We write
$$
\frac1{1+2\cosh x}=\frac{e^{-x}}{1+e^{-x}+e^{-2x}}\,,\quad t=e^{-x}\in(0,1)\,,
$$
so that
$$
\frac1{1+2\cosh x}=\frac{t}{1+t+t^2}\,.
$$
Expanding as a power series in $t$ gives a repeating pattern. In fact one finds
$$
\frac{1}{1+t+t^2}=1 -t +0\cdot t^2 +t^3 - t^4 +0\cdot t^5 + \cdots,
$$
so that
$$
\frac{1}{1+2\cosh x}=\sum_{n=0}^\infty c_n e^{-(n+1)x},
$$
where the coefficients $c_n$ are periodic of period 3: $c_{3m+1}=1,\;c_{3m+2}=-1,\;c_{3m+3}=0$. (This follows directly by solving the recurrence $c_n + c_{n-1}+c_{n-2}=0$ with $c_0=1,c_1=-1$, etc.) Hence
$$
\int_0^\infty \frac{\cos(ax^2)}{1+2\cosh x}\,dx
=\sum_{n=0}^\infty c_n \int_0^\infty e^{-(n+1)x}\cos(a x^2)\,dx.
$$
Each integral
$$
I(k)=\int_0^\infty e^{-k x}\cos(a x^2)\,dx,\qquad k>0,
$$
can be evaluated by expanding $\cos(a x^2)$ in its Taylor series:
$$
\cos(a x^2)=\sum_{p=0}^\infty \frac{(-1)^p}{(2p)!}(a x^2)^{2p}
=\sum_{p=0}^\infty \frac{(-1)^p a^{2p}}{(2p)!}x^{4p}.
$$
Substituting and integrating termwise (justified by uniform convergence of the exponential series for $x\ge0$) gives
$$
I(k)=\sum_{p=0}^\infty \frac{(-1)^p a^{2p}}{(2p)!}\int_0^\infty x^{4p} e^{-k x}\,dx.
$$
But by the gamma-function identity
$\Gamma(n+1)=\int_0^\infty t^n e^{-t}dt$ we have
$\int_0^\infty x^n e^{-k x}dx = n!/k^{\,n+1}$ for Re$(k)>0$. Hence
$$
\int_0^\infty x^{4p}e^{-k x}\,dx = \frac{(4p)!}{k^{4p+1}}.
$$
Thus
$$
I(k)=\sum_{p=0}^\infty \frac{(-1)^p (4p)!}{(2p)!}\,\frac{a^{2p}}{k^{4p+1}}.
$$
Returning to the original sum, with $k=n+1$, we get
$$
\int_0^\infty \frac{\cos(ax^2)}{1+2\cosh x}\,dx
=\sum_{n=0}^\infty c_n I(n+1)
=\sum_{n=0}^\infty c_n \sum_{p=0}^\infty \frac{(-1)^p (4p)!}{(2p)!}\frac{a^{2p}}{(n+1)^{4p+1}}.
$$
Interchanging sums (again justified by uniform convergence for each fixed $a$) yields
$$
=\sum_{p=0}^\infty \frac{(-1)^p (4p)!}{(2p)!}\,a^{2p}
\sum_{n=0}^\infty\frac{c_n}{(n+1)^{4p+1}}.
$$
Observe that $c_n=a_{n+1}$ where $a_k$ is the periodic arithmetic function $a_{3m+1}=1,\;a_{3m+2}=-1,\;a_{3m}=0$. In other words, $a_k=\chi_3(k)$ is the non-principal Dirichlet character mod 3 (with $\chi_3(3m+1)=1,\;\chi_3(3m+2)=-1$). Therefore
$$
\sum_{n=0}^\infty \frac{c_n}{(n+1)^s}
=\sum_{k=1}^\infty \frac{a_k}{k^s}
=\sum_{k=1}^\infty \frac{\chi_3(k)}{k^s},
$$
which by definition is the Dirichlet $L$-series $L(s,\chi_3)$. (MathWorld defines $L(s,\chi)=\sum_{n=1}^\infty \chi(n)n^{-s}$.) Hence for each $p$ we have
$$
\sum_{n=0}^\infty \frac{c_n}{(n+1)^{4p+1}}
=\sum_{k=1}^\infty \frac{\chi_3(k)}{k^{4p+1}}
=L(4p+1,\chi_3).
$$
Putting everything together gives the final result in series form:
$$
\boxed{\;\int_0^\infty \frac{\cos(ax^2)}{1+2\cosh x}\,dx
=\sum_{p=0}^\infty (-1)^p\,\frac{(4p)!}{(2p)!}\,a^{2p}\;L(4p+1,\chi_3)\,.}
$$
Here $\chi_3$ is the non-principal real character mod 3, so that $L(s,\chi_3)=\sum_{k\equiv1(3)}k^{-s}-\sum_{k\equiv2(3)}k^{-s}$. (For example, at $a=0$ the integral becomes $L(1,\chi_3)=\pi/(3\sqrt3)$, which agrees with direct evaluation.) This completes the derivation without recourse to contour methods.
