\nIn this post, we will prove the following monstrous looking identity from Gradshteyn and Ryzhik (3.255):\n$$ \\int _0^1 \\frac{x^{\\mu+\\frac{1}{2}} (1-x)^{\\mu-\\frac{1}{2}}}{(c+2bx-ax^2)^{\\mu+1}}dx = \\frac{\\sqrt{\\pi}}{\\left\\{a + \\left(\\sqrt{c+2b-a} + \\sqrt{c}\\right)^2\\right\\}^{\\mu+\\frac{1}{2}}\\sqrt{c+2b-a}} \\frac{\\Gamma \\left(\\mu + \\frac{1}{2}\\right)}{\\Gamma\\left(\\mu+1\\right)}$$\nwhere $c+2b-a>0$, $a + \\left(\\sqrt{c+2b-a} + \\sqrt{c}\\right)^2 > 0$ and $\\text{Re }\\mu > -\\frac{1}{2}$\n<\/p>\n
\nLet us denote the integral by $I$. We will first use a M\u00f6bius transformation to get rid of the quadratic in the denominator. Pick $\\rho > 0$ such that $$ c\\rho^2 + 2b\\rho – a = 0 \\; \\implies \\; \\rho = \\frac{-b+\\sqrt{b^2+ac}}{c} > 0 $$\nNow, set $x=\\frac{u}{1+\\rho u}$ and $dx = \\frac{du}{(1+\\rho u)^2}$ to get:\n$$I = \\frac{1}{c^{\\mu+1}}\\int_0^U \\frac{u^{\\mu+\\frac{1}{2}} (1+(\\rho-1)u)^{\\mu-\\frac{1}{2}}}{(1+\\kappa u)^{\\mu+1}}du$$\nwhere $U=\\frac{1}{1-\\rho}$ and $\\kappa = 2\\left(\\rho+\\frac{b}{c}\\right)=\\frac{2\\sqrt{b^2+ac}}{c}$.\n<\/p>\n
\nWe can now scale back this integral to $[0, 1]$ using the substitution $u = U v$ and $du = U dv$:\n$$I = \\frac{U^{\\mu+\\frac{3}{2}}}{c^{\\mu+1}} \\int_0^1 \\frac{v^{\\mu+\\frac{1}{2}} (1-v)^{\\mu-\\frac{1}{2}}}{(1+\\lambda v)^{\\mu+1}}dv$$\nwhere $\\lambda = U \\kappa$.\n<\/p>\n
\nWe can immediately recognize the above expression as Euler\u2019s integral for the Gauss hypergeometric function:\n$$I = \\frac{U^{\\mu+\\frac{3}{2}}}{c^{\\mu+1}}B\\left(\\mu+\\frac{3}{2}, \\mu+\\frac{1}{2}\\right) \\;_{2}F_{1}\\left(\\mu+\\frac{3}{2},\\mu+1;2(\\mu+1);-\\lambda\\right)$$\n<\/p>\n
\nApply Pfaff Transformation<\/a>:\n$${\\;}_{2}F_{1}\\left(a,b;c;z\\right) = (1-z)^{-a}{\\;}_{2}F_{1}\\left(a,c-b;c;\\frac{z}{z-1}\\right)$$\nwith $a=\\mu+\\frac{3}{2}$, $b=\\mu+1$, $c=2(\\mu+1)$, $z=-\\lambda$ to get:\n$${\\;}_{2}F_{1}\\left(\\mu+\\frac{3}{2},\\mu+1;2(\\mu+1);-\\lambda\\right) = (1+\\lambda)^{-\\mu-\\frac{3}{2}} {\\;}_{2}F_{1}\\left(\\mu+\\frac{3}{2},\\mu+1;2(\\mu+1);\\frac{\\lambda}{1-\\lambda}\\right)$$\nFor this precise configuration there is a classical quadratic reduction:\n$${\\;}_{2}F_{1}\\left(b+\\frac{1}{2},b;2b;t\\right)=\\frac{1}{\\sqrt{1-t}}\\left(\\frac{1+\\sqrt{1-t}}{2}\\right)^{1-2b}, \\quad \\text{Re }b>0$$\n<\/p>\n \nPut everything together:\n$$I = \\frac{U^{\\mu+\\frac{3}{2}}}{c^{\\mu+1}} \\frac{\\Gamma\\left(\\mu+\\frac{3}{2}\\right)\\Gamma\\left(\\mu+\\frac{1}{2}\\right)}{\\Gamma\\left(2(\\mu+1)\\right)}\\frac{1}{\\sqrt{1+\\lambda}}\\left(\\frac{1+\\frac{1}{\\sqrt{1+\\lambda}}}{2}\\right)^{-2\\mu-1}$$\n<\/p>\n \nAll that remains is to simplify the above expression. Firstly, we can use $\\Gamma\\left(\\mu+\\frac{3}{2}\\right)=\\frac{2\\mu+1}{2}\\Gamma\\left(\\mu+\\frac{1}{2}\\right)$ and $\\Gamma(2(\\mu+1))=2^{2\\mu+1}\\sqrt{\\pi}\\Gamma(\\mu+1)\\Gamma\\left(\\mu+\\frac{1}{2}\\right)$ to get the $\\sqrt{\\pi} \\frac{\\Gamma\\left(\\mu+\\frac{1}{2}\\right)}{\\Gamma\\left(\\mu+1\\right)}$ factor. The remaining purely algebraic constants simplify, with\n$$\\begin{align*}\nU &= \\frac{c}{b+c-\\sqrt{b^2+ac}} \\\\\n\\lambda &= \\frac{2\\sqrt{b^2+ac}}{b+c-\\sqrt{b^2+ac}}\n\\end{align*}$$\nto \n$$\\frac{1}{\\sqrt{c+2b-a}} \\frac{1}{\\left(a+\\left(\\sqrt{c+2b-a}+\\sqrt{c}\\right)^2\\right)^{\\mu+\\frac12}}$$ \nOne convenient algebraic identity used here is the following: $(c+b)^2 – (b^2+ac) = c(c+2b-a)$.\n<\/p>\n\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" In this post, we will prove the following monstrous looking identity from Gradshteyn and Ryzhik (3.255): $$ \\int _0^1 \\frac{x^{\\mu+\\frac{1}{2}} (1-x)^{\\mu-\\frac{1}{2}}}{(c+2bx-ax^2)^{\\mu+1}}dx = \\frac{\\sqrt{\\pi}}{\\left\\{a + \\left(\\sqrt{c+2b-a} + \\sqrt{c}\\right)^2\\right\\}^{\\mu+\\frac{1}{2}}\\sqrt{c+2b-a}} \\frac{\\Gamma \\left(\\mu + \\frac{1}{2}\\right)}{\\Gamma\\left(\\mu+1\\right)}$$ where $c+2b-a>0$, $a + \\left(\\sqrt{c+2b-a} + \\sqrt{c}\\right)^2 > 0$ … Continue reading