How to prove $$\int^\infty_0 \frac{e^{-t}}{t}\left[\frac1{t^2}-\frac1{(1-e^{-t})^2}+\frac1{1-e^{-t}}-\frac1{12}\right]dt=\frac34-\zeta'(-1)+\zeta'(0)$$ ?
This integral appeared in my answer, and according to an arXiv paper and the OP's conjecture this equality is very likely to be true. This is also supported by numerical evidence.
You cannot find the integral in the arXiv paper, as the integral arises in my lengthy proof of a statement (whose proof is omitted) in the paper. Thus I think it is not very useful to provide the link.
Real or complex approaches are welcomed. Thanks in advance.
I will first evaluate $$I(s) = \int_{0}^{\infty} t^{s-1}e^{-t} \left(\frac{1}{t^{2}} - \frac{1}{(1-e^{-t})^{2}} + \frac{1}{1-e^{-t}} - \frac{1}{12} \right) \, \mathrm dt \tag{1}$$ for $\Re(s)>2$.
With the restriction $\Re(s)>2$, we can break up the integral into four convergent integrals and evaluate each integral separately.
The first integral is just $\Gamma(s-2)$, the third integral is $\Gamma(s) \zeta(s)$, and the fourth integral is $\frac{1}{12} \, \Gamma(s)$.
The second integral is $\Gamma(s) \zeta(s-1)$, which as Szeto stated in comments below can be derived by preforming integration by parts on the representation $$\Gamma(s) \zeta(s) = \int_{0}^{\infty} \frac{t^{s-1} e^{-t}}{1-e^{-t}} \, \mathrm dt.$$
Combing all 4 integrals, we get $$I(s) = \left(\Gamma(s-2) -\Gamma(s) \zeta(s-1) + \Gamma(s)\zeta(s) - \frac{1}{12} \,\Gamma(s)\right) \tag{2}$$ for $\Re(s) >2 $.
The Mellin transform, like the Laplace transform, defines an analytic function where the integral converges absolutely. Since integral $(1)$ behaves like a constant times $t^{s+1}$ near $t=0$, the integral defines an analytic function for $\Re(s)>−2$.
And expression $(2)$ is also an analytic function for $\Re(s) >-2$ if we assign it its limit values at $s=2$,$1$,$0$, and $-1$.
It then follows from the identity theorem that $$I(s) =\left(\Gamma(s-2) -\Gamma(s) \zeta(s-1) + \Gamma(s)\zeta(s) - \frac{1}{12} \,\Gamma(s)\right) $$ for $\operatorname{Re}(s) >-2$.
To find $I(0)$, we'll expand the terms in Laurent series at $s=0$.
The gamma function has simple poles at zero and the negative integers with residue $\frac{(-1)^{n}}{n!}$.
So at $s=-2$, $\Gamma(s) = \frac{1}{2(s+2)} + \mathcal{O}(1)$.
The constant term of the Laurent series of $\Gamma(s)$ at $s=-2$ is then $$ \begin{align} \lim_{s \to -2} \left(\Gamma(s) - \frac{1}{2(s+2)} \right) &= \lim_{s \to -2} \left(\frac{\Gamma(s+3)}{s(s+1)(s+2)} - \frac{1}{2(s+2)}\right) \\ &= \lim_{s \to -2} \frac{2 \Gamma(s+3)-s(s+1)}{2s(s+1)(s+2)} \\ &= \frac{1}{2}\lim_{ s \to -2} \frac{2 \Gamma'(s+3)-2s-1}{(s+1)(s+2)+s(s+2)+s(s+1)}\\ &= \frac{2\Gamma'(1)+3}{4}. \end{align}$$
Since the Laurent series of $\Gamma(s)$ at $s=-2$ has the same coefficients as the Laurent series of $\Gamma(s-2)$ at $s=0$, we get $$ \Gamma(s-2) = \frac{1}{2s} +\frac{\Gamma'(1)}{2} + \frac{3}{4} + \mathcal{O}(s).$$
Similarly, $$\Gamma(s) = \frac{1}{s} + \Gamma'(1) + \mathcal{O}(s).$$
Therefore, $$\begin{align} I(0) = \small \lim_{s \to 0} \Big[\frac{1}{2s} &+ \small \frac{\Gamma'(1)}{2} + \frac{3}{4} + \mathcal{O}(s) - \left(\frac{1}{s} + \Gamma'(1) + \mathcal{O}(s) \right) \left(\zeta(-1) - \zeta(0) + \zeta'(-1)s - \zeta'(0)s+ \mathcal{O}(s^{2}) \right) \\ &- \small \frac{1}{12} \left(\frac{1}{s} + \Gamma'(1) + \mathcal{O}(s) \right)\Big] \end{align} $$
where $\zeta(0) = - \frac{1}{2}$ and $\zeta(-1) = -\frac{1}{12}$.
This results in mass cancellation, and we end up with $$I(0) = \lim_{ s \to 0} \left( \frac{3}{4} - \zeta'(1) + \zeta'(0) + \mathcal{O}(s)\right) = \frac{3}{4} - \zeta'(1) +\zeta'(0). $$