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Let $f$ be integrable over the interval $(0, 1)$, and

$$I_n = \int_0^{1} x^n f(x) \, \mathrm{d}x.$$

Suppose $f(x) = f(1-x)$; we can then show that

$$I_n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k \, I_{k}, $$

by letting $u = 1-x$ and using the binomial theorem.

Thanks to a paper, I was able to show that $I_0, I_2, ... , I_{2n}$ are linearly independent, and thus form a basis of the space containing all $I_n$; this would imply that

$$I_{2n+1} = \sum_{k=0}^{n} a_k I_{2k}, \; a_k \in \mathbb{Q}. $$

Why this has become of interest to me was when I found out the existence of the following integral:

$$\int_0^1 \sin(\pi x)x^x (1-x)^{1-x} \mathrm{d} x = \frac{\pi e}{24}.$$

(You can find the link here for further details, including the proof.)

It can be shown that the integrand satisfies the symmetric property using the substitution, and, as a consequence, we see that

$$\int_0^1 x \sin(\pi x)x^x (1-x)^{1-x} \mathrm{d} x = \frac{1}{2} \int_0^1 \sin(\pi x)x^x (1-x)^{1-x} \mathrm{d} x = \frac{\pi e}{48}.$$

Since the even cases are linearly independent, this means the only known (as far as I know) evaluation of the integral

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

is when $n = 0$.

I am curious whether the integral would be of the following form

$$\int_0^1 x^{2n} \sin(\pi x)x^x (1-x)^{1-x} \mathrm{d} x = c_n \pi e, \; c_n \in \mathbb{Q} $$

although numerical methods seem to suggest that this is not the case. Can this problem be attacked using current methods, such as applying residue theory (as seen in the case $n = 0$)?

I should state that my fluency in the topic of residues still needs some polishing, and even then, my gut instinct tells me this may not be the best route. I'd like to be proven wrong, of course.

Thank you for your support.

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    $\begingroup$ What do you mean by $I_0,I_2,...,I_{2n}$ are linearly independent ? If you take for instance $f(x)=x(1-x)$, all $I_k$ are rational. So they are at least not lineraly independent over the rationals. $\endgroup$ – Todd Leason Aug 28 '15 at 8:46
  • $\begingroup$ @ToddLeason I should've specified that. Thank you for the correction! $\endgroup$ – Brian Diaz Aug 28 '15 at 22:19
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High precision numerical computations suggest:

$$c_1=73/5760 $$ $$c_2=3625/580608$$ $$c_3=5233001/1393459200$$

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  • $\begingroup$ What program did you use to derive them? Mathematica? Maple? $\endgroup$ – Brian Diaz Aug 28 '15 at 22:22
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    $\begingroup$ @BrianJ.Diaz Sage and pari (pari is part of sage). Computed the integral numerically and looked for linear integer relation with the constant with high precision. $\endgroup$ – joro Aug 29 '15 at 4:50
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    $\begingroup$ I got the same in Maple (using 200 digit integration). I also got $c_4=927777937/367873228800$, $c_5=43791735453787/24103053950976000$, $c_6=158996102434867/115694658964684800$, $c_7=782501215247703271/726206474732175360000$. The denominators have only small factors. $\endgroup$ – Brendan McKay Aug 29 '15 at 8:42
  • $\begingroup$ @BrendanMcKay Thanks for confirming. I checked with higher precision the conjectured with lower precision constants. $\endgroup$ – joro Aug 29 '15 at 8:46

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