Let $\operatorname{sech}(x) = t$ and integrate by parts to obtain
\begin{align}
-I &= \int \limits_0^\infty \frac{x \log(\cosh(x))}{\cosh^3(x)} \, \mathrm{d}x = \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t) t^2}{\sqrt{1-t^2}} \, \mathrm{d} t \\
&= \int \limits_0^1 \sqrt{1-t^2} \frac{\mathrm{d}}{\mathrm{d} t} \left[-\log(t) t \operatorname{arsech}(t)\right] \mathrm{d} t \\
&= \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t) (1-t^2)}{\sqrt{1-t^2}} \, \mathrm{d} t - \int \limits_0^1 \left[\sqrt{1-t^2} \operatorname{arsech}(t) - \log(t)\right] \mathrm{d} t \, .
\end{align}
Averaging the second and the fourth expression yields
$$- I = \frac{1}{2} \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t)}{\sqrt{1-t^2}} \, \mathrm{d} t - \frac{1}{2}\int \limits_0^1 \sqrt{1-t^2} \operatorname{arsech}(t)\, \mathrm{d} t - \frac{1}{2} \equiv J - K - \frac{1}{2}\, .$$
$K$ can be computed by reversing the previous substitution:
$$ K = \frac{1}{2} \int \limits_0^\infty \frac{x \sinh^2(x)}{\cosh^3(x)} \, \mathrm{d} x = \frac{1}{4} \int\limits_0^\infty \frac{\sinh(x) + x \cosh(x)}{\cosh^2(x)} \, \mathrm{d} x = \frac{1}{4}(1+2 \mathrm{G}) = \frac{1}{4} + \frac{\mathrm{G}}{2} \, . $$
For $J$ we can use $t = \frac{2u}{1+u^2}$ to find
\begin{align}
J &= \int \limits_0^1 \frac{\log(u) \log\left(\frac{2u}{1+u^2}\right)}{1+u^2} \, \mathrm{d} u \\
&= \int \limits_0^1 \frac{-\log(u) \log(1+u^2)}{1+u^2} \, \mathrm{d} u + \int \limits_0^1 \frac{\log^2(u)}{1+u^2} \, \mathrm{d} u - \log(2) \int \limits_0^1 \frac{-\log(u)}{1+u^2} \, \mathrm{d} u \\
&= 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) + \mathrm{G} \log(2) - \frac{\pi}{8} \log^2(2) - \frac{3 \pi^3}{32} + \frac{\pi^3}{16} - \mathrm{G} \log(2) \\
&= 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) - \frac{\pi}{8} \log^2(2) - \frac{\pi^3}{32} \, .
\end{align}
The first integral has been calculated here and the others are well-known special values of the Dirichlet beta function. Therefore,
$$ -I = J - K - \frac{1}{2} = 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) - \frac{\pi}{8} \log^2(2) - \frac{\pi^3}{32} - \frac{\mathrm{G}}{2} - \frac{3}{4} \, . $$
The problem with the original approach is that for $b=3$ the series only converges for $a > 1$, which leads to the two divergent series in the final answer. This can be avoided by computing the result for sufficiently large values of $a$ first and then taking the limit $a \to 1^+$, which can be justified by analytic continuation. $\Omega_3, \dots, \Omega_8$ are calculated as before after taking the limit inside the series, but the divergent terms are replaced by the regularised versions
$$ \Omega_1 = \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{(2n+1)^{a-1}} = \lim_{a \to 1^+} [1 - \beta(a-1)] = 1 - \beta (0) = \frac{1}{2}$$
and
\begin{align}
\Omega_2 &= \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1} H_n}{(2n+1)^{a-1}} = \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{(2n+1)^{a-1}} \int \limits_0^1 \frac{1 - x^n}{1-x} \, \mathrm{d} x \\
&= \lim_{a \to 1^+} \int \limits_0^1 \frac{\frac{\operatorname{Ti}_{a-1}(\sqrt{x})}{\sqrt{x}} - \beta(a-1)}{1-x} \, \mathrm{d} x = \int \limits_0^1 \frac{\frac{\operatorname{Ti}_{0}(\sqrt{x})}{\sqrt{x}} - \beta(0)}{1-x} \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{\frac{1}{1+x} - \frac{1}{2}}{1-x} \, \mathrm{d} x = \frac{1}{2}\int \limits_0^1 \frac{\mathrm{d} x}{1+x} = \frac{1}{2} \log(2) \, .
\end{align}
How about this:
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\;
\sum_{n=0}^{\infty}\frac{2}{\sqrt{\pi}}\int_n^{\infty}e^{-t^2}\, dt
$$
In this sum of integrals, the interval $[0,1)$ will be counted only once, in the $n = 0$ term. The interval $[1,2)$ will be counted twice, in the $n = 0$ and $n = 1$ terms. And so on. This means we can write:
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\;
\frac{2}{\sqrt{\pi}}\int_0^{\infty}\, \lfloor t+1\rfloor \,e^{-t^2}\, dt
$$
I don't know if this is the sort of alternative integral representation you were looking for.
