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I have difficulties with this integral, which is is related to Fresnel integrals ($a>0$):

$$ \int dx\, \cos(ax^2 + 2bx + c) = \\ =\sqrt{\frac{\pi}{2a}} \left[ \cos(\frac{ac-b^2}{a}) \mathscr{C} \left( \frac{\sqrt{2} (ax + b)}{\sqrt{a\pi}} \right) - \sin(\frac{ac-b^2}{a}) \mathscr{S}\left( \frac{\sqrt{2} (ax + b)}{\sqrt{a\pi}} \right) \right] +C $$

with $$ \mathscr{C}(x) = \int_0^x dt \cos(\pi t^2 /2), $$ $$ \mathscr{S}(x) = \int_0^x dt \sin(\pi t^2 /2). $$

How do you show this integral?

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  • $\begingroup$ Depends on what you have at your disposal? And perhaps the purpose? Computing at a single value, for an interval...? $\endgroup$ – H. H. Rugh Aug 30 '16 at 10:36
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Start completing the square $$ax^2+2bx+c=a\left(x+\frac b{a}\right)^2-\left(\frac{b^2}{a}-c\right)$$ So, $\cos(ax^2+2bx+c)$ write $$\cos \left(a\left(x+\frac b{a}\right)^2\right)\cos\left(\frac{b^2}{a}-c\right)+\sin \left(a\left(x+\frac b{a}\right)^2\right)\sin\left(\frac{b^2}{a}-c\right)$$ Now, you have to change variable to go to Fresnel since and cosine integrals. Set $$\sqrt a\left(x+\frac b{a}\right)=\sqrt{\frac \pi 2}t \implies x={\sqrt{\frac{\pi }{2a}} }t-\frac{b}{a}\implies dx=\sqrt{\frac{\pi }{2a}}\,dt $$ which make $$\int \cos \left(a\left(x+\frac b{a}\right)^2\right)\,dx=\sqrt{\frac{\pi }{2a}}\int \cos\left(\frac \pi 2 t^2\right)\,dt=\sqrt{\frac{\pi }{2a}}C(t)=\sqrt{\frac{\pi }{2a}}C\left(\frac{\sqrt 2(ax+b)}{\sqrt{\pi a}}\right)$$ $$\int \sin \left(a\left(x+\frac b{a}\right)^2\right)\,dx=\sqrt{\frac{\pi }{2a}}\int \sin\left(\frac \pi 2 t^2\right)\,dt=\sqrt{\frac{\pi }{2a}}S(t)=\sqrt{\frac{\pi }{2a}}S\left(\frac{\sqrt 2(ax+b)}{\sqrt{\pi a}}\right)$$ and hence the formula.

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To demonstrate validity: Take derivatives and recall that $$\frac{d}{dx} \int_0^x f(t)dt=f(x)$$ and use that $$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$ For evaluation that is a different issue.

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  • $\begingroup$ OK, I did wonder ;-) The answer is correct, though, because there is a factor 2 on bx. $\endgroup$ – H. H. Rugh Aug 30 '16 at 10:59
  • $\begingroup$ I missed the $2$ !!! Shame on me. $\endgroup$ – Claude Leibovici Aug 30 '16 at 11:03

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