Closed-form of $\int_{a}^{b}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx$ for some $a<b$

In this question I asked to prove that

$$\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx=\frac{\pi e}{24}.$$

If we take a look at the plot of the integrand, then we could see some symmetry-property.

enter image description here

Because the integrand is symmectic to the $y=1/2$ line, it is also true, that

$$\int_{0}^{1/2}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx=\int_{1/2}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx=\frac{\pi e}{48}.$$

Question. Is there a close-form of $I(a,b)=\int_{a}^{b}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx$ for some else $a<b$? If it is too hard, then is there a closed-form for some $a<b$, just for the real or imaginary part? For example

$$\Re \left[ \int_{1}^{3/2}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx \right],$$
$$\Re \left[ \int_{3/2}^{2}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx \right],$$
or
$$\Im \left[ \int_{1}^{2}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx \right]?$$

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Is there a close-form of $I(a,b)=\displaystyle\int_{a}^{b}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx$ for some other $a<b$?

Yes, of course there is. An infinity of them, in fact. But, in that case, a and b will lack a closed form. 🙂 For instance, all rational and algebraic numbers possess a closed form expression. Now, limiting ourselves to those in between $0$ and $\dfrac{e\pi}{24}$, which themselves are infinite in number, for each of them there are infinitely many solutions a and b within the interval $[0,1]$, so that $I(a,b)$ will be equal to that particular rational or algebraic value, since $f(t)=I(a,t)$ is a monotonous function for any t in $[a,1]$. So I assume that you are interested in triplets $\Big(a,b,I(a,b)\Big)$, other than the ones already mentioned, and where all three values have a closed form, in which case the answer would be that no others are known to exist $($key word here being “known”, since proving this is, to my knowledge, still an open problem$)$.