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Let $x_0,x \in (0,1/10)$ and define: \begin{equation} g(x):= F_{2,1}\left[\frac{1}{13},\frac{1}{17},\frac{1}{5}; 100 x^2 \right] \end{equation} Then the following identity holds true: \begin{eqnarray} &&\int\limits_{x_0}^x \frac{\left(2404500 \xi ^3+13746115 \xi ^2-47073 \xi -161109\right) }{(1-10 \xi )^{5513/7140} \xi ^{116/35} (10 \xi +1)^{5683/7140}} g(\xi )d\xi =\\ && \left.\frac{(1-10 x)^{1627/7140} (10 x+1)^{1457/7140} }{x^{116/35}} \left(-97461 (10 x-1) (10 x+1) x^2 g'(x)-1785 \left(840 x^2-13 x-39\right) x g(x)\right) \right|_{x_0}^x \end{eqnarray}

In[815]:= x =.; x0 =.; xi =.; Clear[g]; Clear[F]; Clear[f]; Clear[f1];
g[x_] := Hypergeometric2F1[1/13, 1/17, 1/5, 100 x^2];
F[x_] := ((1 - 10 x)^(1627/7140) (1 + 10 x)^(1457/7140))/x^(
   116/35) (- 97461 x^2 (-1 + 10 x) (1 + 10 x) g'[x] - 
     1785 x (-39 - 13 x + 840 x^2) g[x]);
f[x_] := NIntegrate[ (1 - 10 xi)^(1627/7140 - 1) (1 + 10 xi)^(
    1457/7140 - 
     1) (-161109 - 47073 xi + 13746115 xi^2 + 2404500 xi^3)/( xi^(
      116/35) ) g[xi], {xi, x0, x}, WorkingPrecision -> 40];

{x, x0} = RandomReal[{0, 1/10}, 2, WorkingPrecision -> 40];
f1[x_] := F[x] - F[x0];
{f[x], f1[x]} // MatrixForm

enter image description here

This identity comes on one hand from applying the inverse gauge transformation (see https://www.math.fsu.edu/~eimamogl/hypergeometricsols/hypergeometricsols.mpl) to the differential operator: \begin{equation} L:=\frac{d^2}{d x^2} + \frac{140500 x^2+663}{1105 x (10 x-1) (10 x+1)} \frac{d}{d x}+ \frac{400}{221 (10 x-1) (10 x+1)} \end{equation} and to the gauge \begin{equation} G:=-\frac{39 \left(1785 \left(100 x^2-1\right) x \frac{d}{d x} -514500 x^2+425 x+5916\right)}{x \left(2404500 x^3+13746115 x^2-47073 x-161109\right)} \end{equation}

On the other hand the identity in question comes from solving the in-homogeneous first order ODE \begin{equation} G \left[f(x) \right] = g(x) \end{equation} by the standard way of variation of constants.

Note 1: The function $g(x)$ satisfies the equation: \begin{equation} L\left[ g(x) \right] =0 \end{equation}

In[855]:= x =.; FullSimplify[(D[#, {x, 2}] + (663 + 140500 x^2)/(
      1105 x (-1 + 10 x) (1 + 10 x)) D[#, x] + 
     400/(221 (-1 + 10 x) (1 + 10 x)) #) & /@ {Hypergeometric2F1[1/13,
     1/17, 1/5, 100 x^2]}]

Out[855]= {0}

Note 2: The gauge $G$ corresponds to the first element of the global integral basis of the operator $L$, a basis which has been normalized at infinity (see https://arxiv.org/pdf/1606.01576.pdf for details).

Update: Here is actually another example of a similar integral identity: Now define: \begin{equation} g(x):= F_{2,1}\left[\frac{1}{11},\frac{1}{3},\frac{1}{2}; 100 x^2 \right] \end{equation} Then the following holds: \begin{eqnarray} &&\!\!\!\!\!\!\!\!\int\limits_{x_0}^x \frac{\sqrt[3]{1-10 \xi } (10 \xi +1)^{2/15} \left(42500 \xi ^3+170850 \xi ^2-4620 \xi -4389\right)}{1155 \xi ^{24/5}} g(\xi) d \xi = \\ &&\!\!\!\!\!\!\!\!\frac{(1-10 x)^{4/3} (10 x+1)^{17/15}}{x^{24/5}}\left.\left( -\frac{5}{14} (10 x-1) (10 x+1) x^2 g'(x)-\frac{1}{77} \left(500 x^2-55 x-77\right) x g(x)\right)\right|_{x_0}^x \end{eqnarray}

