This answer is based on the Abel-Plana formula
$$\sum\limits_{n=0}^\infty f(n)=\int\limits_0^\infty f(x)\,dx+\frac{1}{2}f(0)+i\int\limits_0^\infty\frac{f(i t)-f(-i t)}{e^{2 \pi t}-1}\,dt\tag{1}$$
where
$$f(x)=C(x)+S(x)-\sqrt{\frac{\pi}{2}}\tag{2}$$
$$C(x)=\int\limits_0^x\cos\left(t^2\right)\,dt\tag{3}$$
$$S(x)=\int_\limits0^x\sin\left(t^2\right)\,dt\tag{4}$$
which leads to
$$\sum\limits_{n=0}^\infty\left(C(n)+S(n)-\sqrt{\frac{\pi}{2}}\right)=\int_0^\infty\left(C(x)+S(x)-\sqrt{\frac{\pi}{2}}\right)\,dx+\frac{1}{2}\left(C(0)+S(0)-\sqrt{\frac{\pi}{2}}\right)+i\int_0^\infty\frac{C(i t)+S(i t)-\sqrt{\frac{\pi}{2}}-\left(C(-i t)+S(-i t)-\sqrt{\frac{\pi}{2}}\right)}{e^{2 \pi t}-1}\,dt\ .\tag{5}$$
Based on the identities $C(i t)=i\,C(t)$, $S(i t)=-i\,S(t)$, $C(-i t)=-i\,C(t)$, and $S(-i t)=i\,S(t)$ and noting that $\frac{1}{2}\left(C(0)+S(0)-\sqrt{\frac{\pi}{2}}\right)=-\frac{1}{2}\sqrt{\frac{\pi }{2}}$ formula (5) above simplifies to formula (6) below.
$$\sum\limits_{n=0}^\infty\left(C(n)+S(n)-\sqrt{\frac{\pi}{2}}\right)=\int\limits_0^\infty\left(C(x)+S(x)-\sqrt{\frac{\pi}{2}}\right)\,dx-\frac{1}{2}\sqrt{\frac{\pi}{2}}+2\int\limits_0^\infty\frac{S(t)-C(t)}{e^{2 \pi t}-1}\,dt\tag{6}$$
Observational convergence of $\int_0^y\left(C(x)+S(x) -\sqrt{\frac \pi 2}\right) dx\to -\frac{1}{2}$ as $y\to\infty$ leads to formula (7) below.
$$\underset{K\to \infty}{\text{lim}}\left(\sum\limits_{n=0}^K\left(C(n)+S(n)-\sqrt{\frac{\pi}{2}}\right)\right)=-\frac{1}{2}-\frac{1}{2}\sqrt{\frac{\pi}{2}}+2\int\limits_0^\infty\frac{S(t)-C(t)}{e^{2 \pi t}-1}\,dt\tag{7}$$
The integral in formula (7) above can be evaluated using the identity
$$\frac{1}{e^{2 \pi t}-1}=\sum\limits_{n=1}^{\infty} e^{-2 \pi n t}\tag{8}$$
which leads to the following series representation.
$$2\int\limits_0^\infty\frac{S(t)-C(t)}{e^{2 \pi t}-1}\,dt=\underset{N\to \infty}{\text{lim}}\left(\frac{1}{\pi}\sum\limits_{n=1}^N\frac{\left(\cos\left(\pi^2 n^2\right)-\sin\left(\pi^2 n^2\right)\right)S(\pi n)-\left(\cos\left(\pi^2 n^2\right)+\sin\left(\pi^2 n^2\right)\right)C(\pi n)+\sqrt{\frac{\pi}{2}} \sin\left(\pi^2 n^2\right)}{n}\right)\tag{9}$$
The following table illustrates evaluation of the right side of formula (7) above using formula (9) above for several values of the upper evaluation limit $N$. The evaluations in the table below are consistent with evaluation of the right side of formula (7) above using numerical integration.
Table (1): Evaluation of the right side of formula (7) using formula (9)
$$\begin{array}{cc}
\text{N} & \text{Formula (7)} \\
10 & -1.20211 \\
100 & -1.20643 \\
1000 & -1.20688 \\
10000 & -1.20693 \\
100000 & -1.20693 \\
\end{array}$$
However there seems to be an unresolved discrepancy between evaluation of the right side of formula (7) above using numerical integration (or using the series representation of the integral defined in formula (9) above) and the following table which illustrates evaluation of the left side of formula (7) above using several values of the upper evaluation limit $K$.
