# On $\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\frac{n}{6^6}\,x\right)\,dx$

Reshetnikov gave the remarkable evaluation,

\begin{align} I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\,dx \\ &=\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha^5-\alpha^4+1\right)\arctan\frac1\alpha-6\ln\left(1+\alpha^2\right)\right)\\ &=0.7857194\dots \end{align} where $\alpha$ is a quartic root. However, it seems this can be simplified a bit and generalized.

I. $p=5$: Given,

$$I(n)=\int_0^1\arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{n}{4^4}\,x\right)\,\mathrm dx$$ Is it true that, in general, $$\frac{12n}{5^5}\, I(n) = -12\Big(1\color{blue}-\frac{n}{5^5}\Big)\arctan\frac1\alpha+6\ln\Big(\frac{2}{1+\alpha^2}\Big)+3\pi+P(\alpha)$$ where $P(\alpha)=(1-\alpha)^3(5+3\alpha)$ and $\alpha$ is the largest positive root of the quintic, $$\alpha^5-\alpha^4+\frac{n}{5^5}=0$$ provided real number $n<4^4\,$? (Note: By sheer coincidence, the choice of $n=4$ in the other post caused the quintic to factor.)

II. $p=7$: Given,

$$J(n)=\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\,\frac{n}{6^6}\,x\right)\,\mathrm dx$$ is it true that, $$\frac{60n}{7^7}\, J(n) = 60\Big(1\color{blue}+\frac{n}{7^7}\Big)\arctan\frac1\alpha-30\ln\Big(\frac{2}{1+\alpha^2}\Big)-15\pi-P(\beta)$$ where $P(\beta) = (1-\beta)^2(27-6\beta-9\beta^2+8\beta^3+10\beta^4)$ and $\beta$ is the largest positive root of, $$\beta^7-\beta^6+\frac{n}{7^7}=0$$ provided real $n<6^6\,$?

Questions:

1. How do we prove the two conjectured equalities?
2. What's the formula for $p=3$? (Mathematica takes too long to evaluate the integral that I couldn't use an integer relations subroutine on it.)
• But Reshetnikov didn't show a complete solution. Just the final answer. – Zaid Alyafeai Jan 3 '17 at 5:56
• @ZaidAlyafeai: Yes, but it's a great final answer. He used an $\alpha$ that turns out to be the quartic factor of the quintic in this post. How he came across the correct root and the correct form is remarkable. I just found the $p=7$ by analogy. – Tito Piezas III Jan 3 '17 at 6:04
• It looks like one can solve the general problem here for an arbitrary $k$. If $$I_k(n) = \int_{0}^{1}\arctan\left._k\right.F_{k-1}\left(\frac{1}{k+1},\frac{2}{k+1}\cdots,\cdots,\frac{k}{k+1},\frac{k+1}{k}; \frac{nx}{k^k}\right){\rm d}x$$ then we should have something like $$I_k(n) = \arctan\left(\frac{1}{z_0}\right) + \frac{(k+1)^{k+1}}{n}\int_{z_0}^1\frac{w^{k+1}-w^k}{1+w^2}{\rm d}w$$ where $z_0$ is a solution to $z_0^{k+1} - z_0^k + \frac{n}{(k+1)^{k+1}} = 0$. – Winther Jan 3 '17 at 22:38

I think you should start by the following general form

$${}_{k}F_{k-1}\left(\frac{1}{k+1} ,\cdots ,\frac{k}{k+1};\frac{2}{k} \cdots ,\frac{k-1}{k},\frac{k+1}{k};\left( \frac{m(1-m^k)}{f_k}\right)^k \right) = \frac{1}{1-m^k}$$

Where

$$f_k \equiv \frac{k}{(1+k)^{1+1/k}}$$

Put $k=4$

$${}_{4}F_{3}\left(\frac{1}{5} ,\cdots ,\frac{4}{5};\frac{2}{4} \cdots ,\frac{3}{4},\frac{5}{4};\left( \frac{m(1-m^4)}{f_4}\right)^4 \right) = \frac{1}{1-m^4}$$

The argument simplifies to

$$\left(\frac{m(1-m^4)}{\frac{4}{5^{1+1/4}}}\right)^4 = \frac{5^5 \,m^4(1-m^4)^4}{4^4}$$

Hence we have

$${}_{4}F_{3}\left(\frac{1}{5} ,\cdots ,\frac{4}{5};\frac{2}{4} \cdots ,\frac{3}{4},\frac{5}{4};\frac{5^5 (1-m)m^4}{4^4} \right) = \frac{1}{m}$$

Now suppose we want to find

$$\int \arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{n}{4^4}\,x\right)\,\mathrm dx$$

Use $nx = 5^5m^4(1-m)$

$$\frac{5^5}{n}\int (4m^3(1-m)-m^ 4)\arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{5^5 (1-m)m^4}{4^4}\right)\,\mathrm dm$$

This simplifies to

$$\frac{5^5}{n}\int (4m^3(1-m)-m^ 4)\arctan\,\frac{1}{m}\,\mathrm dm$$

The anti-derivative of this is elementary.

