# Compute convolution

Let $a>0$ and $a_1>0$ be real numbers. Using the convolution theorem I have shown a following identity: $$\sum\limits_{l=1}^n \binom{a_1}{n-l} \frac{1}{l} \binom{a}{l} = -\gamma \left[\binom{a_1}{n} - \binom{a+a_1}{n}\right] - \sum\limits_{l=1}^n \binom{a}{l} (-1)^l \binom{a+a_1-l}{n-l} \psi(0,l+1)$$ Here $\gamma$ is the Euler's constant and $\psi$ is the digamma function. Now, the question is how do I generalize the result to $n$ being a real number. Is it also possible to provide a combinatorial derivation of that identity ?

We compute the Z transform of $\binom{a}{l}/l$. We have: \begin{eqnarray} S(x) = \sum\limits_{l=1}^\infty \binom{a}{l} \frac{x^l}{l} = \int\limits_0^x \left\{(1+\xi)^a - 1\right\} \frac{d\xi}{\xi} \end{eqnarray} Now we integrate by parts: \begin{eqnarray} S(x) &=& \log(x) \left\{(1+x)^a-1\right\} - a \int\limits_0^x (1+\xi)^{a-1} \log(\xi) d\xi \\ &=& \log(x) \left\{(1+x)^a-1\right\} - a (1+x)^{a-1} \left(x \log(x)-x\right) + a_{(2)} \int\limits_0^x (1+\xi)^{a-2} \cdot \left(\xi \log(\xi) - \xi\right) d\xi \end{eqnarray} Note that since $x>0$ and $a>0$ after successive integrations the integral term is smaller and smaller. Integrating by parts several times we have: \begin{eqnarray} S(x) &=& \log(x) \left\{(1+x)^a-1\right\} + \sum\limits_{l=1}^\infty (-1)^l a_{(l)} (1+x)^{a-l} \cdot \int\limits_0^x \frac{(x-\xi)^{l-1}}{(l-1)!} \log(\xi) d\xi \end{eqnarray} Here the integral term is the $l$th primary function of the logarithm. We have: \begin{eqnarray} \int\limits_0^x \frac{(x-\xi)^{l-1}}{(l-1)!} \log(\xi) d\xi &=& \frac{x^l}{l!} \left[\log(x) + \sum\limits_{m=1}^l \binom{l}{m} \frac{(-1)^m}{m} \right] = \frac{x^l}{l!} \left[\log(x) -\gamma -\psi(0,l+1)\right] \end{eqnarray} Therefore the Z transform reads: $$S(x) = -\gamma\left[1-(1+x)^a\right] - \sum\limits_{l=1}^\infty \binom{a}{l} (-1)^l (1+x)^{a-l} \psi(0,l+1) x^l$$ Multiplying the above by $(1+x)^{a_1}$ we get: $$S(x) (1+x)^{a_1} = -\gamma\left[(1+x)^{a_1}- (1+x)^{a+a_1}\right] - \sum\limits_{l=1}^\infty \binom{a}{l} (-1)^l (1+x)^{a+a_1-l} \psi(0,l+1) x^l$$ Now, inverting the Z transform from the above expression is pretty straightforward and gives the expression in the formulation of the question.