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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 hypergeometric functions. I suspect these functions have elementary closed forms.

Here I wanted to ask if there is such a closed form for the following hypergeometric function

$$\mathcal{H}_0 = {}_3F_2 \left(\frac34,\frac34,\frac54;\,\frac74,\frac74;\,\frac12\right) =1.152557194266486\cdots$$

Turns out there is a nice integral representation for $\mathcal{H}_0$

$$\mathcal{H}_0 = -9\cdot 2^{-11/4} \int_0^1\left(1-x\right)^{-1/4}\left(1+x\right)^{-5/4}\log(1-x)\mathrm{d}x $$

In this answer achille hui tackles a similar hypergeometric function with a very clever elliptical substitution. Might such a method work for this function as well?

I should note that there are alternative integral representations for $\mathcal{H}_0$. One of them is

$$\mathcal{H}_0 = -9 \int_0^1 x^2 \ln x \left(1 - \frac{x^4}{2} \right)^{-5/4} \mathrm{d}x$$

Nikitan
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    Why did you remove the info about $\mathcal{J}(s)$? Did you determine that approach is a red herring? – teadawg1337 Mar 01 '25 at 16:12
  • @teadawg1337, I did think so, since if we were to find $\mathcal{J}(s)$, we would be able to find closed forms for all integrals of the form $$\int_0^1 x^2 \ln^n(x) \left(1 - \frac{x^4}{2} \right)^{-5/4} \mathrm{d}x$$ which seems unlikely... If you think it's promising I'll put it back. – Nikitan Mar 01 '25 at 16:40
  • I'm unsure... I found $\mathcal{J}(s)=2^{s+2}{}_2F_1\left(s-\frac14,-\frac14;\frac34;-1\right)+\frac{2^s\Gamma(s)\Gamma\left(-\frac14\right)}{\Gamma\left(s-\frac14\right)}$ with my method here, and this identity gives $_2F_1\left(-\frac14,\frac12;\frac34;-1\right)=\frac1{\sqrt2}+\frac{\pi^{3/2}}{\Gamma(1/4)^2}$, but evaluating $-9/16\mathcal{J}'(3/4)$ still seems tricky. – teadawg1337 Mar 01 '25 at 16:52

1 Answers1

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I'll start by saying that the solution isn't graphically nice to look at, but on the other hand the integral is quite complicated anyway, anyway respect the fact that you asked for a closed form of the integral (incidentally, even without the presence of the logarithm it would have been necessary to use hypergeometric functions).

Using Feynman trick, let $$I(x)=\int_0^1\frac{t^{2+x}}{\left(1+\frac{t^4}{2}\right)^{5/4}}\mathrm{d}t\qquad\Rightarrow\qquad I'(0)=\int_0^1\frac{t^2\ln(t)}{\left(1+\frac{t^4}{2}\right)^{5/4}}\mathrm{d}t$$ $I(x)$ has a closed form $$I(x)=\frac{1}{x+3}{}_2F_1\left(\left.{\frac{5}{4},\frac{x+3}{4}\atop\frac{7+x}{4}}\right|-\frac{1}{2}\right)$$

Where ${}_2F_1\left(\left.{a,b\atop c}\right|z\right)$ is the hypergeometric function

Deriving $I(x)$ and calculating it at $x=0$ you get \begin{align*} \int_0^1\frac{t^2\ln(t)}{\left(1+\frac{t^4}{2}\right)^{5/4}}\mathrm{d}t=&\frac{1}{12}{}_2F_1^{\left({0,0\atop 1},0\right)}\left(\left.{\frac{5}{4},\frac{3}{4}\atop\frac{7}{4}}\right|-\frac{1}{2}\right)+\frac{1}{12}{}_2F_1^{\left({0,1\atop 0},0\right)}\left(\left.{\frac{5}{4},\frac{3}{4}\atop\frac{7}{4}}\right|-\frac{1}{2}\right)\\ -&\frac{1}{9}{}_2F_1\left(\left.{\frac{3}{4},\frac{5}{4}\atop\frac{7}{4}}\right|-\frac{1}{2}\right) \end{align*}

I used a non-standard notation of derivatives with multi-indexes, however the meaning is

$${}_2F_1^{\left({0,0\atop 1},0\right)}\left(\left.{\frac{5}{4},\frac{3}{4}\atop\frac{7}{4}}\right|-\frac{1}{2}\right):=\left.\frac{\partial}{\partial c}{}_2F_1\left(\left.{\frac{5}{4},\frac{3}{4}\atop c}\right|-\frac{1}{2}\right)\right|_{c=\frac{7}{4}}$$

and

$${}_2F_1^{\left({0,1\atop 0},0\right)}\left(\left.{\frac{5}{4},\frac{3}{4}\atop\frac{7}{4}}\right|-\frac{1}{2}\right):=\left.\frac{\partial}{\partial b}{}_2F_1^{\left({0,1\atop 0},0\right)}\left(\left.{\frac{5}{4},b\atop\frac{7}{4}}\right|-\frac{1}{2}\right)\right|_{b=\frac{3}{4}}$$

For brevity you can define

$$\psi(x,y):={}_2F_1\left(\left.{\frac{5}{4},x\atop y}\right|-\frac{1}{2}\right)$$

so you can write

$$\int_0^1\frac{t^2\ln(t)}{\left(1+\frac{t^4}{2}\right)^{5/4}}\mathrm{d}t=\frac{1}{12}\psi^{(1,0)}\left(\frac{3}{4},\frac{7}{4}\right)+\frac{1}{12}\psi^{(0,1)}\left(\frac{3}{4},\frac{7}{4}\right)-\frac{1}{9}\psi\left(\frac{3}{4},\frac{7}{4}\right)$$

Math Attack
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    After toiling with the $\wp$-function for a little bit, I get that the value of the integral without the logarithm is

    $$\int_0^1 x^2 \left(1-\frac{x^4}{2}\right)^{-5/4} \mathrm{d}x = 2^{1/4}\cdot\left(1-\frac{\sqrt{\pi}\cdot\Gamma\left( \frac34 \right)}{\Gamma\left( \frac14 \right)} \right) = 0.476789\cdots$$

    I wonder if something similar could be done with the integral in question.

     – Nikitan Mar 04 '25 at 18:44