Fresnel Sine Integral's Maclaurin Series Expansion
Formula
$$ S(x) = \sqrt{{2} \over {\pi}} \int_{0}^{x} \sin (w^2) dw = \sqrt{{2} \over {\pi}} \sum_{n=0}^{\infty} {{(-1)^{n}} \over {(2n+1)! (4n+3)}} x^{4n+3} $$
Description
Fresnel was a physicist who studied optics, and his name is attached to most results involving trigonometric functions. Likely, trigonometric functions are deeply related to wave functions, which explains why he felt the need to vigorously study them, even if it meant creating formulas where none existed.
Such studies may or may not have directly contributed to optics, but they certainly left a significant mark on the calculus of trigonometric functions.
Proof
The Maclaurin series 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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