In the paper E. Mehlum, Appell and the apple (nonlinear splines in space), Technical Report No. 1676 (1981), Central institute for industrial research, Oslo (reproduced in the book Mathematical Methods for Curves and Surfaces, pages 365–384, Vanderbilt University Press, 1995) we can find the following interesting identity $$S^2(x)+C^2(x)=\sum\limits_{n=0}^\infty \frac{(1)^n2^{2n}x^{4n+2}}{(2n+1)(4n+1)!!},$$ involving Fresnel integrals $$S(x)=\int\limits_0^x \sin{(t^2)}dt,\;\;\; C(x)=\int\limits_0^x \cos{(t^2)}dt.$$ In the paper the identity is proved indirectly as a by product of a solution of an intricate problem. The author also mentions that the identity can be deduced directly from the definition of the Fresnel integrals, but does not provide any details of such deduction. How this identity can be proved? I suspect we can use $$\int\limits_0^x dy\int\limits_0^y dz\;\cos{(y^2z^2)}=\frac{1}{2}(C^2(x)+S^2(x))$$ and the TaylorMacLaurin series representation of the cosine function.

$\begingroup$ It seems can be generalized in the same way for higher degrees, $t^2 \to t^3, t^n$? What about general polynome $t^2 \to P(t)$ ? $\endgroup$ – Sergei Oct 30 '14 at 6:34
It seems easier to integrate over the unit square, giving
$$ C^2(x) + S^2(x) = \int_0^x \int_0^x \cos(y^2z^2) \;\mathrm{d}y\;\mathrm{d}z. $$
Using the Taylor series for cosine, and then normalizing the integral by setting $s = xy$, $t = xz$, gives
$$ C^2(x) + S^2(x) = \sum_{n=0}^\infty \frac{(1)^n}{(2n)!} x^{4n+2} \int_0^1 \int_0^1 (s^2t^2)^{2n} \; \mathrm{d}s \; \mathrm{d}t. $$
So it is sufficient to prove that
$$ \int_0^1 \int_0^1 (s^2t^2)^{2n} \;\mathrm{d}s\; \mathrm{d}t = \frac{2^{2n}(2n)!}{(2n+1)(4n+1)!!}. $$
This can be done immediately by Mathematica (although not Maple, to my surprise). Alternatively, here is a short proof using binomial coefficients. Expand the lefthand side using the binomial theorem and perform both integrations to get the equivalent identity
$$ \sum_{m=0}^{2n} \binom{2n}{m} \frac{(1)^m}{(2m+1)(4n2m+1)} = \frac{2^{2n}(2n)!}{(2n+1)(4n+1)!!}. $$
Using partial fractions on the lefthand side this is equivalent to
$$ \sum_{m=0}^{2n}\binom{2n}{m} \frac{(1)^m}{2m+1} = \frac{2^{2n}(2n)!}{(4n+1)!!} $$
which follows from a more basic integral, namely
$$ \int_{0}^1 (1u^2)^{2n} \;\mathrm{d} u = \frac{2^{4n}}{4n+1} \binom{4n}{2n}^{1}. $$

2$\begingroup$ The binomial identity from the proof has the form $$\sum\limits_{m=0}^n\binom{n}{m}\frac{(1)^m}{2m+1}=\frac{(2n)!!}{(2n+1)!!}.$$ I was not able to find this identity in I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series and Products, 7th ed., Academic Press, Oxford, 2007. However, it follows from more general result (Melzak's Formula) cited in Henry W. Gould's manuscript Vol.5.PDF on his home page math.wvu.edu/~gould The identity in the form $$\sum\limits_{m=0}^n\binom{n}{m}\frac{(1)^m}{2m+1}=\frac{2^{2n}}{(2n+1)\binom{2n}{n}}$$ appears also (formula 9.3) in Vol.4.PDF $\endgroup$ – Zurab Silagadze Apr 3 '13 at 10:04

1$\begingroup$ That's interesting. I stopped at the final integral because it can be done in a routine way with a reduction formula: write $(1u^2)^{2n} = (1u^2)^{2n1}  2u(1u^2) u/2$ and use integration by parts on the second term. I don't know of a combinatorial proof. $\endgroup$ – Mark Wildon Apr 3 '13 at 11:30