Finding the leading order contribution to a certain integral. I am trying to compute the leading order term of the following expression in the small $\epsilon$ limit;
$$
I
=
\frac{d}{ds}\biggr|_{s\rightarrow 0}\frac{1}{\Gamma(s)}
\int_{0}^{\infty} dt 
\frac{t^{s-1}e^{itx}}{(1-e^{i\epsilon_{1}t}) (1-e^{i\epsilon_{2} t})}
$$
First of all I tried expanding the exponentials with $\epsilon$'s in them, leading to 
$$
-\frac{1}{\epsilon_{1}\epsilon_{2}}
\frac{d}{ds}\biggr|_{s\rightarrow 0}\frac{1}{\Gamma(s)}
\int_{0}^{\infty} 
t^{s-3}e^{itx}
dt
$$
I'm not sure how valid this is given that $\epsilon t=\mathcal{O}(1)$ in the large $t$ region, but it's all I could think of doing for now.
From here I noticed that the integral looked very similar to the gamma function. I tried changing variables to convert it to something involving the gamma function, but the integration limits were giving me issues.
Next I tried taking the derivative inside the integral. I found that $\left(\frac{t^{s}}{\Gamma(s)}\right)'\biggr|_{s=0}=-1$, resulting in the following expression:
$$
\frac{1}{\epsilon_{1}\epsilon_{2}}
\int_{0}^{\infty} 
t^{-3}e^{itx}
dt
$$
This looks relatively simple, but evaluating the antideriavtive gave an expression involving triginometric integrals which are divergent at zero. This makes me think that maybe one of my approximations is invalid.
I am quite certain that the resulting expression should be $\frac{1}{2\epsilon_{1}\epsilon_{2}}x^{2}(\log(x)-\frac{3}{2})$, and would really appreciate some help with proving it.
 A: The OP said in the comments 

I am starting to think that is the case. Without the $i$ the integral is easily transformed into a gamma function giving the result.

Regarding this:


*

*According to arXiv:hep-th/0306238 p.78 the $i$ is absent, therefore I interpret you question as 



How to prove $$\frac{d}{ds}\biggr|_{s\rightarrow 0}\frac{1}{\Gamma(s)}
\int_{0}^{\infty} dt 
\frac{t^{s-1}e^{-tx}}{(e^{at}-1) (e^{bt}-1)}\stackrel{a,b\to0^+}{\sim}\frac{x^2}{ab}\left(\frac34-\frac{\log x}{2}\right)\qquad(\star)$$ under zeta-regularization?

(Yes, the OP missed a minus sign in the conjectured form. Also, the mentioned arXiv paper states that the definition is true only under zeta-regularization.)


*
Without the $i$ the integral is easily transformed into a gamma function giving the result.

This is true if you completely neglect convergence issues:
$$\begin{align}
\frac{1}{\Gamma(s)}
\int_{0}^{\infty} dt 
\frac{t^{s-1}e^{-tx}}{(e^{at}-1) (e^{bt}-1)}
&\sim \frac{1}{\Gamma(s)}\int_{0}^{\infty} dt \frac{t^{s-1}e^{-tx}}{at\cdot bt} \\
&= \frac{1}{ab\Gamma(s)}\int_{0}^{\infty} t^{s-3}e^{-tx}dt \\
&= \frac{1}{ab\Gamma(s)}\frac{\Gamma(s-2)}{x^{s-2}} \\
&\stackrel{\frac{d}{ds},s\to0}{=}\frac1{ab}\left(\frac34-\frac{\log x}{2}\right)
\end{align}
$$
However the first and third line is not rigorous.
Below is my attempt to rigorously prove $(\star)$, and it is nearly complete - I only failed to solve a definite integral that numerically equals the expected $\frac34$.

