I need to compute the following integral

$\begin{equation*} I(t)\triangleq \int_0^t\, \tau\,\cos\left(c+b\tau+a\,\tau^2\right)\text{ d}\tau \end{equation*},$ where $t,c,b,a$ are given parameters.


First of all, by completing the square I write the cosine argument as $\begin{equation*}c+b\tau+a\tau^2=q_1(\tau-q_2)^2+q_3\end{equation*}$ where the parameters of the quadratic form on the LHS are $\begin{equation*} q_1\triangleq a \qquad q_2\triangleq -\frac{b}{2a} \qquad q_3\triangleq c-\frac{b^2}{4a} \end{equation*}$ hence the integral gets the form $\begin{align*} I(t)&=\int_0^t \tau \cos[q_1(\tau-q_2)^2+q_3]\text{ d}\tau\\ &=\underbrace{\left[\int_0^t \tau \cos[q_1(\tau-q_2)^2]\text{ d}\tau\right]}_{\triangleq I_1(t)}\cos(q_3)- \underbrace{\left[\int_0^t \tau \sin[q_1(\tau-q_2)^2]\text{ d}\tau\right]}_{\triangleq I_2(t)}\sin(q_3) \end{align*}$

where I have exploited the trigonometric identity $\begin{equation*} \cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \end{equation*}$.

Now, in order to compute the integrals $I_1$, $I_2$ above I define the change of integration variable \begin{equation*}\theta(\tau)\triangleq q_1(\tau-q_2)^2\end{equation*} which implies \begin{equation*} \text{d}\theta=2q_1\tau\text{ d}\tau \qquad \text{and} \qquad \tau\text{ d}\tau=\frac{\text{d}\theta}{2 q_1} \end{equation*} and so \begin{equation*}\begin{aligned} I_1 =\int_{\theta(0)}^{\theta(t)} \cos(\theta)\frac{\text{d}\theta}{2 q_1}=\frac{\sin(\theta)}{2 q_1}\bigg|_{\theta(0)}^{\theta(t)}=\frac{\Delta s}{2q_1}\\ I_2=\int_{\theta(0)}^{\theta(t)} \sin(\theta)\frac{\text{d}\theta}{2 q_1}=-\frac{\cos(\theta)}{2 q_1}\bigg|_{\theta(0)}^{\theta(t)}=-\frac{\Delta c}{2q_1}\\ \end{aligned} \end{equation*} where \begin{equation*}\begin{aligned} \Delta s &\triangleq \sin(\theta(t))-\sin(\theta(0))= \sin[q_1(t-q_2)^2]-\sin(-q_1q_2^2)\\ \Delta c &\triangleq \cos(\theta(t))-\cos(\theta(0))= \cos[q_1(t-q_2)^2]-\cos(-q_1q_2^2)\\ \end{aligned} \end{equation*} my conclusion (in terms of the quadratic parameters $q_1,q_2,q_3$) is thus \begin{equation*}I(t)=\frac{\Delta s}{2q_1}\cos(q_3)+\frac{\Delta c}{2q_1}\sin(q_3)\end{equation*}


Consider the following indefinite integral $\begin{equation*}I_{\text{ind}}\triangleq \int \tau\cos[q_1(\tau-q_2)^2]\text{ d}\tau\end{equation*}$ according to the procedure above, its explicit expression (neglecting the generic integration constant) is $\begin{equation*} \frac{\sin(\theta)}{2q_1}=\frac{\sin[q_1(\tau-q_2)^2]}{2 q_1} \end{equation*}.$

Now, if one tries to compute the derivative of such result, due to the chain rule, gets $\begin{equation*} \frac{\text{d}}{\text{d}\tau}\frac{\sin[q_1(\tau-q_2)^2]}{2 q_1}= \cos[q_1(\tau-q_2)^2](\tau-q_2) \end{equation*},$ which is not the integrand of the indefinite integral $I_{\text{ind}}$.

Hence, I have the strong suspect that I'm committing some error somewhere in my attempt above. However, I cannot spot any erroneous passage in my derivation.

Question Is my derivation correct? If not, which with high probability is the case, where did I commit an error?

  • 1
    $\begingroup$ I find $d\theta = 2q_1(\tau - q_2) d \tau$ $\endgroup$ Commented Jan 26, 2022 at 21:08
  • 1
    $\begingroup$ I have not read the entire thing yet... I think this result must include some $\int \cos(t^2)$ or so... but your method is correct, first make it into square. $\endgroup$
    – JetfiRex
    Commented Jan 26, 2022 at 21:12
  • 1
    $\begingroup$ Wolfram does not seem to find a closed form solution to your integral, that's not very encouraging $\endgroup$
    – Lelouch
    Commented Jan 26, 2022 at 21:36
  • $\begingroup$ @ Lelouch Probably because are involved the Fresnel integrals in the solutions @ Gribouillis I think you are right $\endgroup$
    – matteogost
    Commented Jan 26, 2022 at 21:40

1 Answer 1


My screen teller having problems, I shall restart the calculations for $$I=\int\tau\cos(c+b\tau+a\tau^2)d\tau$$ Completing the square as you did $$c+b\tau+a\tau^2=a \left(\tau+\frac{b}{2 a}\right)^2+c-\frac{b^2}{4 a}$$ $$a \left(\tau+\frac{b}{2 a}\right)^2=x^2 \implies \tau=\frac{2 \sqrt{a} x-b}{2 a}\implies d\tau=\frac{dx}{\sqrt{a}}$$ $$\cos(c+b\tau+a\tau^2)=\cos \left(x^2+\frac{4ac-b^2}{4 a}\right)=\alpha \cos(x^2)-\beta \sin(x^2)$$ where $$\alpha=\cos \left(\frac{4ac-b^2}{4 a}\right)\qquad \text{and} \qquad \beta=\sin \left(\frac{4ac-b^2}{4 a}\right)$$ So, we face four integrals $$I_1=\int x \cos(x^2) \,dx=\frac 12 \sin(x^2)\qquad\qquad I_2=\int x \sin(x^2) \,dx=-\frac 12 \cos(x^2)$$ $$I_3=\int \cos(x^2) \,dx=\sqrt{\frac{\pi }{2}} C\left(\sqrt{\frac{2}{\pi }} x\right)\qquad\qquad I_4=\int \sin(x^2) \,dx=\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} x\right)$$

Back to $\tau$ and using the bounds

$$4 a^{3/2}\,I(t)=2 \sqrt{a} \Big[\sin (c+bt+at^2)-\sin (c)\Big]+$$ $$b\sqrt{2 \pi } \cos \left(\frac{b^2-4ac}{4 a}\right) \left(C\left(\frac{b}{ \sqrt{2 \pi a}}\right)-C\left(\frac{2 a t+b}{ \sqrt{2 \pi a }}\right)\right)+$$ $$b\sqrt{2 \pi } \sin \left(\frac{b^2-4ac}{4 a}\right) \left(S\left(\frac{b}{ \sqrt{2 \pi a}}\right)-S\left(\frac{2 a t+b}{ \sqrt{2 \pi a }}\right)\right)$$

  • 1
    $\begingroup$ thank you Claude $\endgroup$
    – matteogost
    Commented Jan 27, 2022 at 6:26
  • 2
    $\begingroup$ @matteogost. You are very welcome. I am glad when I can help. Cheers :-) $\endgroup$ Commented Jan 27, 2022 at 6:30

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