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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.

RGS
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2 Answers2

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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:

enter image description here

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

anderstood
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7

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.

B. Goddard
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  • How about the closed form @B.Goddard –  Jan 20 '17 at 05:45
  • I looked at the Taylor series for the integrand, centered at $0$, $1$ and $1/2$. All three series are a mess. I don't think you can get a series form, let alone a closed form. Even tricks like changing $\sin(\pi x)$ to $e^x$ don't help a bit. – B. Goddard Jan 20 '17 at 14:09
  • What I mean here is that I don't think there is a closed form for the anti-derivative. The cites in the comments above show that the definite integral can be evaluated fairly nicely with contour integrals. – B. Goddard Jan 20 '17 at 19:37
  • If I give you the closed form can you provid the proof it @B.Goddard? – gymbvghjkgkjkhgfkl Jan 21 '17 at 07:44
  • Following the links above, one gets here: http://www.artofproblemsolving.com/community/c7h501365p2817263 Where nearly the same integral is done. The comment above asserts that somewhere on that site this integral is done similarly. The residues seem easy to calculate. So the answer to your question is "Yes", because I could just copy and paste. – B. Goddard Jan 21 '17 at 12:53