Quadratic-trigonometric integral -- part 3 Problem
This is a continuation of these other two posts: Quadratic-trigonometric integral, Quadratic-trigonometric integral -- part 2.
I'm studying the following integral
\begin{equation*}
I_1(t)\triangleq\int_0^{t}\cos\left(c+b\tau+a\tau^2\right)\text{ d}\tau
\end{equation*}
where $c,b,a$ and $t$ are given parameters. Note that with respect to the linked posts I've changed the notation for the coefficients of the quadratic polynomial inside the cosine (I've switched $a$ and $c$) and also now I'm considering a simpler problem where the lower bound of the integration is null.
For the sake of simplicity, I restrict the attention to the simpler case $a\ge0$.
On one hand I know the solution of the integral above in terms of the Fresnel integrals, which is
\begin{equation*}\begin{aligned}
I_1(t)&=\left[\textrm{C}\left(\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}\right)-\textrm{C}\left(\frac{b}{2\sqrt[+]{a}}\right)\right]\frac{\textrm{cos}\left(c-\frac{b^2}{4a}\right)}{\sqrt[+]{a}}\\
&\qquad\qquad-\left[\textrm{S}\left(\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}\right)-\textrm{S}\left(\frac{b}{2\sqrt[+]{a}}\right)\right]\frac{\textrm{sin}\left(c-\frac{b^2}{4a}\right)}{\sqrt[+]{a}}
\end{aligned}
\end{equation*}
where $\sqrt[+]{\cdot}$ denotes the positive root (I chose it instead of the negative one just for simplicity), while $\textrm{C}(\cdot)$ and $\textrm{S}(\cdot)$ are the Fresnel cosine and sine integrals, i.e.
\begin{equation*}
\textrm{C}(h)\triangleq \int_0^h \cos\left(\theta^2\right)\text{ d}\theta \qquad
\textrm{S}(h)\triangleq \int_0^h \sin\left(\theta^2\right)\text{ d}\theta 
\end{equation*}
On the other hand I know also the solution of the simpler integral
\begin{equation*}I_2(t)\triangleq \int_0^t \cos\left(c+b\tau\right)\text{ d}\tau\end{equation*}
which is
\begin{equation*}
I_2(t)=\sin(bt)\frac{\cos(c)}{b}-\left[1-\cos(bt)\right]\frac{\sin(c)}{b}
\end{equation*}
Since the second integral is the limit case where $a=0$, I expect that
\begin{equation*}\lim_{a\to0}I_1(t)=I_2(t)\end{equation*}
but actually I'm not able to compute the limit $\lim_{a\to0}I_1(t)$ over the solution presented above.

Question
1) My solutions $I_1(t)$ and $I_2(t)$ are correct?
2) Is it true that $\lim_{a\to0}I_1(t)=I_2(t)$? If yes, is it possible to prove it by computing the limit
\begin{equation*}
\lim_{a\to0}\left[\textrm{C}\left(\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}\right)-\textrm{C}\left(\frac{b}{2\sqrt[+]{a}}\right)\right]\frac{\textrm{cos}\left(c-\frac{b^2}{4a}\right)}{\sqrt[+]{a}}-\left[\textrm{S}\left(\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}\right)-\textrm{S}\left(\frac{b}{2\sqrt[+]{a}}\right)\right]\frac{\textrm{sin}\left(c-\frac{b^2}{4a}\right)}{\sqrt[+]{a}}
\end{equation*}

Derivation of the solution of $I_2(t)$
\begin{equation*}
\begin{aligned}
I_2(t)&=\int_0^t \cos(c)\cos(b\tau)-\sin(c)\sin(b\tau) \text{d}\tau\\
&=\left[\int_0^{bt}\cos(h)\frac{\text{d}h}{b}\right]\cos(c)-\left[\int_0^{bt}\sin(h)\frac{\text{d}h}{b}\right]\sin(c)\\
&=\left[\sin(h)|_0^{bt}\right]\frac{\cos(c)}{b}-\left[-\cos(h)|_0^{bt}\right]\frac{\sin(c)}{b}\\
&=\sin(bt)\frac{\cos(c)}{b}-\left[1-\cos(bt)\right]\frac{\sin(c)}{b}\\
\end{aligned}
\end{equation*}

