# Find the closed form of $_3F_2(\frac{1}{4},\frac{3}{4},\frac{5}{4};\frac{3}{2},\frac{7}{4};1)$

Context

Some investigation suggests that the following identity is true: \begin{align} _3F_2(\frac{1}{4},\frac{3}{4},\frac{5}{4};\frac{3}{2},\frac{7}{4};1)=\frac{3\sqrt{2}\sqrt{\pi}\left(2\log({1+\sqrt{2}})+\pi-2\sqrt{2} \right)}{\Gamma(\frac{1}{4})^2}\tag{1} \end{align} In this question (Prove $\,_3F_2\left ( \frac12,\frac12,1;\frac34,\frac34;-\frac18 \right ) =\frac23+\frac{\Gamma\left ( \frac34 \right )^2}{3\sqrt{\pi} }$) there is an answer proving a more difficult identity with the WZ method.

Question

Do you think that we can prove $$(1)$$ with the mentioned method? Thanks for your cooperation.

Let $$\mathcal{H}$$ denote the value of the following generalized hypergeometric function at unity:

$$\mathcal{H}:={_3F_2}{\left(\frac14,\frac34,\frac54;\frac32,\frac74;1\right)}.$$

Using Euler's integration formula for higher-order hypergeometric functions, we can express $$\mathcal{H}$$ as

\begin{align} \mathcal{H} &={_3F_2}{\left(\frac14,\frac34,\frac54;\frac32,\frac74;1\right)}\\ &=\frac{1}{\operatorname{B}{\left(\frac54,\frac12\right)}}\int_{0}^{1}\mathrm{d}t\,t^{1/4}\left(1-t\right)^{-1/2}\,{_2F_1}{\left(\frac14,\frac34;\frac32;t\right)}.\\ \end{align}

It can be shown that the Gauss hypergeometric function obeys the following quadratic transformation:

$${_2F_1}{\left(a,b;2b;z\right)}=\left(1-\frac{z}{2}\right)^{-a}{_2F_1}{\left(\frac{a}{2},\frac{a+1}{2};b+\frac12;\left(\frac{z}{2-z}\right)^{2}\right)};~~~\small{0

In particular, for $$a=\frac12\land b=1$$ we have the following identity:

$${_2F_1}{\left(\frac14,\frac34;\frac32;\left(\frac{z}{2-z}\right)^{2}\right)}=\left(1-\frac{z}{2}\right)^{1/2}\,{_2F_1}{\left(\frac12,1;2;z\right)};~~~\small{z\le1}.$$

Using Euler's integration formula for the Gauss hypergeometric function, we obtain the following elementary closed form expression for $${_2F_1}{\left(\frac12,1;2;z\right)}$$ for $$z\le1$$:

\begin{align} {_2F_1}{\left(\frac12,1;2;z\right)} &=\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{1-zt}}\\ &=\frac{2\left(1-\sqrt{1-z}\right)}{z}\\ &=\frac{2}{1+\sqrt{1-z}}.\\ \end{align}

We then find

$${_2F_1}{\left(\frac14,\frac34;\frac32;\left(\frac{z}{2-z}\right)^{2}\right)}=\frac{2\sqrt{1-\frac{z}{2}}}{1+\sqrt{1-z}};~~~\small{z\le1},$$

$$z=1-y\implies {_2F_1}{\left(\frac14,\frac34;\frac32;\left(\frac{1-y}{1+y}\right)^{2}\right)}=\frac{2\sqrt{\frac{1+y}{2}}}{1+\sqrt{y}};~~~\small{0\le y},$$

$$y=\frac{1-x}{1+x}\implies {_2F_1}{\left(\frac14,\frac34;\frac32;x^{2}\right)}=\frac{2\sqrt{\frac{1}{1+x}}}{1+\sqrt{\frac{1-x}{1+x}}};~~~\small{-1

$$\implies {_2F_1}{\left(\frac14,\frac34;\frac32;x^{2}\right)}=\frac{2}{\sqrt{1+x}+\sqrt{1-x}};~~~\small{-1

$$\implies {_2F_1}{\left(\frac14,\frac34;\frac32;x^{2}\right)}=\frac{2}{\sqrt{\left(\sqrt{1+x}+\sqrt{1-x}\right)^{2}}};~~~\small{-1

$$\implies {_2F_1}{\left(\frac14,\frac34;\frac32;x^{2}\right)}=\frac{\sqrt{2}}{\sqrt{1+\sqrt{1-x^{2}}}};~~~\small{-1

$$x=\sqrt{t}\implies {_2F_1}{\left(\frac14,\frac34;\frac32;t\right)}=\frac{\sqrt{2}}{\sqrt{1+\sqrt{1-t}}};~~~\small{0\le t\le1}.$$

