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In this question I asked about a closed form for a certain harmonic sum. That sum was reduced to an expression involving two nearly-poised hypergeometric functions $\mathcal{H}_1$ and $\mathcal{H}_2$. The second one, $\mathcal{H}_2$, is given by

$$\mathcal{H}_2 = {}_3F_2 \left(\frac14,\frac12,\frac12;\,\frac32,\frac32;2(\sqrt{2}-1)\right) = \frac{1}{\Gamma \left(\frac{1}{4}\right)}\sum_{k=0}^\infty \frac{\Gamma \left(k+\frac{1}{4}\right) 2^k (\sqrt{2}-1)^k}{k! (2k+1)^2}$$ In this question I wanted to ask about a polylogarithmic closed form for $\mathcal{H}_2$ which could be expressed as the following integral

$$ \mathcal{H}_2 = -\int_0^1 \ln x\, \left(1-2(\sqrt{2}-1)x^2\right)^{-1/4} \mathrm{d}x = 1.02980730882955079390328611231 \cdots $$

Might such a closed form exist? I could find some papers on nearly-poised series but nothing for the non-unit value of the argument

Nikitan
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    We have a whole industry of similar hypergeometric expressions. You have a question. Do you have a reason for the existence of a polylogarithmic + closed form for this value of an ${}_3F_2$-hypergeometric expression computed at the place $2(\sqrt 2-1)$? Please provide more context for the question. Which is the motivation for the question? – dan_fulea Mar 02 '25 at 19:37
  • @dan_fulea, it's not a very good reason but this hypergeometric function is present in the evaluation of this unit square integral. Many such integral have "nice" closed forms (I'm not sure if I'm using the term closed form correctly), so I wanted to make a vain guess about this particular one as well. I suppose by a "polylogaritmic closed form" I informally mean an expression that contains no special functions other than polylogarithmic values. – Nikitan Mar 02 '25 at 19:56
  • I still allow highly speculative questions, this is not my problem. The problem is the question. It is a yes-or-no-question. In case somebody can find (after a very hard word) a formula for $\mathcal H_2$ involving only polylogarithm values at algebraic integers (say), then yes, we have a wonderful situation. But my comment is about the question, which must come with a suitable context. In the above linked unit square integral not few answers are searching for any formula that should somehow make things explicit. After the effort we have not explicit insight. You give two formulas... – dan_fulea Mar 03 '25 at 00:24
  • ... for $\mathcal H_2$. They are completely different in nature. Can you show how one formula follows from the other one? Which formula is better supposed to lead to polylog-values? Any similar formula for a nearly-poised explicit value at an algebraic integer like $2(\sqrt 2-1)$ of norm $4$ or $1/4$? Which quadratic formula may be involved? There are a lot of questions, any one with a good argument? Examples of how to add sensitive context: https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question - Among them, "include your work" should best accompany this question. – dan_fulea Mar 03 '25 at 00:30
  • @dan_fulea, to derive the integral I used a rather common formula $$\frac{1}{(2k+1)^2} = - \int_0^1 \ln(x)x^{2k} \mathrm{d}x$$ And as for other questions, I just wanted to know if someone might know a trick to solve an integral like that :)

    See Efim Mazhnik's answer to my previous question for a similar ${}_2F_1$ evaluation

     – Nikitan Mar 03 '25 at 04:35

1 Answers1

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Define $$\mathcal J(s):=\int_0^1x^{s-1}\big(1-2(\sqrt 2-1)x^2\big)^{-1/4}dx,\space s>0.\\\implies I=\mathcal J'(1)=\int_0^1x^{s-1}\log (x)\big(1-2(\sqrt2-1)x^2\big)^{-1/4}dx\space\space\space\space \textbf{(I)}$$

$$\mathcal J(s)\stackrel{t=x^2}{=}\frac12\int_0^1 t^{s/2-1}\big(1-2(\sqrt2-1)t\big)^{-1/4}dt\\\stackrel{u=2(\sqrt2-1)t}{=}\frac{1}{2^{s/2+1}(\sqrt2-1)^{s/2}}\int_0^{2(\sqrt2-1)}u^{s/2-1}(1-u)^{-1/4}du$$

Notice that the integral is of the form

$$\int_0^a v^{\alpha-1}(1-v)^{\beta-1}dv =\frac{a^\alpha}{\alpha}\;_2F_1\Bigl(\alpha,1-\beta;\alpha+1;a\Bigr),\quad |a|<1,$$

thus ,

$$\mathcal J(s)=\frac{1}{2^{\frac{s}{2}+1}(\sqrt2-1)^{\frac{s}{2}}}\cdot \frac{\Bigl[2(\sqrt2-1)\Bigr]^{\frac{s}{2}}}{\frac{s}{2}}\; _2F_1\Bigl(\frac{s}{2},\tfrac{1}{4};\frac{s}{2}+1;2(\sqrt2-1)\Bigr)$$

$$=\frac{1}{s}\;_2F_1\Bigl(\frac{s}{2},\tfrac{1}{4};\frac{s}{2}+1;2(\sqrt2-1)\Bigr).$$

And, $$\mathcal J'(s)=-\frac{1}{s^2}F(s)+\frac1sF'(s)\stackrel{\textbf{(I)}}{\implies} I=-F(1)+F'(1)$$

With $F(s)=\,_2F_1\Bigl(\frac{s}{2},\tfrac{1}{4};\frac{s}{2}+1;2(\sqrt{2}-1)\Bigr).$

Antony Theo.
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    +1 Really interesting! I do wonder how tricky finding a closed form for $F(s)$ would be. These formulas seem quite relevant. – Nikitan Feb 26 '25 at 21:00
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    The correct identity, derived previously by Efim Mazhnik, is $_2F_1\left(\frac14,\frac12;\frac32;2(\sqrt2-1)\right)=\frac1{\sqrt{\sqrt2-1}} \left(\frac{\pi^{3/2}}{\Gamma(1/4)^2}+1-\frac1{\sqrt2}\right).$ Computing $-F(1)+F'(1)$ directly results in cancellations that give the tautology $-I={}_3F_2\left(\frac14,\frac12,\frac12;\frac32,\frac32;2(\sqrt2-1)\right)$, but using a combination of identities like the ones (1/2) – teadawg1337 Feb 28 '25 at 03:37
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    here and here may be fruitful. For instance, I derived $F(s)=(2\sqrt2-2)^{-s/2}\left(\frac{\Gamma\left(\frac34\right)\Gamma\left(\frac{s}2+1\right)}{\Gamma\left(\frac{s}2+\frac34\right)}-\frac{2s}3(\sqrt2-1)^{3/2}{}2F_1\left(1-\frac{s}2,\frac34;\frac74;3-2\sqrt2\right)\right),$ which gives a simpler derivative evaluation but requires computing $\sum{n=0}^\infty\frac{\Gamma(n+1/2)\psi(n+1/2)(3-2\sqrt2)^n}{(4n+3)n!}$. (2/2) – teadawg1337 Feb 28 '25 at 03:37
  • The question asks about a polylogarithmic closed form for $\mathcal H_2$. Ist this an answer to the question? A partial one (and in this case which is the expected path to continue so that in the end there are still polylogarithms values explicitly to be seen)? – dan_fulea Mar 02 '25 at 19:35
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    @dan_fulea, I have trouble understanding what you're saying. This is certainly an interesting idea but it still requires computing the digamma sum teadawg gives. If you have any ideas on how that sum could be evaluated, please let me know! – Nikitan Mar 03 '25 at 06:24