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 $$\int\limits\_0^x dy\int\limits\_0^y dz\;\cos{(y^2-z^2)}=\frac{1}{2}(C^2(x)+S^2(x))$$ and the Taylor-MacLaurin series representation of the cosine function.



Identity involving Fresnel integrals
It seems easier to integrate over the unit square, giving


$$ C^2(x) + S^2(x) = \int\_0^x \int\_0^x \cos(y^2-z^2) \;\mathrm{d}y\;\mathrm{d}z. $$


Using the Taylor series for cosine, and then normalizing the integral by setting $s = xy$, $t = xz$, gives


$$ C^2(x) + S^2(x) = \sum\_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{4n+2} \int\_0^1 \int\_0^1 (s^2-t^2)^{2n} \; \mathrm{d}s \; \mathrm{d}t. $$


So it is sufficient to prove that


$$ \int\_0^1 \int\_0^1 (s^2-t^2)^{2n} \;\mathrm{d}s\; \mathrm{d}t = \frac{2^{2n}(2n)!}{(2n+1)(4n+1)!!}. $$


This can be done immediately by Mathematica (although not Maple, to my surprise). Alternatively, here is a short proof using binomial coefficients. Expand the left-hand side using the binomial theorem and perform both integrations to get the equivalent identity


$$ \sum\_{m=0}^{2n} \binom{2n}{m} \frac{(-1)^m}{(2m+1)(4n-2m+1)} = \frac{2^{2n}(2n)!}{(2n+1)(4n+1)!!}. $$


Using partial fractions on the left-hand 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


$$ \int\_{0}^1 (1-u^2)^{2n} \;\mathrm{d} u = \frac{2^{4n}}{4n+1} \binom{4n}{2n}^{-1}. $$