Table (2): Evaluation of the left side of formula (7)
$$\begin{array}{cc}
\text{K} & \text{Formula (7)} \\
10 & -1.35553 \\
100 & -1.52682 \\
1000 & -1.52359 \\
10000 & -1.51426 \\
100000 & -1.51101 \\
\end{array}$$
Separating formula (2) for $f(x)$ into the two functions
$$f_c(x)=C(x)-\frac{1}{2}\sqrt{\frac{\pi}{2}}\tag{10}$$
$$f_s(x)=S(x)-\frac{1}{2}\sqrt{\frac{\pi}{2}}\tag{11}$$
where $f(x)=f_c(x)+f_s(x)$ leads to
$$\sum\limits_{n=0}^\infty\left(C(n)-\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)=-\frac{1}{4}\sqrt{\frac{\pi}{2}}-2\int\limits_0^\infty\frac{C(t)}{e^{2 \pi t}-1}\,dt\tag{12}$$
$$\sum\limits_{n=0}^\infty\left(S(n)-\frac{1}{2}\sqrt{\frac{\pi }{2}}\right)=-\frac{1}{2}-\frac{1}{4}\sqrt{\frac{\pi}{2}}+2\int\limits_0^\infty\frac{S(t)}{e^{2 \pi t}-1}\,dt\tag{13}$$
where the two integrals can be evaluated as
$$-2\int\limits_0^\infty\frac{C(t)}{e^{2 \pi t}-1}\,dt=\underset{N\to \infty}{\text{lim}}\left(\frac{1}{\pi}\sum\limits_{n=1}^N\frac{\sin\left(\pi^2 n^2\right)\left(\frac{1}{2}\sqrt{\frac{\pi}{2}}-C(\pi n)\right)+\cos\left(\pi^2 n^2\right)\left(S(\pi n)-\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)}{n}\right)\tag{14}$$
$$2\int\limits_0^\infty\frac{S(t)}{e^{2 \pi t}-1}\,dt=\underset{N\to \infty}{\text{lim}}\left(-\frac{1}{\pi}\sum\limits_{n=1}^N\frac{\cos\left(\pi^2 n^2\right)\left(C(\pi n)-\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)+\sin\left(\pi^2 n^2\right)\left(S(\pi n)-\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)}{n}\right)\tag{15}$$
Note the sum of formulas (12) and (13) above is consistent with formula (7) above, and the sum of formulas (14) and (15) above is consistent with formula (9) above.
The following table illustrates evaluations of the right sides of formulas (12) and (13) above using formulas (14) and (15) above for several values of the upper evaluation limit $N$. The evaluations in the table below are consistent with evaluations of the right sides of formulas (12) and (13) above using numerical integration. Note the right column of the table below is consistent with the evaluations in the corresponding Table (1) above.
Table (3): Evaluations of the right sides of formulas (12) and (13) using formulas (14) and (15)
$$\begin{array}{cccc}
\text{N} & \text{Formula (12)} & \text{Formula (13)} & \text{Sum of Formulas (12) and (13)} \\
10 & -0.391473 & -0.810639 & -1.20211 \\
100 & -0.39579 & -0.810638 & -1.20643 \\
1000 & -0.396243 & -0.810638 & -1.20688 \\
10000 & -0.396289 & -0.810638 & -1.20693 \\
100000 & -0.396293 & -0.810638 & -1.20693 \\
\end{array}$$
However there seems to be an unresolved discrepancy between evaluation of the right sides of formulas (12) and (13) above using numerical integration (or using the series representations of the integrals defined in formulas (14) and (15) above) and the following table which illustrates evaluations of the left sides of formula (12) and (13) above using several values of the upper evaluation limit $K$. Note the right column of the table below is consistent with the evaluations in the corresponding Table (2) above.
Table (4): Evaluations of the left sides of formulas (12) and (13)
$$\begin{array}{cccc}
\text{K} & \text{Formula (12)} & \text{Formula (13)} & \text{Sum of Formulas (12) and (13)} \\
10 & -0.63943 & -0.716102 & -1.35553 \\
100 & -0.645717 & -0.881106 & -1.52682 \\
1000 & -0.654904 & -0.86869 & -1.52359 \\
10000 & -0.653089 & -0.861175 & -1.51426 \\
100000 & -0.652134 & -0.858876 & -1.51101 \\
\end{array}$$
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
One way is to use the power series trick: For $1\leq m<n$, we have
\begin{align} S(m,n)&:=\sum_{k=0}^\infty\frac{(-1)^k}{nk+m}\\[.4em] &=\sum_{k=0}^\infty\int_0^1(-1)^kx^{nk+m-1}dx\\[.4em] &=\int_0^1x^{m-1}\sum_{k=0}^\infty(-x^n)^kdx\\[.4em] &=\int_0^1\frac{x^{m-1}}{x^n+1}dx. \end{align}
Now for your generalisation, taking $(m,n)=(m+1,2m)$ leads to
$$S(m+1,2m)=\int_0^1\frac{x^m}{x^{2m}+1}dx=\frac1m\int_0^{\pi/4}(\tan u)^\frac1mdu$$
where the second step is by the substitution $x=(\tan u)^\frac1m$.
If $1\leq m<n$ are integers, we can evaluate the integral using digamma function $\psi$, that is,
\begin{align} S(m,n)&=\frac1{2n}\left(\psi\left(\frac{m+n}{2n}\right)-\psi\left(\frac{m}{2n}\right)\right)\\ &=\frac\pi{2n}\csc\left(\frac{m\pi}n\right)+\frac2n\sum_{k=1}^{n-1}\ln\left(\csc\left(\frac{k\pi}{2n}\right)\right)\sin\left(\frac{mk\pi}n+\frac{k\pi}2\right)\sin\left(\frac{k\pi}2\right). \end{align}
For the first equality you can find a proof in this post, and for the second we use Gauss's digamma theorem with a few trig identities.