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.

  • $\begingroup$ do you mean $$\sin(y^2)$$ or $$\sin^2(y)$$ $\endgroup$
    – clathratus
    Oct 20, 2018 at 21:31
  • $\begingroup$ @clathratus, the first one. Fresnel integral is well known, thus I didn't clarify $\endgroup$
    – Yuriy S
    Oct 20, 2018 at 21:33
  • $\begingroup$ Right. I wasn't thinking. Thanks. $\endgroup$
    – clathratus
    Oct 20, 2018 at 21:34
  • $\begingroup$ This related question has a lot of nice proofs for Fresnel integral math.stackexchange.com/q/2051273/269624 and this one too math.stackexchange.com/q/2739835/269624 $\endgroup$
    – Yuriy S
    Oct 20, 2018 at 21:49
  • $\begingroup$ Thank you, these are great. $\endgroup$
    – clathratus
    Oct 20, 2018 at 21:55

1 Answer 1


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.


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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