How Do You Solve a Complex Integral Using Cauchy-Goursat's Theorem?

[Terrell]

Homework Statement

$\int_{0}^{2\pi} cos^2(\frac{pi}{6}+2e^{i\theta})d\theta$. I am not sure if I am doing this write. Help me out. Thanks!

Homework Equations

Cauchy-Goursat's Theorem

The Attempt at a Solution

Let $z(\theta)=2e^{i\theta}$, $\theta \in [0,2\pi]$. Then the complex integral above becomes
\begin{align}
\int_{c}cos^2(z+\frac{\pi}{6})dz \quad \text{s.t. C:}z(\theta)=2e^{i\theta}, \quad \theta \in [0,2\pi]
\end{align}
Since $cos^2(z)$ is an entire function and $C$ is a simple closed contour, then by the Cauchy-Goursat Theorem the integral evaluates to zero.

 

[Orodruin]

$d\theta$ is not the same as $dz$.

 

[Terrell]

$d\theta$ is not the same as $dz$.

Can you please help me out? The class I got into went through the course too brashly, prof had lots of personal emergencies.

 

[Orodruin]

This is not restricted to complex analysis. Whenever you make a change of variables in an integral you must not only change the integrand but also you get a Jacobian determinant for the change in the integral measure. In particular, for a single variable integral, if $x = f(t)$, then $dx = (dx/dt)dt = f’(t) dt$.

 

[Orodruin]

See also https://en.wikipedia.org/wiki/Integration_by_substitution

 

[Terrell]

This is not restricted to complex analysis. Whenever you make a change of variables in an integral you must not only change the integrand but also you get a Jacobian determinant for the change in the integral measure. In particular, for a single variable integral, if $x = f(t)$, then $dx = (dx/dt)dt = f’(t) dt$.

*faceslap*... commonly called u-substitution. What was I thinking.

 

[Terrell]

dθdθd\theta is not the same as dzdzdz.

Can you at least give me hints as to how I should go about this? I feel like I should already know this stuff but my math background is very nontraditional.

 

[Orodruin]

Can you at least give me hints as to how I should go about this? I feel like I should already know this stuff but my math background is very nontraditional.

How would you do the substitution $x = t^2$ in the integral
$$
\int t^3 dt?
$$

 

[Terrell]

How would you do the substitution $x = t^2$ in the integral
$$
\int t^3 dt?
$$

\begin{align}
\int t^3 dt = \frac{1}{2}\int t^2\cdot 2dt &= \frac{1}{2}\int x dx \quad \text{s.t.}\quad dx=2dt\\
&=\frac{1}{2}\frac{x^2}{2}=\frac{x^2}{4}+K \quad\text{such that K is some constant.}
\end{align}
With the original post the most "natural" start for me is let $z=2e^{i\theta}+\frac{\pi}{6}$. So I then have $dz=2ie^{i\theta}d\theta$. Which gives me the integral
\begin{align}
\int_{c}\frac{\cos^2(z)}{2ie^{i\theta}}dz
\end{align}
Then I don't know what's next.

 

[Orodruin]

You still have $\theta$ left in your integral. Once you have made your substitution you should only have functions of $z$ (or rather, $\theta$ should be considered as a function of $z$). I would also suggest that your previous $z = e^{2i\theta}$ was easier, but it should not matter for your end result.

 

[Ray Vickson]

\begin{align}
\int t^3 dt = \frac{1}{2}\int t^2\cdot 2dt &= \frac{1}{2}\int x dx \quad \text{s.t.}\quad dx=2dt\\
&=\frac{1}{2}\frac{x^2}{2}=\frac{x^2}{4}+K \quad\text{such that K is some constant.}
\end{align}
With the original post the most "natural" start for me is let $z=2e^{i\theta}+\frac{\pi}{6}$. So I then have $dz=2ie^{i\theta}d\theta$. Which gives me the integral
\begin{align}
\int_{c}\frac{\cos^2(z)}{2ie^{i\theta}}dz
\end{align}
Then I don't know what's next.

