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Inspired by this integral

Complex contour solution

1. Do the substitution: $$\int_{0}^{1}\frac{\sin\pi x}{\left(1-x\right)^{x}x^{1-x}}dx=\left|x=\frac{1}{1+e^{t}};\ \ dx=-\frac{e^{t}dt}{\left(1+e^{t}\right)^{2}}\right|=$$ $$=\int_{-\infty}^{+\infty}\frac{\sin\left(\frac{\pi}{1+e^{t}}\right)}{\left(1-\frac{1}{1+e^{t}}\right)^{\frac{1}{1+e^{t}}}\left(\frac{1}{1+e^{t}}\right)^{1-\frac{1}{1+e^{t}}}}\frac{e^{t}dt}{\left(1+e^{t}\right)^{2}}=\int_{-\infty}^{+\infty}\sin\left(\frac{\pi}{1+e^{t}}\right)\frac{e^{t-\frac{t}{1+e^{t}}}}{1+e^{t}}dt=$$

$$=\lim_{R\to\infty}\ \operatorname{Im}\int_{-R}^{+R}\exp\left(\frac{i\pi-t}{1+e^{t}}\right)\frac{dt}{1+e^{-t}}$$

2. Build the contour and analyse the special points:

$L_1=t,~~ L_2=t + 2\pi i,~~t\in[-R,+R];\quad\quad l_1=R+it,~~l_2=-R+it,~~t\in [0, 2\pi]$

There is only one simple pole in the contour $z=i\pi$

3. Residue and integral estimates:

$$\oint _C \exp\left(\frac{i\pi-z}{1+e^{z}}\right)\frac{dz}{1+e^{-z}}=2\pi i\cdot \lim_{z\to i\pi}\exp\left(\frac{i\pi-z}{1+e^{z}}\right)\frac{z-i\pi}{1+e^{-z}}=2\pi ei\cdot \lim_{z\to i\pi}\frac{z-i\pi}{1+e^{-z}}=2\pi ei$$

$$\int_{L_{1}}^{\ }=\int_{-R}^{+R}\exp\left(\frac{i\pi-t}{1+e^{t}}\right)\frac{dt}{1+e^{-t}}$$

$$\int_{L_{2}}^{\ }=\int_{-R}^{+R}\exp\left(\frac{-i\pi-t}{1+e^{t}}\right)\frac{dt}{1+e^{-t}}$$

In the limit term $\int_{l_{1}}^{\ }=2\pi i$ and $\int_{l_{2}}^{\ }=-2\pi i$

4. Result:

$$\int_{-R}^{+R}\exp\left(\frac{i\pi-t}{1+e^{t}}\right)\frac{dt}{1+e^{-t}}+2\pi i-\int_{-R}^{+R}\exp\left(\frac{-t-\pi i}{1+e^{t}}\right)\frac{dt}{1+e^{-t}}+2\pi i=2\pi ei$$

$$2i\int_{-R}^{+R}\left[\frac{\exp\left(\frac{i\pi}{1+e^{t}}\right)-\exp\left(\frac{-\pi i}{1+e^{t}}\right)}{2i}\right]\exp\left(\frac{-t}{1+e^{t}}\right)\frac{dt}{1+e^{-t}}=2\pi ei-4\pi i$$

$$R\to \infty:~~\int_{-\infty}^{+\infty}\sin\left(\frac{\pi}{1+e^{t}}\right)\exp\left(\frac{-t}{1+e^{t}}\right)\frac{dt}{1+e^{-t}}=\pi\left(e-2\right)$$

Via step 1:

$$\int_{0}^{1}\frac{\sin\pi x}{\left(1-x\right)^{x}x^{1-x}}dx=\pi e - 2\pi$$

The main question

Is it possible to find the value of the integral without complex numbers, or at least without contour integration, without resorting to the methods presented on the linked above page? I have no idea how to approach the solution.

