I am wondering if someone could tell me whether or not the following integral has a closed form representation:

$$\int_0^1 t^n\log\Gamma(t+a)dt$$

In Srivastava's and Choi's wonderful book Zeta and q-Zeta Functions and Associated Series and Integrals, they give a closed form for the following Digamma integral:

$$\int_0^1 t^n\psi(t)dt$$

and the also list specific cases for the integral I am searching for, but not the general case.

The formulae for the integrals containing $n>2$ become extremely complex, which makes me wonder whether a closed form exists but I was hoping someone could settle this question for me.

  • $\begingroup$ Almost certainly not. $\endgroup$ – Alex M. Jan 6 '16 at 16:34
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    $\begingroup$ The 1998 paper by Adamchik "Polygamma functions of negative order" may interest you. The answer is given in function of $\psi^{(-n)}(t)$ (integrals of polygamma or "Negapolygamma" in Adamchik 'parlance'). An example using Alpha. Another paper by Espinosa and Moll. $\endgroup$ – Raymond Manzoni Jan 6 '16 at 17:02
  • $\begingroup$ Raymond: Thank you so much for the references, I will take a look! $\endgroup$ – FofX Jan 6 '16 at 17:18
  • $\begingroup$ Glad it interested you @FofX. Note that Gosper's 'negapolygamma' is in fact defined by $$\psi^{(-n-2)}=(z):=\frac 1{n!}\int_0^z (z-t)^{n}\log\Gamma(t)\,dt$$ $\endgroup$ – Raymond Manzoni Jan 6 '16 at 17:26

This is purely guessing. For integer $n$, we seem to have something like $$ \int_0^1 t^n\log\Gamma(t+a)dt = -n! \psi^{(-n-2)}(a)+n!\psi^{(-n-2)}(1+a)-\frac{n!}{1!}\psi^{(-n-1)}(1+a)+\frac{n!}{2!}\psi^{(-n)}(1+a)-\frac{n!}{3!}\psi^{(-n+1)}(1+a)+\frac{n!}{4!}\psi^{(-n+2)}(1+a)-\cdots $$ so I suspect that $$ I(a,n)=\int_0^1 t^n\log\Gamma(t+a)dt = n!(-1)^{n+1}\left( \psi^{(-n-2)}(a) -\sum_{k=0}^n \frac{(-1)^k}{k!}\psi^{(-n-2+k)}(1+a) \right) $$ this seems to work numerically for integer $n\ge 0$ and any $a$ I have tried. This gives some sort of closed forms, for example $$ I(1,1) = -\frac{1}{4} - 2 \ln A + \psi^{(-3)}(1)\\ I(1,2) = \frac{1}{4}\ln\left(\frac{2\pi}{A^4}\right)\\ $$ with $A$ the Glaisher–Kinkelin constant.

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    $\begingroup$ Yes that looks good to me. Awhile after asking this question, I came across a semi-closed form in Henri Cohen's Number Theory Vol 2 (pg. 130). It is essentially expressible in terms of the Hurwitz zeta function and it's derivative at negative integers. $\endgroup$ – FofX Jul 26 '17 at 0:54

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