I'm working through Vaughan's book on the Hardy Littlewood circle method, which uses the following lemma:
Suppose that $\alpha \geq \beta$ are positive real numbers, and that $\beta \leq 1$. Then:
$ \sum \limits_{m=1}^{n-1} m^{\beta-1}(n-m)^{\alpha-1} = n^{\beta+\alpha-1} (\frac{ \Gamma(\beta) \Gamma(\alpha) }{ \Gamma(\beta+\alpha) } + O(n^{-\beta}) $
For the proof they define the function $\phi(\gamma) = \gamma^{\beta-1}(n-\gamma)^{\alpha-1}$ and proceed as follows:
On $(0,n)$ it has at most one stationary point, so we can divide it into two intervals $(0,X),(X,n)$, one of which may be empty, such that $\phi$ is increasing on one and decreasing on the other. Therefore \begin{equation} \begin{split} \sum_{m=1}^{n-1} \phi(m) & = \int_0^{n} \phi(\gamma)d\gamma + O(n^{\alpha-1}) \\ & = \frac{ \Gamma(\beta) \Gamma(\alpha)}{\Gamma(\beta+\alpha)} + O(n^{\alpha-1}) \end{split} \end{equation}
How does the gamma function appear? I tried evaluating the integral of $\phi$ via integration by parts but I don't see the gamma function popping up. Any help is appreciated.