Then,

\begin{align} \mathcal{H} &={_3F_2}{\left(\frac14,\frac34,\frac54;\frac32,\frac74;1\right)}\\ &=\frac{1}{\operatorname{B}{\left(\frac54,\frac12\right)}}\int_{0}^{1}\mathrm{d}t\,t^{1/4}\left(1-t\right)^{-1/2}\,{_2F_1}{\left(\frac14,\frac34;\frac32;t\right)}\\ &=\frac{3}{\operatorname{B}{\left(\frac14,\frac12\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt[4]{t}}{\sqrt{1-t}}\cdot\frac{\sqrt{2}}{\sqrt{1+\sqrt{1-t}}}\\ &=\frac{6\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt[4]{t}}{\sqrt{1-t}\sqrt{1+\sqrt{1-t}}}\\ &=\frac{6\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\int_{0}^{1}\mathrm{d}u\,\frac{\sqrt[4]{1-u}}{\sqrt{u}\sqrt{1+\sqrt{u}}};~~~\small{\left[1-t=u\right]}\\ &=\frac{12\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\int_{0}^{1}\mathrm{d}v\,\frac{\sqrt[4]{1-v^{2}}}{\sqrt{1+v}};~~~\small{\left[\sqrt{u}=v\right]}\\ &=\frac{12\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\int_{0}^{1}\mathrm{d}v\,\sqrt[4]{\frac{1-v}{1+v}}\\ &=\frac{24\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\int_{0}^{1}\mathrm{d}w\,\frac{\sqrt[4]{w}}{(w+1)^{2}};~~~\small{\left[\frac{1-v}{1+v}=w\right]}\\ &=\frac{24\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\int_{0}^{1}\mathrm{d}x\,\frac{4x^{4}}{(x^{4}+1)^{2}};~~~\small{\left[\sqrt[4]{w}=x\right]}\\ &=\frac{24\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\int_{0}^{1}\mathrm{d}x\,\left[\frac{4}{x^{4}+1}-\frac{4}{(x^{4}+1)^{2}}\right].\\ \end{align}

Consider the following derivative:

$$\frac{d}{dx}\left[\frac{x}{x^{4}+1}\right]=\frac{1-3x^{4}}{(x^{4}+1)^{2}}=\frac{4}{(x^{4}+1)^{2}}-\frac{3}{x^{4}+1}.$$

Integrating both sides, we obtain

$$\int_{0}^{1}\mathrm{d}x\,\frac{4}{(x^{4}+1)^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{3}{x^{4}+1}=\int_{0}^{1}\mathrm{d}x\,\frac{d}{dx}\left[\frac{x}{x^{4}+1}\right]=\frac12,$$

$$\implies\int_{0}^{1}\mathrm{d}x\,\frac{4}{(x^{4}+1)^{2}}=\frac12+\int_{0}^{1}\mathrm{d}x\,\frac{3}{x^{4}+1},$$

$$\implies\int_{0}^{1}\mathrm{d}x\,\frac{4}{x^{4}+1}-\int_{0}^{1}\mathrm{d}x\,\frac{4}{(x^{4}+1)^{2}}=-\frac12+\int_{0}^{1}\mathrm{d}x\,\frac{1}{x^{4}+1}.$$

Hence, $$\mathcal{H}$$ can be reduced to a single elementary integral, leading to the conjectured result:

\begin{align} \mathcal{H} &=\frac{24\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\int_{0}^{1}\mathrm{d}x\,\left[\frac{4}{x^{4}+1}-\frac{4}{(x^{4}+1)^{2}}\right]\\ &=\frac{24\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\left[\int_{0}^{1}\mathrm{d}x\,\frac{4}{x^{4}+1}-\int_{0}^{1}\mathrm{d}x\,\frac{4}{(x^{4}+1)^{2}}\right]\\ &=\frac{24\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\left[-\frac12+\int_{0}^{1}\mathrm{d}x\,\frac{1}{x^{4}+1}\right]\\ &=\frac{24\sqrt{\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\left[-\frac12+\frac{2\ln{\left(1+\sqrt{2}\right)}+\pi}{4\sqrt{2}}\right]\\ &=\frac{3\sqrt{2\pi}}{\left[\Gamma{\left(\frac14\right)}\right]^{2}}\left[-2\sqrt{2}+2\ln{\left(1+\sqrt{2}\right)}+\pi\right].\blacksquare\\ \end{align}

• Beautiful, congrats
– Zima
Apr 3 at 11:51
• C'est prodigieux. Apr 3 at 16:38
• Thanks David H. I have found another identity: math.stackexchange.com/questions/4892574/… Best.
– User
Apr 3 at 18:06

In the book on special functions by Askey, Roy and Gasper, you will find the Thomae formula about $$_3F_2(a,b,c;d,e;1).$$ I have checked and your $$abcde$$ leads through this formula to $$K\times \ _3F_2(A,B,C;D,E;1)$$ where $$K$$ is a product of gamma functions. Miraculously with your $$abcde$$ we get $$A=E=-1/4$$, landing on a quite computable $$K \times \ _2F_1(B,C;D;1).$$