g[x_] := Hypergeometric2F1[1/11, 1/3, 1/2, 100 x^2];
F[x_] := ((1 - 10 x)^(4/3) (1 + 10 x)^(17/15))/x^(
   24/5) (-(5/14) x^2 (-1 + 10 x) (1 + 10 x) g'[x] - 
     1/77 x (-77 - 55 x + 500 x^2) g[x]);
f[x_] := (NIntegrate[( (1 - 10 xi)^(1/3) (1 + 10 xi)^(
      2/15) (-4389 - 4620 xi + 170850 xi^2 + 42500 xi^3))/(
     1155 xi^(24/5)) g[xi], {xi, x0, x}, WorkingPrecision -> 40]);
{x, x0} = RandomReal[{0, 1/10}, 2, WorkingPrecision -> 40];
{f[x], F[x] - F[x0]} // MatrixForm

enter image description here

The identity above has been derived in exactly the same way as the one before except that now we started from an abscissa transformed $x \rightarrow 100 x^2$ Gaussian hypergeometric operator with parameters $(a,b,c)=(1/11,1/3,1/2)$ and then used the first element of the infinity-normalized global integral basis to construct the relevant inverse gauge transformation.

Update 1: Here are two other examples which I generated this morning.

Firstly we define: \begin{equation} g(x):=F_{2,1}\left[-\frac{1}{4},-\frac{1}{6},\frac{1}{2};9 x^2\right] \end{equation} Then the following holds true: \begin{eqnarray} &&\int\limits_{x_0}^x \frac{\sqrt[12]{1-9 \xi ^2} e^{-4 \sqrt{\frac{2}{97}} \tan ^{-1}\left(\sqrt{\frac{2}{97}} (5 \xi +8)\right)}}{\left(10 \xi ^2+32 \xi +45\right)^{3/4}} \frac{45 \left(2400 \xi ^3+10547 \xi ^2+8640 \xi +5625\right)}{\left(90 \xi ^4+288 \xi ^3+395 \xi ^2-32 \xi -45\right)} g(\xi) d\xi=\\ &&\left.\frac{\sqrt[12]{1-9 x^2} e^{-4 \sqrt{\frac{2}{97}} \tan ^{-1}\left(\sqrt{\frac{2}{97}} (5 x+8)\right)}}{x \left(10 x^2+32 x+45\right)^{3/4}} \left(450 x^2 g(x)-90 x \left(10 x^2+32 x+45\right) g'(x)\right)\right|_{x_0}^x \end{eqnarray}

g[x_] := Hypergeometric2F1[-1/4, -1/6, 1/2, 9 x^2];
F[x_] := (1 - 9 x^2)^(1/12)/(
   E^(4 Sqrt[2/97] ArcTan[Sqrt[2/97] (8 + 5 x)])
     x (45 + 32 x + 10 x^2)^(
    3/4)) (-90 x (45 + 32 x + 10 x^2) g'[x] + 450 x^2 g[x]);
{x0, x} = RandomReal[{0, 1/3}, 2, WorkingPrecision -> 40];
X1 = NIntegrate[ ((1 - 9 xi^2)^(1/12))/(
   E^(4 Sqrt[2/97]
      ArcTan[Sqrt[2/97] (8 + 5 xi)]) (45 + 32 xi + 10 xi^2)^(
    3/4)) (45  (5625 + 8640 xi + 10547 xi^2 + 2400 xi^3))/(-45 - 
      32 xi + 395 xi^2 + 288 xi^3 + 90 xi^4) g[xi], {xi, x0, x}, 
  WorkingPrecision -> 40]
X2 = F[x] - F[x0]

