1
$\begingroup$

(EDIT: It works as I described it, I just made mistakes along the way)

I want to evaluate \begin{align} \int\limits_{0}^{1}\mathrm{d}u\:u^{-1-\varepsilon}F_{21}(1,1-\varepsilon;2-2\varepsilon;1-uz) \end{align} assuming $z\in[0,1]$. My own attempt was to use \begin{align} F_{21}(a,b;c;1-x)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}F_{21}(a,b;a+b-c+1;x)+\frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}F_{21}(c-a,c-b;c-a-b+1;x) \end{align} and then just use the integral representation of the hypergeometric function. This however yields hypergeometric functions with totally different indices than what Mathematica gives me. How does Mathematica arrive at this result: \begin{align} \frac{1}{\varepsilon^2} \left(- \frac{4^{-\varepsilon} \varepsilon (1 - z)^{2 \varepsilon}z^{-\varepsilon} \Gamma(3/2 - \varepsilon) \Gamma(\varepsilon)}{\sqrt{\pi}} + (1 - 2 \varepsilon) F_{21}(1, -\varepsilon, 1 + \varepsilon; z)\right) \end{align} assuming $\mathrm{Re}(\varepsilon)<0$. The step-by-step solution function of WolframAlpha does not help me here, since the WolframAlpha engine seemingly cannot compute this integral.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.