Upper and lower bounds of integral after substitution

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!

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The substitution introduced a factor of $-1$ in the integrand ($du=-\lambda_c \,d\tau$). This was dropped and the upper and lower limits of integration were interchanged. – David Mitra Jan 17 '13 at 14:01
@DavidMitra: This should be an answer :) – Stefan Hansen Jan 17 '13 at 14:04
Thanks a lot! Yeah, I knew I must have overlooked something basic ! – Kristian Jan 17 '13 at 14:10

1 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.

$$\int_{0}^{t_c} \frac{\sin(\lambda_c (t - \tau))}{\pi (t - \tau)} d \tau \ \ \buildrel{u=\lambda_c(t-\tau)}\over={1\over\pi}\ \ \int_{u=\lambda_c\tau}^{u=\lambda_c(t-t_c)}{\sin (u)\over u/\lambda_c}\cdot\underbrace{{-du\over\lambda_c}}_{d\tau} ={1\over\pi}\int_{\lambda_c(t-t_c)}^{\lambda_ct}{\sin(u)\over u}\,du$$

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Ah! Of course! I was sure it was something basic I had overlooked. Thanks a lot! – Kristian Jan 17 '13 at 14:10
@Kristian You're welcome; glad to help. – David Mitra Jan 17 '13 at 14:16