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I am currently struggling to find out where the following equation comes from. The authors of the article (Bayer, Friz, Gatheral: Pricing under rough volatility, p.12) where I got it from just wrote it down like it is obvious. Let $\gamma\in(0,\frac{1}{2})$ and $x>1$. Then

$\int\limits_0^1(1-s)^{-\gamma}(x-s)^{-\gamma}ds=\frac{x^\gamma}{1-\gamma}\text{ }_2F_1(1,\gamma,2-\gamma,x)$

where $_2F_1$ denotes the hypergeometric function.

user3236841
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Benedikt Heiland
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    Small note: 2F1 is not a confluent hypergeometric function. – Aaron Hendrickson Jun 24 '20 at 20:44
  • $x$ on the rhs should be $1/x$, then it's Euler's integral representation of ${_2 F_1}$: $$\int_0^1 (1 - s)^{-\gamma} (x - s)^{-\gamma} ds = x^{-\gamma} \int_0^1 (1 - s)^{-\gamma} (1 - x^{-1} s)^{-\gamma} ds = \ \frac {x^{-\gamma}} {1 - \gamma} , {_2 F_1}(1, \gamma; 2 - \gamma; x^{-1}).$$ – Maxim Jul 01 '20 at 14:32
  • @Maxim It's not clear to me how this is equivalent to Euler's integral formula. Wouldn't it be $${}_2F_1(1,\gamma;2-\gamma\mid x)=\frac{\Gamma(2-\gamma)}{\Gamma(\gamma)\Gamma(2\gamma+2)}\int_0^1 t^{\gamma-1}(1-t)^{1-2\gamma}(1-xt)^{-1}\mathrm dt$$ ? – K.defaoite Mar 12 '24 at 19:13

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I think you mean to integrate from $0$ to $x$, and you also got the exponent of $x$ wrong. Maple says $$ \int_0^x (1-s)^{-\gamma} (x-s)^{-\gamma} ds = \frac{x^{1-\gamma}}{1-\gamma} {}_2F_1(1,\gamma; 2-\gamma; x)$$ which it apparently gets using the "meijerg" method.

Robert Israel
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