enter image description here

Secondly we define: \begin{equation} g(x):=x^{1/3} F_{2,1}\left[-\frac{2}{15},-\frac{1}{12},\frac{2}{3};49 x^2\right] \end{equation} Then the following holds true:
\begin{eqnarray} &&\int\limits_{x_0}^x \left(-49 \xi +2 \sqrt{246}+2\right)^{\frac{\sqrt{246}-1599}{1845}} \left(49 \xi +2 \sqrt{246}-2\right)^{-\frac{13}{15}-\frac{\sqrt{\frac{2}{123}}}{15}} \cdot\\ && \left(1-49 \xi ^2\right)^{7/60} \frac{20 \left(-1323 \xi ^3-13798 \xi ^2+210 \xi +700\right)}{2401 \xi ^4-196 \xi ^3-1029 \xi ^2+4 \xi +20} g(\xi) d\xi=\\ && \left.\frac{\left(-49 x+2 \sqrt{246}+2\right)^{\frac{\sqrt{\frac{2}{123}}}{15}-\frac{13}{15}} \left(49 x+2 \sqrt{246}-2\right)^{\frac{-1599-\sqrt{246}}{1845}} \left(1-49 x^2\right)^{7/60}}{x} \left(\frac{75}{98} \left(441 x^2-20 x-100\right) g(x)-\frac{1125}{98} x \left(49 x^2-4 x-20\right) g'(x)\right)\right|_{x_0}^x \end{eqnarray}

g[x_] := x^(1/3) Hypergeometric2F1[-(2/15), -(1/12), 2/3, 49 x^2];
F[x_] := (1 - 49 x^2)^(
   7/60)/((2 + 2 Sqrt[246] - 49 x)^(13/15 - Sqrt[2/123]/15)
     x (-2 + 2 Sqrt[246] + 49 x)^((1599 + Sqrt[246])/
    1845)) (-(1125/98) x (-20 - 4 x + 49 x^2) g'[x] + 
     75/98 (-100 - 20 x + 441 x^2) g[x]);
{x0, x} = RandomReal[{0, 1/7}, 2, WorkingPrecision -> 40];
NIntegrate[(1 - 49 xi^2)^(
  7/60)/((2 + 2 Sqrt[246] - 49 xi)^((1599 - Sqrt[246])/
   1845) (-2 + 2 Sqrt[246] + 49 xi)^(13/15 + Sqrt[2/123]/15)) (
  20     (700 + 210 xi - 13798 xi^2 - 1323 xi^3))/(
  20 + 4 xi - 1029 xi^2 - 196 xi^3 + 2401 xi^4) g[xi], {xi, x0, x}, 
 WorkingPrecision -> 40]

F[x] - F[x0]

enter image description here

Update 2: In order to avoid confusion let me write down in a concise way how all those results were generated. Let $L:=d^2/d x^2 + a_1(x) d/d x + a_0(x)$ and $G:=r_1(x) d/d x + r_0(x)$ be a pair of differential operators or order two and order one respectively with coefficients being rational functions in $x$. Then let $Q(x):= \int r_0(x)/r_1(x) dx$ and let $(R_0(x),R_1(x))$ be the inverse gauge of the operator $L$ and the gauge $G$. In other words the functions $(R_0(x),R_1(x))$ are such that: \begin{equation} \left( r_0(x) + r_1(x) \frac{d}{d x}\right)\left(R_0(x) + R_1(x) \frac{d}{d x}\right) f(x) = L f(x) \quad (i) \end{equation} Note that the inverse gauge can always be found by making a polynomial ansatz for $R_0(x),R_1(x)$ inserting that ansatz to the equation above and then by matching the coefficients at respective derivatives solving for the unknown polynomial coefficients.

Now let the function $g(x)$ solve the ODE below: \begin{equation} L g(x) = 0 \quad (ii) \end{equation}

Then the following identity holds true: \begin{equation} \int\limits_{x_0}^x \frac{\exp(Q(\xi)}{r_1(\xi)} g(\xi) d\xi = \exp(Q(x)) \left.\left[ R_0(x) g(x) + R_1(x) g^{'}(x) \right] \right|_{x_0}^x \end{equation}

Note that the proof of this statement follows in a straightforward way from differentiating the right hand side and then using the properties $(i)$ and $(ii)$. As such this identity is not astounding . However the tricky parts is to find the inverse gauge and also to solve the ODE $(ii)$.

Having said all this my final question is now how do you prove such identities as above in a different way?

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  • $\begingroup$ I would appreciate if somebody explained the reason for down-voting my question? Is there anything unclear in here? I just show an example of a highly non-trivial integral that can be done in closed form. Moreover this is not an isolated example but it is clear that one can construct a whole family of such identities by playing around with parameters. $\endgroup$ – Przemo Jan 10 at 16:31

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