I feel a bit stupid asking this question since I passed Calculus I and II several years ago with solid A's, but I have encountered an integral example in a text book where a substitution yields upper and lower bounds which I don't fully understand.
The example is as follows. We have:
$$\int_{0}^{t_c} \frac{\sin(\lambda_c (t – \tau))}{\pi (t – \tau)} d \tau$$
Using the substitution $u = \lambda_c(t – \tau)$, the integral becomes:
$$\frac{1}{\pi} \int_{\lambda_c(t – t_c)}^{\lambda_c t} \frac{\sin(u)}{u} du$$
However, as far as I can tell – shouldn't the lower and upper bounds be switched here? After all, for the lower bound, we originally have $\tau = 0$, so I would then assume that the $u$ would be:
$$u = \lambda_c (t – \tau) = \lambda_c (t-0) = \lambda_c t$$
Again, I feel dumb asking this since this is so basic, but if I'm doing something wrong here in my reasoning, I would greatly appreciate it if anyone could help me out!
Best Answer
The upper and lower limits of integration were switched, as the substitution demanded. However, a simplification was made: The substitution introduced a factor of $-1$ in the integrand ($du=-\lambda_c \,d\tau$). The negative was dropped and the upper and lower limits of integration were interchanged.
$$ \int_{0}^{t_c} \frac{\sin(\lambda_c (t - \tau))}{\pi (t - \tau)} d \tau \ \ \buildrel{u=\lambda_c(t-\tau)}\over={1\over\pi}\ \ \int_{u=\lambda_c\tau}^{u=\lambda_c(t-t_c)}{\sin (u)\over u/\lambda_c}\cdot\underbrace{{-du\over\lambda_c}}_{d\tau} ={1\over\pi}\int_{\lambda_c(t-t_c)}^{\lambda_ct}{\sin(u)\over u}\,du $$