A proof of GR 3.255 · Integrals and Series

A quadratic denominator can make a beta-type integral look forbidding. For entry 3.255 in Gradshteyn and Ryzhik, a change of variables exposes an inversion symmetry and reduces the calculation to a standard beta integral. We will prove

\[ \int _0^1 \frac{x^{\mu+\frac{1}{2}} (1-x)^{\mu-\frac{1}{2}}}{(c+2bx-ax^2)^{\mu+1}}dx = \frac{\sqrt{\pi}}{\left\{a + \left(\sqrt{c+2b-a} + \sqrt{c}\right)^2\right\}^{\mu+\frac{1}{2}}\sqrt{c+2b-a}} \frac{\Gamma \left(\mu + \frac{1}{2}\right)}{\Gamma\left(\mu+1\right)} \]

Assume that \(a,b,c\) are real and

\[ c>0,\qquad d:=c+2b-a>0,\qquad D:=a+(\sqrt d+\sqrt c)^2>0, \qquad \operatorname{Re}\mu>-\frac12. \]

All powers of positive real numbers use the real logarithm. These hypotheses ensure that the denominator stays positive on \([0,1]\) and that the endpoint singularities are integrable.

Move the endpoints to zero and infinity

Write the integral as \(I\) and set \(t=(1-x)/x\), so that \(x=1/(1+t)\). The powers of \(1+t\) cancel, leaving

\[ I=\int_0^\infty\frac{t^{\mu-\frac12}}{\bigl(ct^2+2(c+b)t+d\bigr)^{\mu+1}}\,dt. \]

Now put \(r=\sqrt{d/c}\), \(t=rs\), and \(h=(c+b)/\sqrt{cd}\). Then

\[ I=\frac{r^{\mu+\frac12}}{d^{\mu+1}}J, \qquad J=\int_0^\infty\frac{s^{\mu-\frac12}}{(s^2+2hs+1)^{\mu+1}}\,ds. \]

Notice that \(D=2(c+b+\sqrt{cd})\), so \(D>0\) is equivalent to \(h>-1\). Consequently \(s^2+2hs+1=(s-1)^2+2(h+1)s>0\) for \(s>0\).

Use the inversion symmetry

Replacing \(s\) by \(1/s\) in \(J\) gives a second representation:

\[ J=\int_0^\infty\frac{s^{\mu+\frac12}}{(s^2+2hs+1)^{\mu+1}}\,ds. \]

Average the two representations and factor a power of \(s\) out of the denominator:

\[ J=\int_0^\infty \frac{s+1}{2s^{3/2}}\bigl(s+s^{-1}+2h\bigr)^{-\mu-1}\,ds. \]

The factor in front is exactly the derivative of \(u=\sqrt s-1/\sqrt s\). This substitution maps \((0,\infty)\) onto the real line and satisfies \(s+s^{-1}=u^2+2\). Hence

\[ J=\int_{-\infty}^{\infty}(u^2+2+2h)^{-\mu-1}\,du. \]

Finish with the beta integral

For \(A>0\) and \(\operatorname{Re}\mu>-1/2\), substituting \(u=\sqrt A\,v\) and then using \(v^2\) as a variable gives

\[ \int_{-\infty}^{\infty}(u^2+A)^{-\mu-1}\,du =A^{-\mu-\frac12}B\left(\frac12,\mu+\frac12\right) =\frac{\sqrt\pi\,\Gamma(\mu+\frac12)}{\Gamma(\mu+1)}A^{-\mu-\frac12}. \]

Take \(A=2+2h=D/\sqrt{cd}\) and restore the prefactor. The constants simplify as

\[ \frac{r^{\mu+\frac12}}{d^{\mu+1}} \left(\frac{D}{\sqrt{cd}}\right)^{-\mu-\frac12} =\frac{1}{\sqrt d\,D^{\mu+\frac12}}, \]

which is the claimed evaluation. The symmetry step is the key: averaging before substituting produces the exact differential needed for a single beta integral.