Note that when we use substittution we will need the roots of

$$m^5-m^4+\frac{n}{5^5}= 0$$

As conjectured by the OP.

Using the value $n=4$ we can verify Reshetnikov result

\begin{align}\int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\,dx &= \frac{5^5}{4}\int^{\alpha}_1 (4x^3(1-x)-x^ 4)\arctan \left(\frac{1}{x} \right)\,\mathrm dx\\&=0.7857194\dots \end{align}

Where $\alpha$ is the largest positive root of

$$m^5-m^4+\frac{4}{5^5}= 0$$

• Ok, but perhaps you should change your variable $x$ to the more generic $m$, as it will confuse/conflate the use of $F(x)dx$ in the post. – Tito Piezas III Jan 3 '17 at 6:51
• @TitoPiezasIII, yes sure. – Zaid Alyafeai Jan 3 '17 at 6:56
• Ah, beautiful! Mathematica can integrate $$\int^{\alpha}_1 (4x^3(1-x)-x^ 4)\arctan\left(\frac{1}{x} \right)\,\mathrm dx$$ for generic $\alpha$. I can now see why using $\ln(1+\alpha^2)$ and the other expressions make sense. – Tito Piezas III Jan 3 '17 at 12:29
• do you have a proof for your first eqn.? – tired Jan 3 '17 at 13:24
• @tired, I never worked with the generalized hyper geometric function. I will try to provide a proof for k=2. – Zaid Alyafeai Jan 3 '17 at 18:36

The result does generalize. We will derive a closed formula for the integral $$I_k(n) = \int_0^1\arctan\left[_{k}F_{k-1}\left(\frac{1}{k+1},\cdots, \frac{k}{k+1},\frac{2}{k},\cdots,\frac{k+1}{k}; \frac{nx}{k^k}\right)\right]\,{\rm d}x$$ valid for $|n| \leq k^k$.

For this particular combination of arguments the hypergeometrical function simplifies greatly, see property $(25)$ here$^{(1)}$. Using this we can perform a substitution $\frac{nx}{(k+1)^{k+1}} = y^k(1-y^k)^k$ and the integrand simplifies to $\frac{(k+1)^{k+1}}{n}\frac{1}{1-y^k}\frac{d}{dy}\left(y^k(1-y^k)^k\right)$. Now performing integration by parts and the substitutions $y^k\to y \to 1-y$ we arrive at

$$I_k(n) = \frac{\pi}{2} - \arctan(z_0) + \frac{(k+1)^{k+1}}{n}\int_{z_0}^1\frac{w^{k+1}-w^k}{1+w^2}{\rm d}w$$

where $z_0$ is the largest positive root of $z_0^{k+1} - z_0^k + \frac{n}{(k+1)^{k+1}} = 0$. The integral above can be evaluated in closed form using

$$\matrix{\int \frac{w^{2k}}{1+w^2}{\rm d}w &=& (-1)^k\left[\arctan(w) + F_k(w)\right] + C\\\int \frac{w^{2k+1}}{1+w^2}{\rm d}w &=& (-1)^k\left[\frac{1}{2}\log(1+w^2) + G_k(w)\right] + C}$$ where $F_k(x) = \sum_{i=1}^{k}\frac{(-1)^i x^{2i-1}}{2i-1}$ and $G_k(x) = \sum_{i=1}^{k}\frac{(-1)^i x^{2i}}{2i}$. These relations are easily proven by differentation and summing a geometrical series. This leads to

$$I_{2k}(n) = \frac{\pi}{2} - \arctan(z_0) + \frac{(-1)^k(2k+1)^{2k+1}}{n}\left[\frac{1}{2}\log\left(\frac{2}{1+z_0^2}\right) - \frac{\pi}{4} + \arctan(z_0)\right] + \frac{(-1)^k(2k+1)^{2k+1}}{n}\left[G_k(1) - G_k(z_0) - F_k(1) + F_k(z_0)\right]$$ $$I_{2k+1}(n) = \frac{\pi}{2} - \arctan(z_0) + \frac{(-1)^{k}(2k+2)^{2k+2}}{n}\left[-\frac{1}{2}\log\left(\frac{2}{1+z_0^2}\right) - \frac{\pi}{4} + \arctan(z_0)\right] + \frac{(-1)^{k}(2k+2)^{2k+2}}{n}\left[- G_k(1) + G_k(z_0) -F_{k+1}(1) + F_{k+1}(z_0)\right]$$

