# What's the difference betwen parameterizations and variable substitution for solving integrals?

Asumming I have the following integral to solve in the complex plane: $$\int \frac{dz}{z+1}$$ while $|z|=5$ which means a contour of radius 5 around zero. Is it possible to solve this integral using: $$z+1=5e^{it}$$ while $t$ runs from $0$ to $2\pi$ ?

if this move is legit, what excactly am I doing? is it a parameterization or is it a varbile substitution?

## 1 Answer

That doesn't seem right.

$$z+1=5e^{it}$$ isn't a correct parameterization of the circle with radius 5 and the center as origin.

A correct parameterization would be: $z=z(t) = 5e^{it}$ which would lead to the integral:

$$\int_0^{2\pi}\frac{1}{5e^{it}+1}\cdot ie^{it} \text{d}t$$

### Is it a variable substitution?

No, then you would need to apply the 'transformation of contour integration'-theorem:

If $w$ holomorphic on $\Gamma$ and $f$ continous on $w(\Gamma)$ then $$\int\limits_{w(\Gamma)} f(w) \operatorname dw = \int\limits_\Gamma f(w(z)) w'(z) \operatorname dz$$

Which would imply that you need to change the contour itself when transforming $z+1=5e^{it}$

### How do I solve it?

There are various ways of solving this. You could use the residue theorem (which seems a bit overkill here).

Another possibility is to use the transformation of contour integrals and transforming $z +1 =w$, where the contour would shift 1 to the left.

Since the new contour still encloses the origin, you probably know the result: $2\pi i$

• If I make a transformation like you suggested for $z+1=w$ what would my curve shift into? – user3921 Jan 10 '15 at 12:36
• A circle with center $-1$ and radius 5. – dietervdf Jan 10 '15 at 13:58