Derivation of the solution of $I_1(t)$
Let
\begin{equation*}
q_1\triangleq a \qquad q_2\triangleq -\frac{b}{2a} \qquad q_3\triangleq c-\frac{b^2}{4a}
\end{equation*}
so
\begin{equation*}\begin{aligned}
I_2(t)&=\int_0^t \cos\left[q_1(\tau-q_2)^2+q_3\right]\text{ d}\tau\\
&=\left[\int_0^t \cos\left[q_1\left(\tau-q_2\right)^2\right]\text{d}\tau\right]\cos(q_3)-\left[\int_0^t \sin\left[q_1\left(\tau-q_2\right)^2\right]\text{d}\tau\right]\sin(q_3)\\
&=\left[\int_{-\sqrt[+]{q_1}q_2}^{\sqrt[+]{q_1}(t-q_2)} \cos(h^2)\frac{\text{d}h}{\sqrt[+]{q_1}}\right]\cos(q_3)-\left[\int_{-\sqrt[+]{q_1}q_2}^{\sqrt[+]{q_1}(t-q_2)} \sin(h^2)\frac{\text{d}h}{\sqrt[+]{q_1}}\right]\sin(q_3)\\
&=\left[\textrm{C}\left(\sqrt[+]{q_1}(t-q_2)\right)-\textrm{C}\left(-\sqrt[+]{q_1}q_2\right)\right]\frac{\cos(q_3)}{\sqrt[+]{q_1}}
-\left[\textrm{S}\left(\sqrt[+]{q_1}(t-q_2)\right)-\textrm{S}\left(-\sqrt[+]{q_1}q_2\right)\right]\frac{\sin(q_3)}{\sqrt[+]{q_1}}\\
&=\left[\textrm{C}\left(\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}\right)-\textrm{C}\left(\frac{b}{2\sqrt[+]{a}}\right)\right]\frac{\textrm{cos}\left(c-\frac{b^2}{4a}\right)}{\sqrt[+]{a}}\\
&\qquad\qquad-\left[\textrm{S}\left(\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}\right)-\textrm{S}\left(\frac{b}{2\sqrt[+]{a}}\right)\right]\frac{\textrm{sin}\left(c-\frac{b^2}{4a}\right)}{\sqrt[+]{a}}
\end{aligned}\end{equation*}

Attempt to compute the limit
Honestly, I don't have a clue on how to compute the limit. The only thing that I'm able to write is the following
\begin{equation*}
\lim_{a\to0} I_1(t)=\lim_{a\to0}\left\{\frac{\Delta\textrm{C}}{\sqrt[+]{a}}\left[\cos(c)\cos\left(\frac{b^2}{4a}\right)+\sin(c)\sin\left(\frac{b^2}{4a}\right)\right]-
\frac{\Delta\textrm{S}}{\sqrt[+]{a}}\left[\sin(c)\cos\left(\frac{b^2}{4a}\right)-\cos(c)\sin\left(\frac{b^2}{4a}\right)\right]
\right\}\end{equation*}
where
\begin{equation*}\begin{aligned}
\Delta\textrm{C}&\triangleq\textrm{C}\left(\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}\right)-\textrm{C}\left(\frac{b}{2\sqrt[+]{a}}\right)\\
\Delta\textrm{S}&\triangleq\textrm{S}\left(\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}\right)-\textrm{S}\left(\frac{b}{2\sqrt[+]{a}}\right)\\
\end{aligned}
\end{equation*}
here the problems that arise when $a\to0$ are the following:

*

*$\Delta \textrm{C}/\sqrt[+]{a}$, $\Delta \textrm{S}/\sqrt[+]{a}$ are undetermined forms $0/0$, which, if my conjecture that the final result is $I_2(t)$, they must converge to a finite limit;

*$\cos(b^2/(4a))$, $\sin(b^2/(4a))$ does not exists, and so, if my conjecture is true, they must disappear from the expression with some trick.

