Maclaurin Expansion of the Fresnel Sine Integral
Formula
$$ S(x) = \sqrt{{2} \over {\pi}} \int_{0}^{x} \sin (w^2) dw = \sqrt{{2} \over {\pi}} \sum_{n=0}^{\infty} {{(-1)^{n}} \over {(2n+1)! (4n+3)}} x^{4n+3} $$
Explanation
Fresnel was a physicist who studied optics, and if you look at the results that bear his name, most of them involve trigonometric functions. Since trigonometric functions are deeply related to wave functions, he probably had to study them even if it meant creating formulas that did not exist.
Whether or not these studies contributed to optics, it is certain that they made a considerable contribution to the calculus of trigonometric functions.
Proof
The Maclaurin expansion of the sine function is $\displaystyle \sin x = x - {{x^{3}} \over {3!}} + {{x^{5}} \over {5!}} + {{x^{7}} \over {7!}} + \cdots$. Therefore,
$$ \sin x^2 = x^2 - {{x^{6}} \over {3!}} + {{x^{10}} \over {5!}} + {{x^{14}} \over {7!}} + \cdots $$
Thus,
$$ \begin{align*} S(x) =& \sqrt{{2} \over {\pi}} \int_{0}^{x} \sin (w^2) dw \\ =& \sqrt{{2} \over {\pi}} \int_{0}^{x} \left( w^2 - {{w^{6}} \over {3!}} + {{w^{10}} \over {5!}} + {{w^{14}} \over {7!}} + \cdots \right) dw \\ =& \sqrt{{2} \over {\pi}} \left( {{1}\over{3}} {{x^{3}} \over {1!}} - {{1}\over{7}}{{x^{7}} \over {3!}} + {{1}\over{11}}{{x^{11}} \over {5!}} + {{1}\over{15}}{{x^{15}} \over {7!}} + \cdots \right) \\ =& \sqrt{{2} \over {\pi}} \sum_{n=0}^{\infty} {{(-1)^{n}} \over {(2n+1)! (4n+3)}} x^{4n+3} \end{align*} $$
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