I've searched for a post about this, to no avail. I am pretty sure I know which method is correct, but I'm having trouble pinning down exactly why the incorrect method is wrong.
I wish to evaluate the following expression: $$\frac{d\ }{dt} \int_{-\infty}^{t} \int_{\tau_{1}}^{t} e^{\pm i a (\tau_{2} - t)} \, d\tau_{2} \, d\tau_{1} \ , $$ where $a$ is a constant. I have done so with three different methods. The first is to use the Leibniz integral rule, which gives $$\frac{d\ }{dt} \int_{-\infty}^{t} \int_{\tau_{1}}^{t} e^{\pm i a (\tau_{2} - t)} \, d\tau_{2} \, d\tau_{1} = \int_{t}^{t} e^{\pm i a (\tau_{2} - t)} \, d\tau_{2} + \int_{-\infty}^{t} \frac{\partial\ }{\partial t} \left[ \int_{\tau_{1}}^{t} e^{\pm i a (\tau_{2} - t)} \, d\tau_{2} \right] \, d\tau_{1} \\ = \int_{-\infty}^{t} \left[ 1 + \int_{\tau_{1}}^{t} \frac{\partial\ }{\partial t} e^{\pm i a (\tau_{2} - t)} \, d\tau_{2} \right] \, d\tau_{1} = \int_{-\infty}^{t} \left[ 1 \mp i a \int_{\tau_{1}}^{t} d\tau_{2} \, e^{\pm i a (\tau_{2} - t)} \right] \, d\tau_{1} \\ = \int_{-\infty}^{t} \left[ 1 - 1 + e^{\pm i a (\tau_{1} - t)} \right] \, d\tau_{1} = \lim_{\Lambda \rightarrow \infty} \int_{-\Lambda}^{t} e^{\pm i a (\tau_{1} - t)} \, d\tau_{1} \\ = \lim_{\Lambda \rightarrow \infty} \mp \frac{i}{a} \left(1 - e^{\mp i a (\Lambda + t)} \right) $$ The second method is brute-force: $$\frac{d\ }{dt} \int_{-\infty}^{t} \int_{\tau_{1}}^{t} e^{\pm i a (\tau_{2} - t)} \, d\tau_{2} \, d\tau_{1} = \frac{d\ }{dt} \left[\lim_{\Lambda \rightarrow \infty} \mp \frac{i}{a} \int_{-\Lambda}^{t} \left(1 - e^{\pm i a (\tau_{1} - t)} \right) \, d\tau_{1} \right] \\ = \lim_{\Lambda \rightarrow \infty} \frac{d\ }{dt} \left[\mp \frac{i}{a} \left(t + \Lambda \pm \frac{i}{a} \left(1 - e^{\mp i a (\Lambda + t)} \right) \right) \right] \\ = \lim_{\Lambda \rightarrow \infty} \mp \frac{i}{a} \left(1 - e^{\mp i a (\Lambda + t)} \right) $$ These two methods agree, and the brute-force method seems safe, so I am confident that these are correct. However, I can also approach this with two u-substitutions, which at least make it appear as though the double integral is independent of $t$: $$\frac{d\ }{dt} \int_{-\infty}^{t} \int_{\tau_{1}}^{t} e^{\pm i a (\tau_{2} - t)} \, d\tau_{2} \, d\tau_{1} = \frac{d\ }{dt} \int_{-\infty}^{t} \int_{\tau_{1}-t}^{0} e^{\pm i a \tau_{2}^{\,\prime}} \, d\tau_{2}^{\,\prime} \, d\tau_{1} \\ = \frac{d\ }{dt} \int_{-\infty}^{0} \int_{\tau_{1}^{\,\prime}}^{0} e^{\pm i a \tau_{2}^{\,\prime}} \, d\tau_{2}^{\,\prime} \, d\tau_{1}^{\,\prime} = 0 \ .$$ This is fixed if, as before, we regulate the infinite lower limit of the $\tau_{1}$ integral and account for the effect of the $\tau_{1}^{\,\prime}$ substitution on that lower limit: $$\frac{d\ }{dt} \int_{-\infty}^{t} \int_{\tau_{1}}^{t} e^{\pm i a (\tau_{2} - t)} \, d\tau_{2} \, d\tau_{1} = \frac{d\ }{dt} \int_{-\infty}^{t} \int_{\tau_{1}-t}^{0} e^{\pm i a \tau_{2}^{\,\prime}} \, d\tau_{2}^{\,\prime} \, d\tau_{1} \\ = \frac{d\ }{dt} \left[ \lim_{\Lambda \rightarrow \infty} \int_{-\Lambda - t}^{0} \int_{\tau_{1}^{\,\prime}}^{0} e^{\pm i a \tau_{2}^{\,\prime}} \, d\tau_{2}^{\,\prime} \, d\tau_{1}^{\,\prime} \right] = \lim_{\Lambda \rightarrow \infty} \int_{- \Lambda - t}^{0} e^{\pm i a \tau_{2}^{\,\prime}} \, d\tau_{2}^{\,\prime} \\ = \lim_{\Lambda \rightarrow \infty} \mp \frac{i}{a} \left(1 - e^{\mp i a (\Lambda + t)} \right) \ . $$ Why do we need to account for a finite translation of the infinite limit in order to get the correct answer?