Edited to add two other ways to write this expression:
First way: The quantity $\lfloor t+1\rfloor$ can be written as
$$
\lfloor t+1\rfloor \;=\; (t+1) \;-\; S(t)\, .
$$
where $S(t)$ is a sawtooth wave of period $1$ with minimum value $0$ and a maximum value $1$. One way we could write this sawtooth is as $S(t) \,=\, t\;\mathrm{mod}\;1$. Substituting this expression for $\lfloor t+1\rfloor$ into the integral above and using the fact that $\int_0^{\infty} dt \, (t+1)\,e^{-t^2} = (1+\sqrt{\pi})/2$ yields
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\;
\frac{2}{\sqrt{\pi}}\left[\frac{1}{2}(1+\sqrt{\pi}) \;-\; \int_0^{\infty}dt\, S(t)\, e^{-t^2}\right].
$$
Evaluating this numerically in Mathematica with 20 digits of precision yields $1.16200283409802758182$, which is greater than Mathematica's estimate for the original sum by roughly $3.8\times {10}^{-6}$. Close enough for the vagaries of numerically evaluating weird sums and integrals.
Second way: Poisson's summation formula states that if $f(x)$ is a function and
$$
\hat{f}(q) \;=\; \int_{-\infty}^{+\infty} dx\, e^{-i q x}\, f(x)
$$
is its Fourier transform, then
$$
\sum_{n = -\infty}^{+\infty} f(n) \;=\; \sum_{n=-\infty}^{+\infty} \hat{f}(2\pi n)\, .
$$
Define the even function $f(x) = \mathrm{erfc}(|x|)$. Since $f$ is even, we can write the original sum as
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\; \frac{1}{2}f(0) \,+\,\frac{1}{2}\sum_{n = -\infty}^{+\infty} f(n)
\;=\; \frac{1}{2} \,+\,\frac{1}{2}\sum_{n = -\infty}^{+\infty} \hat{f}(2\pi n)\, .\hspace{0.5in}\text{(1)}
$$
The Fourier transform of $f$ is:
\begin{align}
\hat{f}(q) &\;=\; \int_{-\infty}^{+\infty} dx\, e^{-i q x}\, \mathrm{erfc}(|x|)\\[0.1in]
&\;=\; 2\int_0^{+\infty} dx\, \cos(q x)\, \mathrm{erfc}(x)\hspace{0.5in}\text{Since $\mathrm{erfc}(|x|)$ is even}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dx\, \cos(q x)\,\int_x^{\infty}dt\, e^{-t^2} \hspace{0.5in}\text{Definition of $\mathrm{erfc}$}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dt\, e^{-t^2}\,\int_0^t dx\, \cos(q x) \hspace{0.5in}\text{Reverse order of integration}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dt\, e^{-t^2}\,\frac{\sin(q t)}{q}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\frac{D(q/2)}{q}
\end{align}
In the last line above, "$D$" is the Dawson function according to the third definition here.
This expression for $\hat{f}(q)$ is valid everywhere except at $q = 0$, where
$\hat{f}(0) = 2/\sqrt{\pi}$. Plugging all of this into (1) results in:
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n)
\;=\; \frac{1}{2}\left[1 + \frac{2}{\sqrt{\pi}} \,+\,\frac{8}{\sqrt{\pi}}\sum_{n = 1}^{\infty} \frac{D(\pi n)}{2\pi n}\right]\, .
$$
Evaluating this numerically in Mathematica yields $1.16199904795$.
Best Answer
This is how I see the theory behind those things, which means there could very well be a simpler and more concrete way to solve it in this particular case.
We can evaluate $$\sum_{m,n}\ \frac{1}{(m+\frac12+ni)^{2k}},\qquad \sum_{(m,n)\ne (0,0)}\ \frac{1}{(m+ni)^{2k}}$$ in term of $\wp_i^{(k-2)}(1/2)$ and $(\wp_i^{(k-2)}-z^{-1})(0)$ where $\wp_i(z) = \sum_{m,n}\frac{1}{(z+m+ni)^2}-\frac{1_{(m,n)\ne (0,0)}}{(m+in)^2}$,
which reduces to show that $\wp'(z)^2-4\wp(z)^3-g_2(i) \wp(z)=0$ (bounded entire function) then find $g_2(i)$ from $1=\int_0^1 dz=\int_0^1 \frac{d\wp(z)}{\sqrt{4\wp(z)^3+g_2(i)\wp(z)}}=\int_C \frac{dx}{\sqrt{4x^3+g_2(i)x}}$ $=A g_2(i)\int_0^1 \frac{du}{\sqrt{u^3-1}}=2A g_2(i)B(1/2,3/4)=C g_2(i)\Gamma(1/4)^2$.
This way we'll obtain some recursive formulas for the derivatives of $\wp_i$ in term of $\wp_i,\wp_i'$ which will give for the values at $0,1/2$ a rational expression in term of $g_2(i)$.
For $f$ a level $N$ modular form with integer coefficients then the coefficients of $F_z(X)=\prod_{\gamma\in SL_2(Z)/\Gamma(N)}(X-f|_k\gamma)$ are level $1$ weight $\le m$ modular forms thus in $\sum_{4a+6b=m} \Bbb{Q} E_4^aE_6^b$ so that $f(i)$ is algebraic over $E_4(i)$ (as $E_6(i)=0$), and the complex multiplication stuffs show that in $F_i(X)$ we can replace $SL_2(Z)/\Gamma(N)$ by an abelian group (the class group of $\Bbb{Z}[Ni]$) so that $f(i)$ is radical over $E_4(i)$ whose $\Gamma(1/4)$ expression is found from the elliptic way.