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)
$$\begin{array}{cccc}
\text{K} & \text{Formula (12)} & \text{Formula (13)} & \text{Sum of Formulas (12) and (13)} \\
10 & -0.63943 & -0.716102 & -1.35553 \\
100 & -0.645717 & -0.881106 & -1.52682 \\
1000 & -0.654904 & -0.86869 & -1.52359 \\
10000 & -0.653089 & -0.861175 & -1.51426 \\
100000 & -0.652134 & -0.858876 & -1.51101 \\
\end{array}$$
Best Answer
Not sure if this counts as "backtracking" since your answer in the first link already mentions this equation
$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}=\frac 2{\sqrt\pi}\sum\limits_{k=0}^{+\infty}(-1)^k\int\limits_0^1t^{k-3/4}(1-t)^{1/2}\,\mathrm dt$$
But it is easier to evaluate the sum by employing the Beta function and integrating the sum of the expression inside the integral. Swapping the order of the integral and summation before applying the geometric series leads to
$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}=\frac 2{\sqrt\pi}\int\limits_0^1\frac {\sqrt{1-t}}{t^{3/4}(1+t)}\,\mathrm dt$$
This resulting integral might seem more convoluted, however the simple substitution $u=\frac {1-t}{1+t}$ reduces it into another form of the beta function which can then be further simplified using Euler's Reflection formula.
\begin{align*} \int\limits_0^1\frac {\sqrt{1-t}}{t^{3/4}(1+t)}\,\mathrm dt & =\sqrt2\int\limits_0^1u^{1/2}\left(1-u^2\right)^{-3/4}\,\mathrm du\\ & =\frac 1{\sqrt2}\int\limits_0^1u^{-1/4}(1-u)^{-3/4}\,\mathrm du\\ & =\frac 1{\sqrt2}\Gamma\left(\frac 14\right)\Gamma\left(\frac 34\right)\\ & =\pi \end{align*}
This proves the result.
$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}\color{blue}{=2\sqrt\pi}$$