Why is $\int_{0}^{1}x^x(1-x)^{2-x}\sin(x\pi)dx=\int_{0}^{1}x^{1+x}(1-x)^{1-x}\sin(x\pi)dx?$

Question

I don't know why these two integrals yield the same results.

$$\int_{0}^{1}x^x(1-x)^{2-x}\sin(x\pi)dx=\int_{0}^{1}x^{1+x}(1-x)^{1-x}\sin(x\pi)dx$$

Any hints, clues and ideas how to go about dealing these integrals(showing that they are the same and determine the closed form)?

$$\int_{0}^{1}x^x(1-x)^{2-x}\sin(x\pi)dx=\int_{0}^{1}x^{1+x}(1-x)^{1-x}\sin(x\pi)dx$$I know it is similar to the sophere's dream integral. I think you can possible use integration by parts. Because there are three functions are involved applying by parts it would be very long.

I try to make a substitution of $u=1-x$

$$-\int_{0}^{1}u^{1+u}(1-u)^{1-u}\sin((u+1)\pi)du$$

Q: show that they are the same and evaluate its closed form.

Edited

$$\int_{0}^{1}u^{1+u}(1-u)^{1-u}\sin(u\pi)du$$

---

Let try and applying by parts

$u=x^x(1-x)^{2-x}$

$du=x^x(1-x)^{2-x}\ln{x\over 2-x}dx$

$v=-{\cos{(x\pi)}\over \pi}$

$$=-{1\over \pi}x^x(1-x)^{2-x}\cos{(x\pi)}+{1\over \pi}\int_{0}^{1}x^x(1-x)^{2-x}\cos{(x\pi)}\ln{x\over 2-x}dx$$

This looked more complicate than before, so ideally, no, it is not a good way of tackling this problem.

Answer 1

The equality can be shown by computing the integral of the difference, and observing a punctual symmetry around $(1/2,0)$.

\begin{align}\int_{0}^{1}x^x(1-x)^{2-x}&\sin(x\pi)dx-\int_{0}^{1}x^{1+x}(1-x)^{1-x}\sin(x\pi)dx \\&= \int_{0}^{1}x^x((1-x)^{2-x}-x(1-x)^{1-x})\sin(x\pi)dx\\ &= \int_{0}^{1}x^x(1-x)^{1-x}(1-2x)\sin(x\pi)dx\end{align}

Let's denote the integrand by $f$: $$f(x)=x^x(1-x)^{1-x}(1-2x)\sin(x\pi).$$

The function $f$ appears to have a punctual symmetry: \begin{align}f(1-x)&=(1-x)^{1-x}(1-(1-x))^{1-(1-x)}(1-2(1-x))\sin((1-x)\pi)\\ &=-x^x (1-x)^{1-x}\sin(\pi x) & \\ &= -f(x)\end{align} so the point $(1/2,0)$ is a point of symmetry of the graph of $f$, hence the integral over $[0,1]$ (where $f$ is defined) is $0$.

---

The punctual symmetry can be guess from the graph of $f$ represented below:

Note that the equality also holds for integral boundaries in $[0,1]$ and centered around $1/2$.

Answer 2

In the second integral, make the substitution $v=1-x$, so that $x=1-v$ and $dx = -dv$. We get

$$\int_0^1 x^{1+x}(1-x)^{1-x}\sin(x\pi) \; dx = \int_1^0 (1-v)^{1+(1-v)} v^v \sin((1-v)\pi) \; (-dv).$$

The minus sign on $dv$ reverses the order of integration, and we have the identity $\sin(\pi -A) = \sin A$, so the above is

$$\int_0^1 (1-v)^{2-v}v^v \sin(v\pi) \; dv,$$

but since $v$ is a dummy variable, we can change it to $x$ and we have the first integral.