)}$$ $$=\frac1\pi\,\int_0^\infty\int_0^\infty du\,dv\,\int_0^1dx\,\sin^2\pi x\,e^{-x\ell(u)+(x-1)\ell(v)}$$ $$=2\pi\int_0^\infty\int_0^\infty du\,dv\,\frac{e^{-\ell(v)}-e^{-\ell(u)}}{(\ell(u)-\ell(v))[(\ell(u)-\ell(v))^2+4\pi^2]} $$ and that it also equals $$ 2\pi\int_0^\infty\int_0^\infty \frac{du\,dv}{uv}\,\frac{e^{-\ell(v)}-e^{-\ell(u)}}{(\ell(u)-\ell(v))[(\ell(u)-\ell(v))^2+4\pi^2]}. $$

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Although not this particular problem, I've looked about evaluating an analogous integral in hopes of furthering my understanding what real methods would be of good use.

Consider the following integral:

$$ \int_0^1 x^x (1-x)^{1-x} \sin(\pi x) \mathrm{d} x$$

Using the method of residues, the integral can be shown to equal $\frac{\pi e}{24}$. However, here's one real attempt. Consider the following Gamma function identity

$$\Gamma(x) = \sqrt{2 \pi } x^{x-1/2} e^{-x} e^{f(x)} $$

where $$f(x) = \int_0^\infty \frac{2\arctan{\frac{t}{x}}}{e^{2\pi t}-1} \mathrm{d}t$$

Using Euler's reflection formula $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin{\pi x}}$, we have

$$ \int_0^1 x^x (1-x)^{1-x} \sin(\pi x) \mathrm{d} x = \frac{e}{2} \int_0^1 \sqrt{x(1-x)}e^{-(f(x)+f(1-x))} \mathrm{d}x$$

Substituting $x = \sin^2(\frac{\pi\theta}{2})$, we have then

$$\frac{e}{2} \int_0^1 \sqrt{x(1-x)}e^{-(f(x)+f(1-x))} \mathrm{d}x = \frac{\pi e}{8}\int_0^1 \sin^2(\pi \theta) e^{-H(\theta)}\mathrm{d} \theta$$

where $$ H(\theta) = f\left(\sin^2\left(\frac{\pi\theta}{2}\right)\right) + f\left(\cos^2\left(\frac{\pi\theta}{2}\right)\right) = \int_0^\infty \frac{2\arctan{\frac{4t}{\sin^2(\pi \theta)-4t^2}}}{e^{2\pi t}-1} \mathrm{d}t$$

I'm still looking towards it, so at least we can claim the original integral is bounded by $\frac{\pi e}{16} $ by using the fact $e^{-H(\theta)}$ is positive, and $\int_0^1 \sin^2(\pi \theta) \mathrm{d} \theta = \frac{1}{2}$. Thus, our integral reduces to showing the following

$$\int_0^1 \sin^2(\pi \theta) e^{-H(\theta)}\mathrm{d} \theta = \frac{1}{3}$$

using real methods.

I'll continue updating as I find ways to approach this problem.

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