Question

Let γ:[0,1]→C be a smooth curve and −γ the curve with the same image as γ, but traversed in the reverse direction (i.e., −γ(t)=γ(1−t) ). Show that for any continuous function f defined along γ, ∫γf(z)dz = −∫−γf(z)dz

Answer 1

To show that ∫γf(z)dz = -∫-γf(z)dz, we can use a change of variables and apply the properties of integrals. By introducing a new variable t = 1 - s and rearranging the integral expressions, we can show that the two integrals are equal but with opposite signs.

Step-by-step explanation:

To show that ∫γf(z)dz = -∫-γf(z)dz, we can use a change of variables. Let's define a new variable t = 1 - s. This means that s = 1 - t. Now, we can rewrite -γ(t) as γ(1 - t). Next, we need to find the differential dt in terms of the original variable ds. Taking the derivative of t with respect to s gives dt/ds = -1. Therefore, dt = -ds. Now, we can substitute this new variable and differential back into the integrals: ∫γf(z)dz = ∫f(γ(s))dγ(s) = ∫f(γ(s))d(γ(s))/ds ds = ∫f(γ(s))γ'(s)ds. Similarly, ∫-γf(z)dz = ∫f(γ(1 - t))γ'(1 - t)(-dt) = -∫f(γ(1 - t))γ'(1 - t)dt. Notice that γ(s) = γ(1 - t). Therefore, we have shown that ∫γf(z)dz = -∫-γf(z)dz, as desired.