I've been trying to prove Fresnel integrals by real methods and encountered an interesting problem.
Let's start with the known result:
$$\int_0^\infty \sin y^2 dy = \sqrt{\frac{\pi}{8}}$$
Can we prove it without complex methods?
I have tried to do the following:
$$\int_0^\infty \sin y^2 dy =\frac{1}{2} \int_0^\infty \sin x \frac{dx}{\sqrt{x}} =\frac{1}{2} \sum_{n=0}^\infty (-1)^n \int_0^\pi \sin x \frac{dx}{\sqrt{x+\pi n}}$$
This directly follows from the properties of the sine, except the series only converges conditionally, not absolutely, so there's a question of if we can bring it inside the integral.
I will do it without proper justification for now, but if anyone has a comment on this, I would appreciate it.
So we obtain, after a simple change of variables:
$$\int_0^\infty \sin y^2 dy = \frac{\sqrt{\pi}}{2} \int_0^1 \sin \pi t ~ dt \sum_{n=0}^\infty \frac{(-1)^n}{\sqrt{n+t}} $$
Wolfram gives for this series:
$$\sum_{n=0}^\infty \frac{(-1)^n}{\sqrt{n+t}}=\frac{1}{\sqrt{2}}\left( \zeta \left(\frac12, \frac{t}{2}\right)-\zeta \left(\frac12, \frac{t+1}{2}\right) \right)$$
Which is easy enough to show using the definition of Hurwitz zeta function.
Surprisingly enough, this brings us exactly the known value of the Fresnel integral as a coefficient:
$$\int_0^\infty \sin y^2 dy =\sqrt{\frac{\pi}{8}} \int_0^1 \sin \pi t ~ \left( \zeta \left(\frac12, \frac{t}{2}\right)-\zeta \left(\frac12, \frac{t+1}{2}\right) \right) dt$$
Which means that we need to prove the identity in the title of the question:
$$\int_0^1 \sin \pi t ~ \left( \zeta \left(\frac12, \frac{t}{2}\right)-\zeta \left(\frac12, \frac{t+1}{2}\right) \right) dt=1$$
Can we prove this by real methods, not using the Fresnel integral?
Mathematica confirms this identity numerically.
Answer
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.
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