Let's say I have a contour integral on a non-closed contour with starting point $z_0$ and ending point $z_1$. Am I allowed to do a substitution like this? And under what assumptions?

$$\displaystyle \int_{z_0}^{z_1} f (z) \, \mathrm d z = \int_{u^{-1}(z_0)}^{u^{-1}(z_1)} f (u (z))u'(z) \, \mathrm d u $$

For example,

$$\int_\epsilon^T t^{1/2 + i}e^{(-4-i)t}\, \mathrm dt \stackrel{?}{\stackrel{u \leftrightarrow (4+i)t}{\longleftrightarrow} }\int_{4\epsilon+i\epsilon}^{4T+iT}\left({\frac{u}{4+i}}\right)^{1/2 +i}e^{-u}\left({\frac{1}{4+i}}\right)\, \mathrm du$$

where $0 < \epsilon < T \in \mathbb R$.

And the integrand is a contour, it's a smooth map from a real interval to the plane, so the integral is a contour integral with the parameterization $t: t \in [\epsilon..T]$ by definition. At least, I'm pretty sure it matches the definition of a contour integral with a parameterization. I even graphed it:

enter image description here

My thoughts:

Since I'd want want the substitution to move the contour somewhere else without changing the value of the integral, then $u$ would have to be a homotopy; thus a necessary condition would be that the old and new contour should each be in a simply connected domain. I'd guess $u$ might have to be analytic and bijective as well.

  • $\begingroup$ Did you try doing the substitution after parameterizing the curve? Certainly you can map one domain to a completely disjoint domain. $\endgroup$
    – GEdgar
    Dec 7, 2016 at 1:12
  • $\begingroup$ @GEdgar I don't have a particular integral in mind; I'm just trying to fill a gap in my education. But you're right, they shouldn't have to be in the same set. Question amended. $\endgroup$
    – GFauxPas
    Dec 7, 2016 at 1:14
  • $\begingroup$ @GEdgar I added an example $\endgroup$
    – GFauxPas
    Dec 8, 2016 at 1:17
  • 1
    $\begingroup$ You'd better check a section, 'change of variables' in real analysis. $\endgroup$
    – kayak
    Dec 11, 2016 at 15:39

1 Answer 1


Suppose $f$ is analytic in the simply connected region $V$ and $\gamma $ is a contour in $V$ from $z_0$ to $z_1.$ Let $U$ be a region (not necessarily simply connected), and assume $u:U \to V$ is an analytic function whose range contains $\{z_0,z_1\}.$ Choose $w_0,w_1\in U$ such that $u(w_0) =z_0, u(w_1) =z_1.$ Let $\delta$ be a contour in $U$ from $w_0$ to $w_1.$ Then

$$\tag 1 \int_\gamma f(z)\,dz = \int_\delta f(u(w))u'(w)\,dw.$$

Proof: Because $V$ is simply connected, $f$ has an antiderivative $F$ in $V.$ Just from the fundamental theorem of calculus (basic one real variable calculus is meant here), the left side of $(1)$ equals $F(z_1)-F(z_0).$ But notice $f(u(w))u'(w)$ also has an antiderivative–in $U$–namely $F(u(w)).$ That follows from the chain rule for analytic functions. Therefore, again by the FTC, the right side of $(1)$ equals $F(u(w_1)) - F(u(w_0)) = F(z_1)-F(z_0)$ and we're done.

If you look at your example, the integrand $z^{1/2+i}e^{(-4-i)z}$ is analytic in the open right half plane, a simply connected region. The map $u(w)= w/(4+i),$ which is entire, takes $w_0 = (4+i)\epsilon, w_1 = (4+i)T$ to $\epsilon,T$ and $(1)$ guarantees your substitution is valid (as would any analytic substitution respecting the end points).

I am not sure what your paragraph "And the integrand is a contour …" is all about. The range of $f\circ \gamma$ isn't the issue here, at least as far as I can see.

  • $\begingroup$ This is excellent. What I mean by the "this is a contour" paragraph is because the integral in question doesn't use the standard notation for a contour integral. Some people I was discussing this with objected that I'm conflating a Riemann integral with a contour integral, but I'm saying they're the same after you choose a parameterization. $\endgroup$
    – GFauxPas
    Dec 11, 2016 at 19:46
  • 1
    $\begingroup$ You are correct that every ordinary Riemann integral of a continuous function is a contour integral. $\endgroup$
    – zhw.
    Dec 11, 2016 at 19:48

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