This may look like homework, but it is not. I've found this identity (using Mathematica):

$$ {}_3F_2 \left( \matrix{1,1,1 \\ 2, e} ; 1 \right) = (e-1) \psi^{\prime}(e-1), $$

valid for $e$ with $\mathcal{R}(e)>0$, where ${}_3F_2$ is the Generalized Hypergeometric Function (as in here) and $\psi^{\prime}$ is the trigamma function (definition here).

It's also in Wolfram's site: http://functions.wolfram.com/

The problem is... I've no idea how to prove it. I've tried using the definition of the Pochhammer symbol and some simplifications to get this:

$$ {}_3F_2 \left( \matrix{1,1,1 \\ 2, e} ; 1 \right) = \sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(e)}{(k+1)\Gamma(e+k)}, $$

but it's not even close to the series for trigamma function:

$$ (e-1) \psi^{\prime}(e-1) = (e-1) \sum_{n=0}^{\infty} \frac{1}{(e-1+n)^2}. $$

Any help/tips/references are appreciated.


Euler's integral transformation at your 1st link allows to write the left side as an integral of an $_2F_1$ function \begin{align} _3F_2\left[\begin{array}{c}1,1,1\\2,e\end{array};1\right]=\frac{\Gamma(e)}{\Gamma(e-1)}\int_0^1 (1-t)^{e-2}{}_2F_1\left[\begin{array}{c}1,1\\2\end{array};t\right]dt, \end{align} which can be itself expressed via elementary functions: $$ {}_2F_1\left[\begin{array}{c}1,1\\2\end{array};t\right]=-\frac{\ln(1-t)}{t}.$$ (The last identity can be easily derived using series expansions of both sides). So $$ _3F_2\left[\begin{array}{c}1,1,1\\2,e\end{array};1\right]=-(e-1)\int_0^1 (1-t)^{e-2}\frac{\ln(1-t)}{t}dt.$$ Note that the integral on the right is perfectly well-defined at $t=0$, since $$\displaystyle\frac{\ln(1-t)}{t}=-1+\frac{t}{2}-\frac{t^2}{3}+\ldots$$ Moreover, this is nothing but the standard integral representation for $-\psi'(e-1)$; see e.g. the 2nd formula at your 2nd link.

  • $\begingroup$ Thanks! I'll see if I can go on from here. $\endgroup$ – Ferdinand.kraft May 3 '13 at 13:10

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.