How to prove that $\displaystyle\int_0^1{\zeta(1-n,1-t)\ln\Gamma{(t)}}dt = \frac{H_n\zeta(-n)+\zeta’(-n)}{n}$?
For integer values Wolfram Alpha gave me solutions to the integral in the form on the right hand side, so I guessed the form.
It appears to break down when $n$ is less than around $-\frac12$. $\zeta(s,a)$ and $\zeta(s)$ are the Hurwitz zeta and Riemann zeta functions. $H_n$ are the harmonic numbers.
How can we prove it?
Additional info
In order to obtain the closed form for integer $n$, in Wolfram Alpha, we can first rewrite it in terms of the Polygamma function.
First change the zeta function to Bernoulli Polynomials for easier manipulation.
$$\begin{align}
&\frac{1}{\left(n-1\right)!}\int_{0}^{1}\zeta\left(1-n,1-t\right)\ln\Gamma\left(t\right)dt
\\=&
\ -\frac{1}{n!}\int_{0}^{1}B_n\left(1-t\right)\ln\Gamma\left(t\right)dt
\\=&
\ \frac{1}{n!}\int_{0}^{1}n\left(1-t\right)^{n-1}\ln\Gamma\left(t\right)dt-\frac{1}{n!}\int_{0}^{1}B_n\left(2-t\right)\ln\Gamma\left(t\right)dt
\end{align}$$
This result can actually be generalized a bit by introducing the variable $a$.
$$\begin{align}
&\frac{1}{n!}\int_{0}^{a}n\left(a-t\right)^{n-1}\ln\Gamma\left(t\right)dt-\frac{1}{n!}\int_{0}^{1}B_n\left(a+1-t\right)\ln\Gamma\left(t\right)dt
\\=&
\ \psi^{(-n-1)}\left(a\right)-\frac{1}{n!}\int_{0}^{1}B_n\left(a+1-t\right)\ln\Gamma\left(t\right)dt
\end{align}$$
where $\psi^{(s)}(a)$ is the polygamma function. Here we are using this definition for negative order. Next we can utilize the definition of the bernoulli polynomials to obtain.
$$\begin{align}
&\psi^{(-n-1)}\left(a\right)-\frac{1}{n!}\int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}B_k\left(a\right)\left(1-t\right)^{n-k}\ln\Gamma\left(t\right)dt
\\=&
\ \psi^{(-n-1)}\left(a\right)-\sum_{k=0}^{n}\frac{B_k\left(a\right)}{k!}\left(\frac{1}{\left(n-k\right)!}\int_{0}^{1}\left(1-t\right)^{n-k}\ln\Gamma\left(t\right)dt\right)
\\=&
\ \psi^{(-n-1)}\left(a\right)-\sum_{k=0}^{n}\frac{B_k\left(a\right)}{k!}\psi^{(-n-2+k)}\left(1\right)
\end{align}$$
This last form is useful for plugging into Wolfram Alpha and generating our desired form. After analyzing patterns for several integers we see it equals
$$\begin{align}
&-H_n\frac{B_{n+1}\left(a\right)}{\left(n+1\right)!}+\frac{\zeta'\left(-n\right)}{n!}+\frac{1}{n!}\sum_{k=1}^{a-1}k^{n}\ln k
\\=&
\ \frac{H_n\zeta\left(-n,a\right)}{n!}+\frac{\zeta'\left(-n,a\right)}{n!}
\end{align}$$
Multiplying by $(n-1)!$ and setting $a=1$ yields the final result.
Best Answer
The equality (under the usual convention $H_s:=\gamma+\psi(1+s)$ for complex $s$) $$ \mathscr{L}(s):=\int_0^1\zeta(1-s,1-t)\ln\Gamma(t)\,dt=\frac1s\big(H_s\zeta(-s)+\zeta'(-s)\big) $$ for, say, $\Re s>1$ can be viewed as an application of Parseval's theorem to the series $$ \color{blue}{\ln\Gamma(t)=\frac{\ln2\pi}2+\sum_{n=1}^\infty\left(\frac1{2n}\cos 2n\pi t+\frac{\gamma+\ln2n\pi}{n\pi}\sin 2n\pi t\right)} $$ for $0<t<1$ (a variation of the known expansion; the formula $$ \ln\Gamma(1-z)=\int_0^\infty\left(\frac{e^{zt}-1}{e^t-1}-ze^{-t}\right)\frac{dt}{t} $$ is a possible way to obtain this expansion) and the series (for $\Re s>1$ said) $$ \color{blue}{\zeta(1-s,1-t)=\frac{2\Gamma(s)}{(2\pi)^s}\sum_{n=1}^\infty\frac1{n^s}\cos\left(\frac{\pi s}2+2n\pi t\right)} $$ obtained (e.g.) as follows. The definition $\zeta(s,a)=\sum_{n=0}^\infty(n+a)^{-s}$ leads to $$ \Gamma(s)\zeta(s,a)=\int_0^\infty\frac{t^{s-1}e^{-at}}{1-e^{-t}}\,dt\qquad(a>0,\Re s>1) $$ which can be used to obtain (another known integral representation) $$ \frac1{2\pi i}\int_\lambda\frac{z^{s-1}e^{az}}{1-e^z}\,dz=\frac{\zeta(s,a)}{\Gamma(1-s)}, $$ where the contour $\lambda$ encircles the negative real axis (but not the poles $z=2n\pi i$, $n\in\mathbb{Z}_{\neq0}$ of the integrand). This is analytically continued (w.r.t. $s$) and, for $\Re s<0$ and $0<a<1$, the integral can be evaluated using residues (and considering, as a contour, a big rectangle with a notch around the negative real axis). This leads to the expansion of $\zeta(1-s,1-t)$ shown above.
Thus, Parseval's theorem yields $$ \mathscr{L}(s)=\frac{\Gamma(s)}{(2\pi)^s}\left[\left(\frac12\cos\frac{\pi s}2-\frac{\gamma+\ln2\pi}\pi\sin\frac{\pi s}2\right)\zeta(1+s)+\frac1\pi\sin\frac{\pi s}2\zeta'(1+s)\right], $$ which simplifies to the announced result, using Riemann's functional equation for $\zeta(1+s)$.
Of course, the result holds analytically continued (from $\Re s>1$ to ???).