I'm studying the following integral \begin{equation*} \int_0^t \tau \cos(c+b\tau+a\tau^2)\text{ d}\tau \end{equation*} in particular, its limit \begin{equation*} \lim_{a\to 0}\int_0^t \tau \cos(c+b\tau+a\tau^2)\text{ d}\tau \end{equation*} and I'm not able to show that it is equal to \begin{equation*} \int_0^t \tau \cos(c+b\tau)\text{ d}\tau \end{equation*} Hence, at this point, I'm wondering if the following passages are true \begin{equation*}\begin{aligned} \lim_{a\to 0} \int_0^t \tau \cos(c+b\tau+a\tau^2)\text{ d}\tau &= \int_0^t \lim_{a\to 0}\tau \cos(c+b\tau+a\tau^2)\text{ d}\tau\\ &=\int_0^t \tau \cos\left[\lim_{a\to 0}(c+b\tau+a\tau^2)\right]\text{ d}\tau\\ &=\int_0^t \tau \cos(c+b\tau)\text{ d}\tau\\ \end{aligned}\end{equation*} I remember (not so well) that there are results, for example the dominated convergence theorem, that say when is possible to switch the order of the limit and integration operation. However, I'm not able to justify the equivalence \begin{equation*} \lim_{a\to 0} \int_0^t \tau \cos(c+b\tau+a\tau^2)\text{ d}\tau = \int_0^t \lim_{a\to 0}\tau \cos(c+b\tau+a\tau^2)\text{ d}\tau \end{equation*}

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    $\begingroup$ Just note that the integrand when seen as a function of two variables $a, \tau$ is continuous on some two dimensional interval of the form $[-h, h] \times [-k, k] $ and then it is possible to pass the limit under integral sign. You may prove this with some amount of $\epsilon, \delta $ gymnastics. $\endgroup$
    – Paramanand Singh
    Jan 30, 2022 at 6:30


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