$\int_0^1 x^{\alpha-1} (1-x)^{\beta-1} (x-A)^{\gamma-1} \mathrm{d} x$, where $\alpha,\beta,\gamma \ge 1$, and $A < 0$.

Question

Is there an analytical expression for this integral?

$$\int_0^1 x^{\alpha-1} (1-x)^{\beta-1} (x-A)^{\gamma-1} \mathrm{d} x$$

where $\alpha,\beta,\gamma \ge 1$, and $A < 0$.

P.S.: By analytical expression, I mean, pragmatically, an expression in terms of functions usually found in numeric computation libraries.

The particular case with $\gamma=1$ coincides with the Beta function. So no. — , Feb 12 '16 at 14:43
@YvesDaoust Why not? The Beta function is an "analytical expression" for me. — a06e, Feb 12 '16 at 14:44
@YvesDaoust Agree. Pragmatically what I mean is an expression in terms of functions usually found in numeric computation libraries. — a06e, Feb 12 '16 at 14:52
That's an excellent criterion, you could have stated it upfront. Then of course, yes. — , Feb 12 '16 at 14:55

score 3 · Answer 1 · answered Jun 28 '14 at 13:30

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What is amazing is that the antiderivative can be analytically expressed on the basis of Appell hypergeometric functions of two variables. For the integral, similar to what O.L. answered, I (a CAS did it !) obtained $$\int_0^1 x^{\alpha-1} (1-x)^{\beta-1} (x-A)^{\gamma-1} \mathrm{d} x= (-A)^{\gamma-1}\Gamma (\alpha) \Gamma (\beta) \, _2\tilde{F}_1\left(\alpha,1-\gamma;\alpha+\beta;\frac{1}{A}\right)$$ where appears the regularized hypergeometric function which can be expressed using, from definition, the classical hypergeometric function

answered Jun 28 '14 at 13:30

Claude Leibovici

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Hi Claude ! I Our answers are consistent according to the relationship between standard- and regularized- hypergeometric2F1. – JJacquelin Jun 28 '14 at 14:56
@JJacquelin. Yes, indeed (by definition of the regularized function). Cheers. – Claude Leibovici Jun 28 '14 at 14:58

score 1 · Answer 2 · answered Jun 28 '14 at 13:19

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Yes, your expression essentially coincides with the Euler's integral representation of the Gauss hypergeometric function: $$_2F_1(a,b;c;t)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1\frac{x^{b-1}(1-x)^{c-b-1}}{(1-tx)^a}dx.$$

answered Jun 28 '14 at 13:19

Start wearing purple

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score 1 · Accepted Answer · answered Jun 28 '14 at 14:02

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The integral can be expressed on a closed form, thanks to the Hypergeometric2F1 function :

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answered Jun 28 '14 at 14:02

JJacquelin

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$\int_0^1 x^{\alpha-1} (1-x)^{\beta-1} (x-A)^{\gamma-1} \mathrm{d} x$, where $\alpha,\beta,\gamma \ge 1$, and $A < 0$.

3 Answers3