Probably there is some property of the Fresnel integrals (that actually I don't know) that can be used to solve the limit above.

EDIT
I've made a progress, but I still have some problems.
I've noticed that when $a\to0$ the ratios $\Delta \text{C}/\sqrt[+]{a}$, $\Delta \text{S}/\sqrt[+]{a}$ highly resembles the derivatives of $\textrm{C}(\cdot)$ and $\textrm{S}(\cdot)$. In particular, if I'm not wrong,
\begin{equation*}\begin{aligned}
\lim_{(\sqrt[+]{a} t)\to0}
\frac{\Delta \textrm{C}}{\sqrt[+]{a}t}&=\lim_{(\sqrt[+]{a} t)\to0}\frac{\textrm{C}\left(\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}\right)-\textrm{C}\left(\frac{b}{2\sqrt[+]{a}}\right)}{\sqrt[+]{a} t}
=\lim_{(\sqrt[+]{a} t)\to0}\frac{\textrm{C}\left(\sqrt[+]{a}t+\frac{bt}{2\sqrt[+]{a}t}\right)-\textrm{C}\left(\frac{bt}{2\sqrt[+]{a}t}\right)}{\sqrt[+]{a} t}
\\
&=\frac{\text{d}\textrm{C}}{\text{d}h}\bigg|_{h=\frac{bt}{2\sqrt[+]{a}t}}
=\cos\left[\left(\frac{bt}{2\sqrt[+]{a}t}\right)^2\right]=
\cos\left(\frac{b^2}{4a}\right)
\end{aligned}
\end{equation*}
and in the same way
\begin{equation*}\begin{aligned}
\lim_{(\sqrt[+]{a} t)\to0}
\frac{\Delta \textrm{S}}{\sqrt[+]{a}t}=
\sin\left(\frac{b^2}{4a}\right)
\end{aligned}
\end{equation*}
note that here this two results are quite doubtful because the point $b/\sqrt[+]{a}t$ where are evaluated the derivatives is not fixed during the computation of the limit $\sqrt[+]{a}t\to0$.
 A: This is not a proof.
I computed $I_1$ using the convention
$$\int \cos(x^2)\,dx=\sqrt{\frac{\pi }{2}}\,\, C\left(x\sqrt{\frac{2}{\pi }} \right)$$ and obtained
$$\sqrt{\frac{2 a}{\pi } }\,\, I_1=\cos \left(\frac{b^2-4ac}{4 a}\right) \left(C\left(\frac{b+2 a t}{\sqrt{2a
   \pi }}\right)-C\left(\frac{b}{ \sqrt{2a \pi }}\right)\right)+$$ $$\sin
   \left(\frac{b^2-4ac}{4 a}\right) \left(S\left(\frac{b+2 a t}{ \sqrt{2 a\pi
   }}\right)-S\left(\frac{b}{ \sqrt{2a \pi }}\right)\right)$$ and
$$I_2=\frac{\sin (b t+c)-\sin (c)}{b}$$
These results are the same as yours.
Now, for the behavior of $I_1$ when $a\to 0$, I faced the same problems but, in my humble opinion, this is not related to Fresnel integrals.
Unable to see any way to prove it, I made a bunch of random runs and $I_1 \to I_2$ is numerically always true.
Using $c=e$, $b=\gamma$, $t=\pi$ and $a=10^{-k}$ some results for the ratio $\frac {I_1}{I_2}$.
$$\left(
\begin{array}{cc}
k & \frac {I_1}{I_2} \\
 2 & 0.96690423 \\
 3 & 0.99673868 \\
 4 & 0.99967437 \\
 5 & 0.99996744 \\
 6 & 0.99999674 \\
 7 & 0.99999967 \\
 8 & 0.99999997 \\
 9 & 1.00000000