Define $$F(s)\Gamma(s):=\int_{0}^{\infty} 
\frac{t^{s-1}e^{-tx}}{(e^{at}-1) (e^{bt}-1)}dt$$
or equivalently $$\alpha\beta x^s F(s)\Gamma(s):=\int_{0}^{\infty} 
t^{s-1}e^{-t}\cdot\underbrace{\frac{\alpha\beta}{(e^{\alpha t}-1) (e^{\beta t}-1)}}_{g(t)}dt \qquad (1)$$
where $\alpha=\frac ax,\beta=\frac bx$.
Note that $(\star)$ can be rewritten as 

$$\alpha\beta F'(0)\stackrel{\alpha,\beta\to0^+}{\sim}\frac34-\frac{\log x}{2}$$

By series expansion, $$g(t)=\frac1{t^2}-\frac{\alpha+\beta}{2}\frac1t+\left(\frac{\alpha^2+3\alpha\beta+\beta^2}{12}\right)+O(t)$$
Since $s$ is near to $0$, the integral is not integrable at $t=0$. We would like to remove $t^{-2}, t^{-1}, t^0$ terms from $g(t)$ to regularize the integral in $(1)$. Noting that 
$$\Gamma(s)\zeta(s)=\int^\infty_0 t^{s-1} e^{-t}\cdot\frac1{1-e^{-t}}dt$$
$$\Gamma(s)\zeta(s-1)=\int^\infty_0 t^{s-1} e^{-t}\cdot\frac1{(1-e^{-t})^2}dt$$
we apply zeta-regularization by subtracting $\frac1{1-e^{-t}},\frac1{(1-e^{-t})^2}$ (which are $\sim t^{-1}$ and $\sim t^{-2}$ respectively) from $g(t)$.
After tedious algebra, we find that
$$H(t):=g(t)-G(t)\in O(t)$$
$$G(t)=\frac1{(1-e^{-t})^2}-\underbrace{\left(1+\frac{\alpha+\beta}2\right)}_{k_1}\frac1{1-e^{-t}}+\underbrace{\left(\frac1{12}+\frac{\alpha+\beta+\alpha\beta}{4}+\frac{\alpha^2+\beta^2}{12}\right)}_{k_2}$$
Then,
$$\begin{align}
\alpha\beta x^s F(s)\Gamma(s)
&=\int^\infty_0 t^{s-1} e^{-t}H(t)dt+\int^\infty_0 t^{s-1} e^{-t}G(t)dt \\
&=\int^\infty_0 t^{s-1} e^{-t}H(t)dt+\Gamma(s)\zeta(s-1)-k_1\Gamma(s)\zeta(s)+k_2\Gamma(s) \\
\alpha\beta F(s)&=\frac1{\Gamma(s)}\int^\infty_0 \left(\frac tx\right)^s \frac{H(t)}{t}e^{-t}dt+[\zeta(s-1)-k_1\zeta(s)+k_2]x^{-s} \\
\end{align}
$$
Differentiating and taking $s\to 0$,
$$
\alpha\beta F'(0)=-(\zeta(-1)-k_1\zeta(0)+k_2)\log x+\zeta'(-1)-k_1\zeta'(0)
+\int^\infty_0 \frac{H(t)}{t}e^{-t}dt
$$
Taking $\alpha,\beta\to 0^+$, we have $k_1\to 1, k_2\to\frac1{12}$, $g(t)\to \frac1{t^2}$, 
$$H(t)\to \frac1{t^2}-\frac1{(1-e^{-t})^2}+\frac1{1-e^{-t}}-\frac1{12}$$
Plugging in $\zeta$ values, 
$$\begin{align}
\alpha\beta F'(0)&\stackrel{\alpha,\beta\to0^+}{\sim} -\frac{\log x}2
+\zeta'(-1)-\zeta'(0) \\
&\,\,\,\,\qquad +\int^\infty_0 \frac{e^{-t}}{t}\left[\frac1{t^2}-\frac1{(1-e^{-t})^2}+\frac1{1-e^{-t}}-\frac1{12}\right]dt \\
\end{align}
$$
Numerically, the constants sum up to $\frac34$, but I have no idea how the integral can be solved analytically.
EDIT: the integral was solved by our integration master @RandomVariable here, and it is equal to $\frac34-\zeta'(-1)+\zeta'(0)$.
Therefore, we arrived at the desired result

$$\alpha\beta F'(0)\stackrel{\alpha,\beta\to0^+}{\sim} -\frac{\log x}2+\frac34$$

