A parametric beta integral with a shifted linear denominator

Recall that Euler's beta function is defined for \(\operatorname{Re} x > 0\) and \(\operatorname{Re} y > 0\) by

\[ B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1}\,dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}. \]

When an extra linear factor \((x+p)^{a+b}\) appears in the denominator with its exponent matching the sum of the two beta parameters, a homographic (Möbius) substitution removes the denominator completely while preserving the unit interval \([0,1]\). We will prove that for \(a, b, p > 0\),

\[ \int_0^1 x^{a-1}(1-x)^{b-1} \frac{dx}{(x+p)^{a+b}} = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\frac{1}{(1+p)^{a} p^b}. \]

The homographic substitution

Before changing variables, rewrite the integrand by grouping factors of \(x+p\) with \(x^{a-1}\), \((1-x)^{b-1}\), and \(dx\):

\[ x^{a-1}(1-x)^{b-1}\frac{dx}{(x+p)^{a+b}} = \left(\frac{x}{x+p}\right)^{a-1}\left(\frac{1-x}{x+p}\right)^{b-1}\frac{dx}{(x+p)^2}. \]

Now perform the change of variables

\[ \frac{x}{x+p} = \frac{t}{p+1}. \]

Notice that \(x=0\) corresponds to \(t=0\), and \(x=1\) corresponds to \(t=1\), so the limits of integration remain \(0\) and \(1\). Differentiating both sides with respect to \(x\) gives

\[ \begin{aligned} \frac{p}{(x+p)^2} &= \frac{1}{p+1}\frac{dt}{dx} \\ \implies \frac{dx}{(x+p)^2} &= \frac{dt}{p(p+1)}. \end{aligned} \]

To express \(\frac{1-x}{x+p}\) in terms of \(t\), observe that

\[ \frac{1+p}{x+p} = 1 + \frac{1-x}{x+p} \quad\text{and}\quad 1 - \frac{x}{x+p} = \frac{p}{x+p} = 1 - \frac{t}{p+1}. \]

Subtracting \(\frac{x}{x+p} = \frac{t}{p+1}\) from \(\frac{1+p}{x+p} = \frac{1+p}{p}\left(1 - \frac{t}{p+1}\right) = \frac{p+1-t}{p}\) yields

\[ \frac{1-x}{x+p} = \frac{1-t}{p}. \]

Reduction to the standard beta integral

Substituting these three relations into the integral, we obtain

\[ \begin{aligned} \int_0^1 x^{a-1}(1-x)^{b-1} \frac{dx}{(x+p)^{a+b}} &= \int_0^1 \left(\frac{t}{p+1}\right)^{a-1}\left(\frac{1-t}{p}\right)^{b-1}\frac{dt}{p(p+1)} \\ &= \frac{1}{p^b(1+p)^a}\int_0^1 t^{a-1}(1-t)^{b-1}\,dt \\ &= \frac{1}{p^b(1+p)^a} B(a,b) \\ &= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\frac{1}{(1+p)^a p^b}. \end{aligned} \]

By analytic continuation, the same identity remains valid for all complex parameters \(\operatorname{Re} a > 0\) and \(\operatorname{Re} b > 0\), and for any \(p \in \mathbb{C} \setminus [-1,0]\) using the principal branches of the powers. For a concrete application of this formula, see an algebraic integral with a fifth root.