\end{array}
\right)$$
A: I believe that I found a proof, the idea is the following: start by observing that as $a\to0$ the arguments of the Fresnel integrals diverge to infinity, i.e.
\begin{equation*}
\lim_{a\to0} \underbrace{\sqrt[+]{a}t+\frac{b}{2\sqrt[+]{a}}}_{\triangleq h_1}=\infty \qquad 
\lim_{a\to0} \underbrace{\frac{b}{2\sqrt[+]{a}}}_{\triangleq h_2}=\infty
\end{equation*}
This means that the Fresnel integrals $\textrm{C}(\cdot)$, $\textrm{S}(\cdot)$ can be replaced with their asymptotic approximations
\begin{equation*}
\textrm{C}(h)\approx\sqrt[+]{\frac{\pi}{2}}\left[\frac{1}{2}+\frac{\sin(h^2)}{h\sqrt[+]{2\pi}}\right] \qquad
\textrm{S}(h)\approx\sqrt[+]{\frac{\pi}{2}}\left[\frac{1}{2}-\frac{\cos(h^2)}{h\sqrt[+]{2\pi}}\right]
\end{equation*}
In order to apply these formulas, note that
\begin{equation*}\begin{aligned}
h_1^2&\approx bt+\frac{b^2}{4a} \qquad h_2^2=\frac{b^2}{4a}\\
h_1&\approx \frac{b}{2\sqrt[+]{a}} \qquad h_2=\frac{b}{2\sqrt[+]{a}}
\end{aligned}
\end{equation*}
where $h_1^2$, $h_1$ are approximated according to the fact $a\to0$. Consequently
\begin{equation*}\begin{aligned}
\Delta\textrm{C} &\approx \phantom{+}\frac{\sqrt[+]{a}}{b}\left[\sin\left(bt+\frac{b^2}{4a}\right)-\sin\left(\frac{b^2}{4a}\right)\right]\\
\Delta\textrm{S} &\approx -\frac{\sqrt[+]{a}}{b}\left[\cos\left(bt+\frac{b^2}{4a}\right)-\cos\left(\frac{b^2}{4a}\right)\right]\\
\end{aligned}
\end{equation*}
Now, in order to quickly carry out the computation of the RHS of
\begin{equation*}I_1(t)=\frac{\Delta \textrm{C}}{\sqrt[+]{a}}\cos\left(c-\frac{b^2}{4a}\right)-\frac{\Delta \textrm{S}}{\sqrt[+]{a}}\sin\left(c-\frac{b^2}{4a}\right)\end{equation*}
define the shorthands
\begin{equation*}
\alpha \triangleq bt \qquad \beta \triangleq \frac{b^2}{4a} \qquad \gamma\triangleq c
\end{equation*}
Now, for the usual trigonometric identities about the sum of angles, holds (hoping to avoid typos)
\begin{equation*}\begin{aligned}
I_1(t)&\approx \frac{1}{b}\left[\sin(\alpha+\beta)-\sin(\beta)\right]\cos(\gamma-\beta)
-\left\{-\frac{1}{b}\left[\cos(\alpha+\beta)-\cos(\beta)\right]\sin(\gamma-\beta)\right\}\\
&=\frac{\sin(\alpha)\cos(\beta)+(\cos(\alpha)-1)\sin(\beta)}{b}(\cos(\gamma)\cos(\beta)+\sin(\gamma)\sin(\beta))\\
&\qquad\qquad+\frac{(\cos(\alpha)-1)\cos(\beta)-\sin(\beta)\cos(\alpha)}{b}(\sin(\gamma)\cos(\beta)-\cos(\gamma)\sin(\beta))\\
&=\frac{\sin(\alpha)\cos(\gamma)+(\cos(\alpha)-1)\sin(\gamma)}{b}\\
&=\frac{\sin(bt)\cos(c)+(\cos(bt)-1)\sin(c)}{b}
\end{aligned}
\end{equation*}
which is the searched result.
