While proving a pretty simple argument about reverse parametrization of a complex-valued function, I have an integral of the form: $$ \int_a^b f\bigl(\gamma(a+b-t)\bigr)\cdot\bigl(-\gamma'(a+b-t)\bigr) \, dt $$

(with $\gamma: [a, b] \to \mathbb{C},\; f: \mathbb{C} \to \mathbb{C}$)

When performing $u$-sub on this, I reason as follows: \begin{align} u &= a+b-t \\ du &= -dt \\ u(a) &= b \\ u(b) &= a \\ \therefore \quad \int_a^b f\bigl(\gamma(a+b-t)\bigr) \cdot \bigl(-\gamma'(a+b-t)\bigr) \, dt &= -\int_{u(a)}^{u(b)} f\bigl(\gamma(u)\bigr) \gamma'(u)(-1) \, du \\ &= \int_b^a f\bigl(\gamma(u)\bigr) \gamma'(u) \, du \\ &= -\int_a^b f\bigl(\gamma(u)\bigr) \gamma'(u) \, du \end{align}

This is just standard $u$-sub stuff right?

In my textbook they reason slightly more 'directly' that $\gamma'(u) = -\gamma'(a+b-t)$ (in addition to $du = -dt$) and leave the limits of integration as they are, I assume this is just a shortcut / more informal way of making the same argument? Just wanna make sure I understand the process completely.

  • 2
    $\begingroup$ I don't really understand what the book is arguing, but your way is correct. $\endgroup$ Apr 2 at 22:36


You must log in to answer this question.