# Evaluating an integral using real methods

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.

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 $$2\pi\int_0^\infty\int_0^\infty \frac{du\,dv}{uv}\,\frac{e^{-\ell(v)}-e^{-\ell(u)}}{(\ell(u)-\ell(v))[(\ell(u)-\ell(v))^2+4\pi^2]}.$$

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.