I need to compute the following interval \begin{equation*}\int_{t_\text{s}}^{t_\text{e}} \cos\left(a+b\tau+c\tau^2\right)\text{ d}\tau\end{equation*} where $t_{\text{s}},t_{\text{e}},a,b,c$ are non-negative given parameters. I have some difficulties in the case where $b$ and $c$ are both non zero. In order to fix the ideas, I will show my solutions in the simpler case where $c=0$ and $a,b>0$.

simpler case $c=0$ and $a,b>0$

The integral above reduce to \begin{equation*}\int_{t_\text{s}}^{t_\text{e}} \cos\left(a+b\tau\right)\text{ d}\tau\end{equation*} the idea is to integrate with respect \begin{equation*}h(\tau)\triangleq a+b\tau\end{equation*} so it turns out \begin{equation*}\begin{aligned}\int_{h(t_\text{s})}^{h(t_\text{e})} \cos\left(h\right)\left(\frac{\text{ d}h}{b}\right)&=\frac{\sin(h)}{b}\bigg|_{h(t_{\text{s}})}^{h(t_\text{e})}\\ &=\frac{\sin(a+bt_{\text{e}})-\sin(a+bt_{\text{s}})}{b} \end{aligned}\end{equation*}

general case $a,b,c>0$

Now seems that the substitution trick does not work. Indeed, now the full integral \begin{equation*}\int_{t_\text{s}}^{t_\text{e}} \cos\left(a+b\tau+c\tau^2\right)\text{ d}\tau\end{equation*} requires the change of variable \begin{equation*}h(\tau)\triangleq a+b\tau+c\tau^2\end{equation*} which causes the following problem: the differential $\text{d}\tau$ is a function of $\tau$ itself because \begin{equation*} \text{d}h=(b+2c\tau)\text{d}\tau \rightarrow \text{d}\tau=\frac{\text{d}h}{b+2c\tau}\end{equation*} and so, due to the presence of $\tau$ in the above expressions, the substitution does not work.


At a first glance the general integral that I'm trying to compute does not seem so problematic: the integrand is just the composition of a cosine with a polynomial function, i.e. two very regular functions. I don't believe that there is no solution, so I'm asking if someone can give me some ideas to carry out the calculation.

  • 1
    $\begingroup$ You can complete the square to get an expression of the form $cos(Kx^2 + L)$. Such an expression integrated is a "Fresnel integral" mathworld.wolfram.com/FresnelIntegrals.html . You may also be interested in Gaussian integrals, to which this is strongly related. $\endgroup$
    – egglog
    Jan 19, 2022 at 10:34
  • $\begingroup$ Thank you so much, your idea works like a charm. Actually that's the first time that I hear about the Fresnel integral. Now, due to your suggestion, I'm wondering if I can express the result in terms of the more familiar $\text{erf}(\cdot)$ function. $\endgroup$
    – matteogost
    Jan 19, 2022 at 15:43

1 Answer 1


As said by @egglog, you can reduce to Fresnel integrals. WLOG, $c>0$ and you can change the variable using

$$a+b\tau+c\tau^2=\upsilon^2+\tfrac\Delta{4c}$$ where $\upsilon=\sqrt c\left(\tau+\tfrac b{2c}\right)$. This gives $$\int_{t_s}^{t_e}\cos(a+b\tau+c\tau^2)\,d\tau=\frac1{\sqrt c}\int_{u_s}^{u_e}\left(\cos\left(\tfrac\Delta{4c}\right)\cos(\upsilon^2)-\sin\left(\tfrac\Delta{4c}\right)\sin(\upsilon^2)\right)\,d\upsilon\\ =\frac{\cos\left(\frac\Delta{4c}\right)}{\sqrt c}\left(C\left(a'+bt_e+ct_e^2\right)-C\left(a'+bt_s+ct_s^2\right)\right) \\-\frac{\sin\left(\frac\Delta{4c}\right)}{\sqrt c}\left(S\left(a'+bt_e+ct_e^2\right)-S\left(a'+bt_s+ct_s^2\right)\right)$$

where $a':=a-\frac\Delta{4c}$.

  • $\begingroup$ thank you, now the answer to my question is very clear. $\endgroup$
    – matteogost
    Jan 19, 2022 at 15:48

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