Line integral along the circle (Correct me)

Question

Given question :

Evaluate

$$\oint_C \dfrac{\mathbb dz}{z-2}$$

Around the circle $|z-1|=5$

Since, the center point at $(1,0)$ and has a radius of $5$, i know $z=1+5e^{i\theta}$ with $0\leq \theta \leq 2\pi$.

Then i get

$$\int_0^{2\pi} \frac{5ie^{i\theta}}{5e^{i\theta}-1} \,\mathbb d\theta$$

Then, i'm trying to substitute $u=5e^{i\theta}$

And it gives me :

$$\int_5^5 \frac{\mathbb du}{u-1}$$

(Is there something wrong in my substitution?)

And we know that integral is $0$.

It contradicts with my book's solution which is the answer is $2\pi i$

And actually i agree with my book's solution answer considering i remembered with something formula or theorem like this:

$$\oint_C \frac{f(z_0)}{z-z_0}\,\mathbb dz=2\pi i f(z_0)$$

I just thought about the answer must be involving $2\pi i$

Why? Where is my mistake?

Edit 1 :

Well, if i really really can't do that integration with fundamental theorem of calculus bcz the function isn't continuous, how do i solve this integration without looking too far using the residual theorem or the cauchy integral theorem? let's say I don't know yet about those two theorems. I mean i want to solve this using basic parametrization. Without any theorem like green's theorem or something.

I'm too confused, bcz my solution manual said we have to Let $z-2=5e^{i\theta}$, But why? Why we have to do that? We know if we have locus $|z-1|=5$ this means $z=1+5e^{i\theta}$ right? And i know $z-2=5e^{i\theta}$ isn't equivalent with $z=1+5e^{i\theta}$ mathematically. My solution manual supposing the contour $C$ to be $\Gamma$ which is a circle of radius $5$ with center $z=a,\,a=2$ and letting $z-2=5e^{i\theta}$.

Does it means, whether we shift it right or left, the contour integral never changes?

Edit 2 :

Actually i know where is my mistake. What about i'm using long division or partial fraction decomposition like this:

$$\int_0^{2\pi} \frac{5ie^{i\theta}}{5e^{i\theta}-1} \,\mathbb d\theta=\int_0^{2\pi} \frac{i}{5e^{i\theta}-1} \,\mathbb d\theta +\int_0^{2\pi}i\,\mathbb d\theta$$

It will works right? It will gives me the correct answer $2\pi i$

Is it allowed???

Answer 1

Suppose that you are computing an integral like $\int_a^bf(x)\,\mathrm dx$. If $g\colon[c,d]\longrightarrow[a,b]$ is differentiable, if $g(c)=a$ and if $g(d)=b$, then we have$$\int_a^bf(x)\,\mathrm dx=\int_c^df\bigl(g(y)\bigr)g'(y)\,\mathrm dy.$$That's simple and I have stated all assumptions needed about the function $g$. In practice, what we do here is to take $x=g(y)$ and $\mathrm dx=g'(y)\,\mathrm dy$. You seem to believe that this is what you did.

But it isn't. You took an integral of the type $\int_a^bf\bigl(g(x)\bigr)g'(x)\,\mathrm dx$ and you did $x=g^{-1}(y)$ and $\mathrm dx=(g^{-1})'(y)\,\mathrm dy$. But your function $g$ is not a bijection and therefore $g^{-1}$ doesn't exist.

Answer 2

The part

And it gives me$$\int_5^5 \frac{\mathbb du}{u-1}$$

is wrong.

You are actually integrating over the contour where the starting point and the end point meets.

Answer 3

The integral

$$f=5i\int_0^{2\pi} \frac{1}{5-e^{-i\theta}} \,\mathbb d\theta$$

has the value $2 \pi i$.

How can we find this result starting from the indefinte integral, i.e. from an antiderivative?

We have

$$a(\phi)=5i\int \frac{1}{5-e^{-i\theta}} \, d\theta = \log \left(-1+5 e^{i \phi }\right)$$

Applying now blindly the fundamental theorem of calculus the integral $f$ should be the difference of the antiderivative at both of the integration interval.

But, alas, we find $a(\phi=0) = a(\phi = 2 \pi) = 0$

which would give $f=0$.

The problem is that that fundamental theorem holds only if the antiderivative is continuous in the integration interval. But this is not the case here.

As can be seen from the graph

the imaginary part of the antiderivative has a jump of $2 \pi$ at $\phi = \pi$.

Hence to make $a(\phi)$ continuous we have to add the amount of the jump to the result $0$ which gives an imaginary part $2 \pi$ as stated.

For completeness, the real part is continuous and hence gives no contribution to the integral $f$.