I usually don't answer my own questions, but I literally just derived the solution and I think it's beautiful.
So we know from the link in the OP that:
$$\zeta(s,a)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{z^{s-1} dz}{e^{a z} (1-e^{-z})}$$
Which makes our expression:
$$\zeta \left(\frac12, \frac{t}{2}\right)-\zeta \left(\frac12, \frac{t+1}{2}\right)=\frac{1}{\sqrt{\pi}} \int_0^\infty \frac{e^{- t z/2}(1-e^{-z/2}) dz}{(1-e^{-z}) \sqrt{z}}=$$
$$=\frac{2}{\sqrt{\pi}} \int_0^\infty \frac{e^{- t u^2/2}(1-e^{-u^2/2}) du}{1-e^{-u^2}}$$
Now we can see that it's very easy to take integral over $t$ (integration by parts):
$$\int_0^1 \sin \pi t~ e^{- t u^2/2} dt=\frac{4 \pi}{4 \pi^2 +u^4} (1+e^{-u^2/2})$$
Now we substitute this into the second integral to get (this is the beautiful part):
$$8 \sqrt{\pi} \int_0^\infty \frac{(1+e^{-u^2/2})(1-e^{-u^2/2}) du}{(1-e^{-u^2})(4 \pi^2 +u^4)}=8 \sqrt{\pi}\int_0^\infty \frac{du}{4 \pi^2 +u^4}$$
After a change of variables we have:
$$\frac{2 \sqrt{2}}{\pi} \int_0^\infty \frac{dv}{1 +v^4}=\frac{2 \sqrt{2}}{\pi} \frac{\pi}{2 \sqrt{2}}=1$$
Just as it was supposed to be.
God, Mathematics is simply perfect sometimes.
Appendix
Trying to prove in a simple way that the integral formula for $\zeta(1/2,a)$ equals the series definition.
$$ \int_0^\infty \frac{z^{-1/2} dz}{e^{a z} (1-e^{-z})}=2 \int_0^\infty \frac{e^{-a u^2}du}{ 1-e^{-u^2}}=2 \sum_{n=0}^\infty \int_0^\infty e^{-(a+n) u^2} du=$$
$$=2 \sum_{n=0}^\infty \frac{1}{\sqrt{a+n}} \int_0^\infty e^{-w^2} dw= \sqrt{\pi} \sum_{n=0}^\infty \frac{1}{\sqrt{a+n}} $$
Formally, this exactly fits the series definition, but it doesn't converge (it's alright, as the function for $s=1/2$ is defined by analytic continuation).
On the other hand, for the particular function in my case, the proof works well, as the series inside the integral becomes alternating due to a factor $(1-e^{-u^2/2})$ in the numerator, so everyting converges.
I would say that all of this constitutes a nice real proof of the Fresnel integrals, especially since the Poisson integral also has a few real proofs.
Though I'm definitely not claiming this is a new result. It was new for me, but a two second google search found me this paper https://www.jstor.org/stable/2320230, and I'm sure there's plenty more.
I will concentrate on this part of your proof, the calculation of this integral (the other steps of your proof are correct):
$$\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx$$
We will make use of the Lobachevsky integral formula:
Let $f(x)$ a $\pi$-periodic function (continous or integrable over its period). with
$\displaystyle f(x+\pi) = f(x)$ and $f(\pi-x) = f(x)$, for $0\leq x <\infty$. Then
$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} f(x) dx = \int_{0}^{\frac{\pi}{2}} f(x) dx. $
Now, for your integral:
If you make the change of variable $\displaystyle x=\frac{w}{2}$
$\displaystyle I=\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx = \int_{0}^{\infty} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{w} dw $
Multiplying and dividing by $\sin w$:
$$I= \int_{0}^{\infty} \frac{\sin w}{w} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w} dw$$
The function
$$f(w) = \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}$$
is $\pi$-periodic
By the Lobachevsky integral formula:
$$\displaystyle I=\int_{0}^{\infty} \frac{\sin w}{w} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w} dw = \int_{0}^{\frac{\pi}{2}}\frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}dw $$
Note that for $w\in\left(0,\frac{\pi}{2}\right)$
$$\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big|= \sin\left(\frac{w}{2}+\frac{\pi}{4}\right)$$
$$\Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|= -\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)$$
Hence
$$\displaystyle\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big| = \sqrt{2}\sin\left(\frac{w}{2}\right)\displaystyle $$
Therefore
$$ I = \int_{0}^{\frac{\pi}{2}}\frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}dw = \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{w}{2}\right)}{\sin w}dw$$
Using $\displaystyle \frac{\sin\left(\frac{w}{2}\right)}{\sin w} = \frac{1}{2}\sec\left(\frac{w}{2}\right)$
we have
$$I = \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{w}{2}\right)}{\sin w}dw = \frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}} \sec\left(\frac{w}{2} \right)dw = \sqrt{2} \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)$$
We can conclude
$$\boxed{\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx = \sqrt{2} \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)}$$
Therefore
$$\boxed{\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+ \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx = \frac{\pi}{2} +\operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)} $$
Some numerical approximations from Wolfram:
The sum is 2.44206
while
$$\frac{\pi}{2} + \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)\approx 2.45216..$$
Best Answer
I think integrating by parts works in most cases where there's a combination of $e^x$ and some other random function and of course we expect an elementary primitive. I will continue what you did: $$I=\frac12 \int_0^\frac{\pi}{6} e^x \frac{1+\cos x}{\color{red}{1-\sin x}}dx=\frac12 \int_0^\frac{\pi}{6} e^x (1+\cos x)\color{red}{\left(\frac{\cos x}{1-\sin x}\right)'}dx$$ $$=\frac12 e^x (1+\cos x)\left(\frac{\cos x}{1-\sin x}\right)\bigg|_0^\frac{\pi}{6}-\frac12 \int_0^\frac{\pi}{6} e^x \frac{1-\sin x+\cos x}{1-\sin x}\cos xdx$$ $$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac12\int_0^\frac{\pi}{6}e^x\left(\cos x +\frac{{\cos^2 x}}{1-\sin x}\right)dx$$ $$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac12\int_0^\frac{\pi}{6}e^x(\cos x+1+\sin x)dx$$ $$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac12 e^x(1+\sin x)\bigg|_0^\frac{\pi}{6}$$$$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac34 e^\frac{\pi}{6}+\frac12=e^{\large \frac{\pi}{6}}\frac{\sqrt 3}{2}-\frac12 $$
Of course we also have: $$\int e^x\frac{1+\cos x}{1-\sin x}dx=e^x(1+\cos x)\frac{\cos x}{1-\sin x}-e^x(1+\sin x)+C=\frac{e^x\cos x}{1-\sin x}+C$$