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Do we know if there is a closed form for $$ I :=\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\ \mathrm{d}x\mathrm{?} $$ Wolfram alpha gives an approximation of $2.66989$ which may be equivalent to: $$10\sqrt{\frac{2\pi}{77\log(\pi)}}.$$ As stated by @Mariusz Iwaniuk, in the comments, we have the equivalent representation of $$I\equiv \int_0^1 \frac{\sin(\pi x)\log^2(1-x)}{\pi x^2(1-x)}.$$Another question, presumably simpler, could be $$\int_0^1 \binom{1}{x}\frac{\log(1-x)}{x} \equiv -\int_0^1 \binom{1}{x}\frac{\mathrm{Li}_1(x)}{x}.$$ I believe I can find a closed form for the latter; if I do, I will edit the post. In general, I am curious as to if we may be able to somehow employ Ramanujan's Beta integral or any of the other Beta integrals. Another approach may be the series representation for $\binom{1}{x}$. Thanks!

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    $\begingroup$ How is $\binom{1}{x}$ defined when $x$ is not an integer? $\endgroup$
    – Pacciu
    May 25, 2020 at 1:38
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    $\begingroup$ Good question. We use the relations of the Gamma functions; in our case, we have $\binom{1}{x} \equiv \frac{1}{x!(1-x)!}= \frac{1}{\Gamma(x+1)\Gamma(2-x)}$; see here: en.wikipedia.org/wiki/Gamma_function $\endgroup$
    – user753116
    May 25, 2020 at 1:39
  • $\begingroup$ The value is about 2.66988745495724134279466825882915370633 and is differ from your closed form suggestion. $\endgroup$
    – dust05
    May 25, 2020 at 3:55
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    $\begingroup$ HINT: $\binom{1}{x}=\frac{\sin (\pi x)}{\pi (1-x) x}$ $\endgroup$ May 25, 2020 at 17:21

2 Answers 2

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In order to approximate the solution, I built for $\binom{1}{x}$ the $$\binom{1}{x}\sim \frac{4}{\pi }+\left(\frac{16}{\pi }-2 \pi \right) \left(x-\frac{1}{2}\right)^2+\left(110-\frac{576}{\pi }+24 \pi -\frac{\pi ^2}{6}\right) \left(x-\frac{1}{2}\right)^4+$$ $$\left(-608+\frac{2816}{\pi }-96 \pi +\frac{4 \pi ^2}{3}\right) \left(x-\frac{1}{2}\right)^6+\left(928-\frac{4096}{\pi }+128 \pi -\frac{8 \pi ^2}{3}\right) \left(x-\frac{1}{2}\right)^8$$ which exactly matches the function, first and second derivatives values for $x=0,\frac12,1$.

Now, concerning the integrals $$I_n=\int_0^1 \left(x-\frac{1}{2}\right)^{2n} \,\,\,\frac{\log^2(1-x)}{x}\, dx$$ we have $$I_0=2 \zeta (3)\qquad I_1=\frac{\zeta (3)}{2}-\frac{1}{4}\qquad I_2=\frac{\zeta (3)}{8}-\frac{71}{864}$$ $$I_3=\frac{\zeta (3)}{32}-\frac{10051}{432000}\qquad I_4=\frac{\zeta (3)}{128}-\frac{116069}{18522000}$$

This finally leads to

$$\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\, dx\sim 2 \zeta (3)-\frac{13129439}{18522000}+\frac{4032568}{1157625 \pi }-\frac{94427 \pi }{2315250}-\frac{45551 \pi ^2}{74088000}$$ which is numerically $2.6698874395$ to be compared to the "exact" $2.6698874550$ given by @dust05 in comments.

We could do much better using a few more terms for the approximation.

Edit

Instead of using the built approximation given above, I used the Taylor expansion of $\binom{1}{x}$ (you can get it using Wolfram Alpha; the problem I faced is that the odd terms are not recognized to be $0$ and I do not see how to tansform the digamma terms into simpler expressions).

So, I computed $$I_k=\sum_{n=0}^k c_n \int_0^1 \left(x-\frac{1}{2}\right)^{2n} \,\,\,\frac{\log^2(1-x)}{x}\, dx$$ As a function of $k$, the numerical results are reported below

$$\left( \begin{array}{cc} k & I_k \\ 0 & 3.06101276824936274 \\ 1 & \color{red} {2.6}4320918239538048 \\ 2 & \color{red} {2.6}7090489976321907 \\ 3 & \color{red} {2.6698}6268034682524 \\ 4 & \color{red} {2.669887}87367277585 \\ 5 & \color{red} {2.6698874}4975680354 \\ 6 & \color{red} {2.66988745}500671797 \\ 7 & \color{red} {2.66988745495}686919 \\ 8 & \color{red} {2.66988745495724}361 \\ 9 & \color{red} {2.6698874549572413}3 \\ 10 & \color{red} {2.66988745495724134} \end{array} \right)$$