metamorphy
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Rezerd Prime
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  • A similar unanswered question was asked on MO : https://mathoverflow.net/q/226870/500489 , indicating this question is a tough one – Quý Nhân Jun 02 '25 at 03:25
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    Absolutely of no use $$\int_{-\infty}^{+\infty}\sin\left(\frac{\pi}{1+e^{t}}\right)\frac{e^t}{e^t+1},dt=\text{Si}(\pi )$$ Very nice and difficult problem ! – Claude Leibovici Jun 02 '25 at 04:10
  • It looks like $-\operatorname{Im}\int_0^1\frac1x(1-\frac1x)^{-x}dx=-2\operatorname{Im}\int_0^1(1-\frac1x)^{-x}dx$, $\operatorname{Im}\int_0^1 (-x\ln(1-\frac1x))^ndx=-\frac{\pi n}2$, and $\operatorname{Im}\int_0^1 \frac1x(-x\ln(1-\frac1x))^ndx=-\pi$. Switch?:$-\int_0^1\operatorname{Im}\sum\limits_{n=0}^\infty \frac1{xn!}\ln^n((1-1/x)^{-x})dx\mathop=\limits^? -\operatorname{Im}\sum\limits_{n=0}^\infty\int_0^1\frac1{xn!}\ \ln^n((1-1/x)^{-x})dx$. Doing so and integrating gives $π(e-1)$. One can integrate from $0$ to $\frac12$ and double, but get a result with derivatives of a special function. – Тyma Gaidash Jun 05 '25 at 00:39
  • @ТymaGaidash looks interesting, i'll try use it somehow – Rezerd Prime Jun 05 '25 at 21:18

1 Answers1

1

$\DeclareMathOperator{Im}{Im}$This answer will have a method that does not mention contour integration nor the residue theorem. Notice: $$\int^1_0 \frac{\sin \pi x}{(1-x)^x x^{1-x}}dx=-\Im\int_0^1\frac1x\left(1- \frac1x\right)^{-x}dx$$

If one were to expand $e^{-x\ln(1-\frac1x)}$ or $\frac1{x-1}e^{(1-x)\ln(1-\frac1x)}$ with $e^x$’s MacLaurin series, the value would be off by $\pi$. However, splitting the integral and expanding like so resolve the problem:

$$-\Im\int_0^1\frac1x\left(1-\frac1x\right)^{-x}dx= \Im\int_0^1\left(1-\frac1x\right)\left(1-\frac1x\right)^{-x}dx-\Im\underbrace{\int_0^1\left(1-\frac1x\right)^{-x}dx}_{x\to1-x}=2\Im\int_0^1\left(1-\frac1x\right)^{1-x}dx=2\Im\int_0^1\frac{x-1}x\left(1-\frac1x\right)^{-x}dx=\Im\sum_{n=0}^\infty\frac1{n!}\int_0^12(-1)^n(1-x)x^{n-1}\ln^n\left(1-\frac1x\right)dx$$

One can see the $n=0$ term is $0$ and use $n$th derivatives and the beta function. Then, one can use beta function properties and get a Pochhammer symbol $(x)_n$ expression

$$\Im\int_0^12(-1)^n(1-x)x^{n-1}\ln^n\left(1-\frac1x\right)dx=\Im \frac{d^n}{da^n}\left.\int_0^12 (-1)^{a+n} x^{n-1-a} (1-x)^{a+1}\right|_{a=0}=2(-1)^n\frac{d^n}{da^n}\left.\sin(\pi a) B(n-a,a+2)\right|_{a=0}= -\frac{2\pi}{(n+1)!}\frac{d^n}{da^n}\left.\prod_{j=-1}^{n-1}(a-j)\right|_{a=0}=-\frac{2\pi}{(n+1)!}\frac{d^n}{da^n}(a-n+1)_{n+1}|_{a=0}$$

$(a-n+1)_{n+1}$ can be a polynomial of degree $n+1$ in $a-n+1$ that contains Stirling numbers of the first kind $S_n^{(k)}$ and the $(n+1)$th and $n$th power terms would survive due to the $n$th derivative. $S_{n+1}^{(n)}=-\frac12n(n+1)$ $$-\frac{2\pi}{(n+1)!}\frac{d^n}{da^n}(a-n+1)_{n+1}|_{a=0}=\frac{2\pi}{(n+1)!}\sum_{k=n}^{n+1}(-1)^{k+n} S_{n+1}^{(k)} \frac{d^n}{da^n}\left.(a-n+1)^k\right|_{a=0}= \frac{2\pi S_{n+1}^{(n)}}{n+1}+2\pi(n-1) S_{n+1}^{(n+1)}=\pi(n-2)$$

Therefore:

$$\int^1_0 \frac{\sin \pi x}{(1-x)^x x^{1-x}}dx=\sum_{n=1}^\infty\frac{\pi(n-2)}{n!}=\boxed{\pi(e-2)}$$

Тyma Gaidash
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  • For some reason I thought I'd have to work with the nth derivative. In my solution, I'm stuck on the proof of $\operatorname{Im}\int_{0}^{1}\frac{1}{x}\left(-x\ln\left(1-\frac{1}{x}\right)\right)^{n}dx=-i\pi$. Thx for solution! – Rezerd Prime Jun 09 '25 at 01:07