I am trying to evaluate the integral \begin{align*} \Omega \equiv \int_0^\infty \frac{\mathrm{Ti}_2(x)}{e^{2 \pi x} - 1}dx\end{align*} where $\mathrm{Ti}_2(x)$ denotes the inverse tangent integral. I first attempted to convert the integral into an infinite series like this: \begin{align*} \Omega &= \int_0^\infty \frac{e^{-2\pi x} \mathrm{Ti}_2(x)}{1-e^{-2 \pi x}}\,dx \\ &= \int_0^\infty \sum_{n=1}^\infty e^{-2 \pi n x} \mathrm{Ti}_2(x)\,dx \\ &= \sum_{n=1}^\infty \int_0^\infty e^{-2 \pi n x} \mathrm{Ti}_2(x)\,dx \end{align*}
From here I am stuck evaluating the integral. I tried looking for whether the inverse tangent integral has a Laplace transform. However, I was not able to find or derive it. I also considered using the Abel-Plana formula alongside the known identity $$\mathrm{Ti}_2(x) = \frac{1}{2i}(\mathrm{Li}_2(ix)-\mathrm{Li}_2(-ix))$$ where $\mathrm{Li}_2(x)$ denotes the dilogarithm, but I did not know how to continue my steps from there. Is there a better way to attack this problem?
Too long for comments.
When I look at the following, I am not surprized by your difficulties
$$I_n=\int_0^\infty e^{-2 \pi n x}\,\, \mathrm{Ti}_2(x)\,dx $$
$$I_n=-\frac{\pi \text{Ci}(2 n \pi )+2 \text{Si}(2 n \pi )\, (\gamma+\log (2 \pi n))}{4 \pi n}+$$ $$\big(\, _3F_3(1,1,1;2,2,2;2 i n \pi )+\, _3F_3(1,1,1;2,2,2;-2 i n \pi )\big)-$$ $$\frac i{8n \pi}\big(\, _1F_1^{(2,0,0)}(0;1;2 i \pi n)-\, _1F_1^{(2,0,0)}(0;1;-2 i \pi n) \big)$$
The partial sums $$S_p=\sum_{n=1}^p I_n$$ converge very slowly because $$\frac {I_{n+1}}{I_n}\sim 1-\frac 2n$$
$$S_1=0.0249575 \qquad S_{10}=0.0388494\qquad S_{100}=0.0410079$$