For the particular case $k = 2$ ($p=3$ in the question) you asked for this reduces to

$$I_2(1) = \frac{27}{2}\log \left(\frac{z_0^2+1}{2}\right) - \frac{27}{2}(z_0-1)^2 - 28\arctan(z_0) + \frac{29 \pi}{4} \simeq 0.795296 \\ \text{where } z_0 = \frac{1}{3} \left(1+2 \cos \left(\frac{\pi }{9}\right)\right)$$

Finlly since the expression has many terms we should check that it gives the correct result. Here is some code used to check this in Mathematica:

(* Working precision *)
ndigits = 100;

Block[{$MinPrecision = ndigits}, (* Pick k and n *) k = 4; n = 1; (* Solve for z0 *) z0 = Max[N[x /. Solve[{x^(k + 1) - x^k + n/(k + 1)^(k + 1) == 0, x > 0}, x], ndigits]]; (* Define the integral *) ak = Table[i/(k + 1), {i, 1, k}]; bk = Table[If[i == k, (i + 1)/k, i/k], {i, 2, k}]; exact = NIntegrate[ArcTan[HypergeometricPFQ[ak, bk, (x n)/k^k]], {x, 0, 1}, WorkingPrecision -> ndigits]; (* Our solution *) result = If[Mod[k, 2] == 0, (* For even k *) \[Pi]/2 - ArcTan[z0] + ((-1)^(k/2) (k + 1)^(k + 1))/ n (1/2 Log[2/(1 + z0^2)] - \[Pi]/4 + ArcTan[z0] + Sum[(-1)^i ((1 - z0^(2 i))/(2 i) - (1 - z0^(2 i - 1))/(2 i - 1)), {i, 1, k/2}]) , (* For odd k *) \[Pi]/2 - ArcTan[z0] + ((-1)^((k - 1)/2) (k + 1)^(k + 1))/n (-(1/2) Log[2/(1 + z0^2)] - \[Pi]/4 + ArcTan[z0] - Sum[(-1)^i ((1 - z0^(2 i))/(2 i)If[i == (k - 1)/2 + 1, 0, 1] + (1 - z0^(2 i - 1))/(2 i - 1)), {i, 1, (k - 1)/2 + 1}]) ]; (* Difference between integral and our formula *) Print["Difference: ", N[result - exact]]; ];  This demonstrates that we get the same result (to numerical precision) for a wide range of$k$and$n$.$^{(1)}$It's easy to check that this holds to arbitrary precisison (checked to a few hundred digits) at least for the smallest values of$k$. Marty Cohen found this reference discussing a very similar type of property which might be useful if anyone wants to try to derive it (which likely is a very messy affair). • +1 Thanks for this highly detailed explanation. If I could accept two answers, I would do so. :) – Tito Piezas III Jan 4 '17 at 5:07 • Unfortunately, the only justification for that amazing equation (25) is "(M. L. Glasser, pers. comm., Sept. 26, 2003)". So, we do not have a proof, just as assertion. – marty cohen Jan 4 '17 at 5:52 • @martycohen, Interesting. Maybe I will post a question about it later. – Zaid Alyafeai Jan 4 '17 at 5:55 • Doing a Google search for "M. L. Glasser hypergeometric functions" results in a number of interesting hits including this one: researchgate.net/profile/Larry_Glasser/publication/… – marty cohen Jan 4 '17 at 6:07 • It's more complicated than I though (and far beyond what I'm interested in doing) to derive the formula. To show it one needs to establish $$\frac{1}{1-x^k} = \sum_{n=0}^\infty \frac{\left(\frac{1}{k+1}\right)_n\cdots \left(\frac{k}{k+1}\right)_n}{\left(\frac{2}{k}\right)_n\cdots \left(\frac{k-1}{k}\right)_n\left(\frac{k+1}{k}\right)_n}\frac{x^{nk}(1-x^k)^{nk}(k+1)^{n(k+1)}}{k^{nk} n!}$$ where$(a)_n\$ is the falling factorial. – Winther Jan 4 '17 at 6:37