Taking into account @Mariusz Iwaniuk 's hint, that is to say $$\binom{1}{x}=\frac{\sin (\pi x)}{\pi (1-x) x}$$ the coefficients $c_n$ can write

$$c_n=i\,\frac{ 2^{2 n+1} }{\pi (2 n)!}\left(\Gamma \left(2 n+1,\frac{i \pi }{2}\right)-\Gamma \left(2 n+1,-\frac{i \pi }{2}\right)\right)$$

Concerning the integrals $$J_n=\int_0^1 \left(x-\frac{1}{2}\right)^{2n} \,\,\,\frac{\log^2(1-x)}{x}\, dx$$ they write $$J_n=2^{1-2 n} \zeta (3)-\frac {a_n} {b_n}$$ $$\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & 4 \\ 2 & 71 & 864 \\ 3 & 10051 & 432000 \\ 4 & 116069 & 18522000 \\ 5 & 52752017 & 32006016000 \\ 6 & 145759321889 & 340800058368000 \\ 7 & 329587937534753 & 2994950912937984000 \\ 8 & 42159304836511 & 1497475456468992000 \\ 9 & 844375600417012397 & 117713550682114523136000 \\ 10 & 11769137630214586888219 & 6459177953028988113518592000 \\ 11 & 11930404954629448855339 & 25836711812115952454074368000 \\ 12 & 18359838608628619185581941 & 157177636309007396754361417728000 \\ 13 & 9275258078308733536880688959 & 314355272618014793508722835456000000 \end{array} \right)$$

Computed up to $n=50$, there is very good correlation $(R^2=0.999994)$ $$\log \left(\frac{a_n}{b_n}\right)=\alpha +\beta \,n$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & +0.507370 & 0.029184 & \{+0.448659,+0.566081\} \\ \beta & -1.378775 & 0.000996 & \{-1.380779,-1.376772\} \\ \end{array}$$

Update

Concerning

$$\int_0^1 \binom{1}{x}\frac{\log(1-x)}{x}\,dx$$ I used my initial approximation and ended with $$\frac{8219}{9800}-\frac{42704}{11025 \pi }+\frac{428 \pi }{11025}-\frac{175933 \pi ^2}{1058400} \approx -1.9128812062$$ while numerical integration gives $-1.9128812187$.

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    $\begingroup$ As soon as I read the first couple lines of the answer, I knew who the poster was right away without looking at the username. You have a very distinctive style of answering! $\endgroup$ May 26, 2020 at 16:13
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Using Mariusz Iwaniuk's hint of $\binom{1}{x}=\frac{\sin(\pi x)}{\pi x(1-x)}$, we can rewrite $I$ as $$\int_0^1 \frac{\ln^2(1-x) \sin(\pi x)}{\pi x^2(1-x)} dx$$

We can split the integrand up as $$\int_0^1 \frac{\ln^2(1-x)}{x^2} \frac{\sin(\pi x)}{\pi (1-x)} dx$$

and then write a Taylor series about $x=1$ for $\frac{\sin(\pi x)}{\pi (1-x)}$ as $$\sum_{k=0}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!} (x-1)^{2k}$$

Plugging this into the integral, I get $$\sum_{k=0}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!} \underbrace{\int_0^1 \frac{\ln^2(1-x)}{x^2} (x-1)^{2k} dx}_{I_k}$$

Other than $I_0 = \frac{\pi^2}{3}$, Mathematica says that $I_k = 2 (\psi^{(1)}(2k) + k\psi^{(2)}(2k)) = \frac{\pi^2}{3} - 2 H_{2k-1}^{(2)}+4k H_{2k-1}^{(3)} - 4k \zeta(3)$

This therefore means that $$I = \frac{\pi^2}{3} + \sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}\left(\frac{\pi^2}{3} - 2 H_{2k-1}^{(2)}+4k H_{2k-1}^{(3)} - 4k \zeta(3)\right)$$

Splitting this up into four distinct sums: $$\frac{\pi^2}{3} + \frac{\pi^2}{3}\underbrace{\sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}}_{S_1} - 2 \underbrace{\sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}H_{2k-1}^{(2)}}_{S_2}+4\underbrace{\sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}k H_{2k-1}^{(3)}}_{S_3} - 4\zeta(3)\underbrace{\sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}k}_{S_4} $$

It is easy to see that $S_1 = -1$, by utilizing the series for $\sin(x)$. Also $S_4 = -\frac{1}{2}$, so it reduces to $-2S_2 + 4S_3 + 2\zeta(3)$. This can be made into $2\zeta(3)$ plus a power series of $\pi$, but I was not able to simplify $S_2$ or $S_3$.

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