An integral involve Guassian Hypergeomstric function.

Question

In handling the following integral: $$I = \int_0^1 {{u^{{m_2} - 1}}{{\left( {1 - u} \right)}^{{m_3} - 1}}{}_2{F_1}\left( {{m_1} + {m_2} + {m_3},{m_2} + {m_3};{m_2} + {m_3} + 1;\left( {q - p} \right)u - q} \right)} du$$ Where all of the parameters are real and positive value, and ${}_2{F_1}\left( {} \right)$ is the Guassian Hypergeomstric function. We decide to use these formula: (1) $$\int_0^y {{x^{\alpha - 1}}{{\left( {y - x} \right)}^{\beta - 1}}{}_2{F_1}} \left( {a,b;c; - \omega x} \right)dx,[y,Re{\text{ }}\alpha {\text{,Re }}\beta {\text{ > 0;}}\left| {{\text{arg}}\left( {1 + \omega y} \right)} \right| < \pi ]$$ (2) $$\eqalign{ & \int_0^y {{x^{\alpha - 1}}{{\left( {y - x} \right)}^{\beta - 1}}{}_2{F_1}} \left( {a,b;c;1 - \omega x} \right)dx, \cr & [y,\operatorname{Re} {\text{ }}\alpha ,\operatorname{Re} {\text{ }}\beta > 0,\operatorname{Re} \left( {c - a - b + \alpha } \right) > 0;\left| {\arg \left( \omega \right)} \right| < \pi ] \cr} $$ But we failed to find the relationship between ${}_2{F_1}\left( {a,b;c;kx + b} \right)$ and ${}_2{F_1}\left( {a,b;c;1 - \omega x} \right),{}_2{F_1}\left( {a,b;c; - \omega x} \right)$, So, is this method really works? Or any other method can work out this integral? Or any approximate method to this problem? Thanks for your attention.

Answer 1

Consider the integral \begin{align} I = \int_{0}^{1} t^{\mu-1} (1-t)^{\nu -1} {}_{2}F_{1}(\lambda+\mu+\nu, \mu+\nu; \mu+\nu+1; (q-p)t-q) dt. \end{align} Expand the hypergeometric function into series form to obtain \begin{align} I &= \sum_{n=0}^{\infty} \frac{(\lambda+\mu+\nu)_{n} (\mu+\nu)_{n} }{n! (\mu+\nu+1)_{n} } \cdot \int_{0}^{1} t^{\mu-1} (1-t)^{\nu -1} (-q + (q-p)t)^{n} dt \\ &= \sum_{n=0}^{\infty} \frac{(\lambda+\mu+\nu)_{n} (\mu+\nu)_{n} (-q)^{n}}{n! (\mu+\nu+1)_{n} } \cdot \int_{0}^{1} t^{\mu-1} (1-t)^{\nu -1} (1 - \frac{(q-p)}{q} t)^{n} dt. \end{align} The remaining integral can be quickly determined by use of the hypergeometric definition \begin{align} B(\beta, \gamma - \beta) \ {}_{2}F_{1}(\alpha, \beta; \gamma; z) = \int_{0}^{1} t^{\beta -1} (1-t)^{\gamma - \beta -1} (1-z t)^{- \alpha} dt \end{align} which is then seen to be \begin{align} I &= B(\mu, \nu) \ \sum_{n=0}^{\infty} \frac{(\lambda+\mu+\nu)_{n} (\mu+\nu)_{n} (-q)^{n}}{n! (\mu+\nu+1)_{n} } \ {}_{2}F_{1}(-n, \mu; \nu-\mu; \frac{q-p}{q}). \end{align} Again, expanding the hypergeometric series yields \begin{align} I &= B(\mu, \nu) \ \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(\lambda+\mu+\nu)_{n} (\mu+\nu)_{n} (-n)_{k} (\mu)_{k} (-q)^{n}}{n! k! (\mu+\nu+1)_{n} (\nu-\mu)_{k}} \left(\frac{q-p}{q} \right)^{k} \\ &= B(\mu,\nu) \ \sum_{n,k=0}^{\infty} \frac{(\lambda+\mu+\nu)_{n+k} (\mu+\nu)_{n+k} (-n-k)_{k} (\mu)_{k} }{ (\mu+\nu+1)_{n+k} (\nu-\mu)_{k}} \cdot \frac{(-q)^{n+k}}{(n+k)!} \cdot \frac{\left(\frac{q-p}{q}\right)^{k}}{k!} \\ &= B(\mu,\nu) \ \sum_{n,k=0}^{\infty} \frac{(\lambda+\mu+\nu)_{n+k} (\mu+\nu)_{n+k} (\mu)_{k} }{ (\mu+\nu+1)_{n+k} (\nu-\mu)_{k}} \cdot \frac{(-q)^{n}}{n!} \cdot \frac{(q-p)^{k}}{k!} \\ &= B(\mu,\nu) \ F_{1:0,1}^{2:0,1} \left[ \begin{array}{cc} \lambda+\mu+\nu, \mu+\nu: -; \mu; \\ \mu+\nu+1: -; \nu-\mu; \end{array} \ -q, q-p \right] \end{align} where the resulting expression is a Kampe de Feriet function.

Hence, \begin{align} & \int_{0}^{1} t^{\mu-1} (1-t)^{\nu -1} {}_{2}F_{1}\left(\begin{array}{cc} \lambda+\mu+\nu, \mu+\nu; \\ \mu+\nu+1; \end{array} \ (q-p)t-q \right) \ dt \\ &= B(\mu,\nu) \ F_{1:0,1}^{2:0,1} \left[ \begin{array}{cc} \lambda+\mu+\nu, \mu+\nu: -; \mu; \\ \mu+\nu+1: -; \nu-\mu; \end{array} \ -q, q-p \right]. \end{align}

The Kamp\'{e} de Feriet function is given by http://en.wikipedia.org/wiki/Kamp%C3%A9_de_F%C3%A9riet_function .