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A Peculiar Integral

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Post Thu Sep 24, 2015 6:24 am

Posts: 2
Currently doing my research, I ran into this pretty little fellow,

$$\int_0^1 x^2 \sin(\pi x) x^x (1-x)^{1-x} \mathrm{d}x.$$

Numerical evidence seems to suggest that

$$ \int_0^1 x^2 \sin(\pi x) x^x (1-x)^{1-x} \mathrm{d}x = \frac{73}{5760} \pi e $$

I am not familiar with contour integration, so I wanted to asked this forum for support. God bless you all!

Post Thu Sep 24, 2015 2:19 pm

Posts: 20
Location: Israel
Very interesting question!
Let's denote $$\mathcal{D}(n)=\int_0^1 x^n \sin(\pi x)x^x(1-x)^{1-x}\,dx$$
the case $n=0$ was discussed in mathSE here and here and was given a solution using complex analysis here (German).
Numerically, it seems that $\mathcal{D}(n)=r\cdot\pi e$, where $r$ is rational, and for $n=0,1,2,3,4$ it seems that
$r=\cfrac1{24},\cfrac1{48},\cfrac{73}{5760},\cfrac{11}{1280},\cfrac{3625}{580608}$.
This is where we employ OEIS and find two sequence corresponding to the numerators and the denominators. It is stated that these fractions appear in "a generalization of Carleman's inequality".
Following the link, we find a simple recurrence relation defining these fractions, which also suggest shifting our function by $2$: $$b_j=-\frac1{j}\sum_{i=1}^{j} \frac{b_{j-i}}{i+1} \,\,\,\,\, , b_0=-1$$
which generates the sequence $b_1=\frac12,b_2=\frac1{24},b_3=\frac1{48},b_4=\frac{73}{5760}....$
I really hope to see a simple solution with real methods...
If we accept $\mathcal{D}(0)=\frac{\pi e}{24}$, the best i can do for now is get $\mathcal{D}(1)$. notice that $$\mathcal{D}(n)=\int_0^1 x^n \sin(\pi x)x^x(1-x)^{1-x}\,dx=\int_0^1 (1-x)^n \sin(\pi x)x^x(1-x)^{1-x}\,dx$$
So, since $(1-x)^2=1-2x+x^2$, $\mathcal{D}(2)=\mathcal{D}(0)-2\mathcal{D}(1)+\mathcal{D}(2)$ and therefore $\mathcal{D}(1)=\frac12\mathcal{D}(0)=\frac{\pi e}{48}.$

Post Sat Aug 06, 2016 10:57 pm

Posts: 2
I will be a publishing a paper on this in arXiv.org, but I think that will be as far as my pursuit will go. This is something I am absolutely cautious to share, but I feel the need unveil anyway. I have lost some will to believe this is a significant result due to doubts expressed by other mathematicians who I have corresponded with, so this led me to construe this might not be important after all. I have read about these integrals supposedly popping up in the work of Ramanujan, though I have found no reliable source, and Bruce Berndt still has yet to get back to me. :(

This project started when I was curious what parametrizations would be needed to encapsulate impressive information about the following integrals:

\begin{align}
&\int_0^1 \sin(\pi x) x^x(1-x)^{1-x} \, dx &= \frac{\pi e}{24} \\
&\int_0^1 \frac{\sin(\pi x)}{x^x(1-x)^{1-x}}\, dx &= \frac{\pi }{e} \\
&\int_0^1 \frac{\sin(\pi x)}{x(1-x)}\frac{1}{x^x(1-x)^{1-x}}\, dx &= 2\pi
\end{align}

However, as it turns out, I was able to show they are related via the following theorem.

$\textbf{Theorem}$ For $m, q \in \mathbb{Z}$, and $m+q+1 \geq 0$,
$$ \int_0^1 x^m \sin\left(\pi q x \right) \left(x^x (1-x)^{1-x}\right)^q\ dx = (-1)^{q+1} \frac{d_{m+q+1}(q)}{(m+q+2)!_\mathbb{P}} \pi e^{q}$$
where $d_n(q)$ is a primitive polynomial of $\mathbb{Z}[x]$ of degree $n$, and $ n!_\mathbb{P}$ is the Bhargava factorial over the set of primes.


In addition, these rational numbers satisfy a neat recurrence relation, of which Carleman's inequality is a [special case][1] of:

$$\frac{d_{n}(q)}{(n+1)!_\mathbb{P}} = -\frac{q}{n} \sum_{k=1}^n \frac{d_{n-k}(q)}{(n-k+1)!_\mathbb{P}} \frac{1}{k+1}; \; d_0(q) = -1,\; \text{if} \,(q\neq 0).$$

Using these results, we can unlock a whole class of crazy stuff:

\begin{align*}\sum_{j=1}^n A_j(1-\alpha_j)^{q\left(1-\frac{1}{\alpha_j}\right)}&= (-1)^q\int_0^1 \frac{\sin\left(\pi q x \right)}{\pi x} \frac{\left[x^x\left(1-x\right)^{1-x}\right]^q}{x^q} \prod_{j=1}^n \frac{1}{1-\alpha_j x}\ dx,
\end{align*}
\begin{align*}
A_j = \prod_{k=1, k\neq j}^n \frac{\alpha_j}{\alpha_j-\alpha_k}, \quad \alpha_j \in (0,1).
\end{align*}

Here are some special values:
\begin{align}
&\int_0^1 \frac{\sin\left( \pi x \right)}{ (1-x)\left[x^x\left(1-x\right)^{1-x}\right]} \ dx = \pi \quad \quad &\int_0^1 \frac{\sin\left( \pi x \right)}{(1-x^2)\left[x^x\left(1-x\right)^{1-x}\right]} \ dx &= \frac{5\pi}{8}
\end{align}

I don't want to reveal too much anyway. Enjoy!
[1]: http://www.people.fas.harvard.edu/~sfin ... ve/crl.pdf
Last edited by Brian J. Diaz on Sun Aug 07, 2016 12:26 am, edited 1 time in total.

Post Sat Aug 06, 2016 11:59 pm

Posts: 38
Brian J. Diaz wrote:
I will be a publishing a paper on this in arVix.org, but I think that will be as far as my pursuit will go. This is something I am absolutely cautious to share, but I feel the need unveil anyway. I have lost some will to believe this is a significant result due to doubts expressed by other mathematicians who I have corresponded with, so this led me to construe this might not be important after all. I have read about these integrals supposedly popping up in the work of Ramanujan, though I have found no reliable source, and Bruce Berndt still has yet to get back to me. :(


Very impressive, I had a brief look at
$$\int_{0}^{1}\sin(\pi x)x^{x}(1-x)^{1-x}\text{d}x=\frac{\pi e}{24}$$
when this thread was first posted but I didn't really get anywhere without using contour integration and residues. Me and LukeSQBacon had another attempt at it last weekend and still didn't get very far, but we have a few ideas that might work. I look forward to your paper although I probably won't read it immediately since I want to find my own method first.


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