If you put $z = e^{i \theta}$ in the above, you will have just about the most famous integral you can ever meet in a complex analysis course.

 

[Orodruin]

If you put $z = e^{i \theta}$ in the above, you will have just about the most famous integral you can ever meet in a complex analysis course.

Note that he used the substitution $z = 2 e^{i\theta} + \pi/6$, so I would not insert $z = e^{i\theta}$ (I would do the substitution $z = e^{i\theta}$ instead ...).

 

[Ray Vickson]

Note that he used the substitution $z = 2 e^{i\theta} + \pi/6$, so I would not insert $z = e^{i\theta}$ (I would do the substitution $z = e^{i\theta}$ instead ...).

I was replying to post #1, where he used $z = e^{i \theta}$ and obtained
$$
\int_{c}\cos^2\left(z+\frac{\pi}{6} \right)dz \quad \text{s.t. C:}z(\theta)=2e^{i\theta}, \quad \theta \in [0,2\pi]
$$
As you noted in #2, $dz$ and $d \theta$ are not the same, so the integral above is not quite correct.

In post #9 the OP switched to $z = \pi/6 + 2 e^{i \theta}$ and obtained the integral in (4). If he were to put back $z$ in place of $\theta$ he would get a simple, standard result and so could easily finish the calculation.

 

[Terrell]

If you put z=eiθz=eiθz = e^{i \theta} in the above, you will have just about the most famous integral you can ever meet in a complex analysis course.

so $dz=ie^{i\theta}d\theta$.
\begin{align}
\frac{1}{ie^{i\theta}}\int_{0}^{2pi}\cos^2(\frac{\pi}{6}+2e^{i\theta})ie^{i\theta}d\theta\\
\frac{1}{iz}\int_{1}^{1}\cos^2(\frac{\pi}{6}+2z)dz
\end{align}
I don't know the famous integral you refer to, my calc background is very minimal - I didn't really enjoy integrals in contrast to proof writing courses. I don't know where to go from here. Thanks.

 

[Orodruin]

You cannot move the z out of the integral. Also, writing 1 and 1 as limits is misleading. You are integrating along a closed path in $\mathbb C$.

 

[Terrell]

You cannot move the z out of the integral.

Yes, I know this. I just had to; to emphasize the substitution. Is it a long way to finish from here or is it close?

 

[Orodruin]

I just had to; to emphasize the substitution.

I cannot approve of this practice. You cannot write something that does not make sense just to emphasize something else. You could have just put it as a denominator for the rest of the integrand.

Is it a long way to finish from here or is it close?

You talked about the integrand being an entire function in your first post. What about now with the additional factor of 1/z?

 

[Terrell]

You talked about the integrand being an entire function in your first post. What about now with the additional factor of 1/z?

Well, z cannot be zero. But I don't think $z$ will ever be zero because $z=e^{i\theta}$. Right? So I still think it's entire.

 

[Orodruin]

Well, z cannot be zero. But I don't think $z$ will ever be zero because $z=e^{i\theta}$. Right? So I still think it's entire.

On the curve, no. But this is not what it means to be entire. The Cauchy integral theorem rests upon the assumption that the integrand is holomorphic everywhere.

 

[Terrell]

On the curve, no.

Why? The curve we speak of is $e^{i\theta}$, right?

But this is not what it means to be entire. The Cauchy integral theorem rests upon the assumption that the integrand is holomorphic everywhere.

Yes, entire is analytic everywhere which is synonymous to holomorphic.

Also, writing 1 and 1 as limits is misleading. You are integrating along a closed path in CC\mathbb C.

So it's better to keep 0 and $2\pi$ as limits?

 

[Orodruin]

Yes, entire is analytic everywhere which is synonymous to holomorphic.

So is your function holomorphic?

So it's better to keep 0 and 2π2π2\pi as limits?

It is better to write $\oint_\Gamma$ and specify what $\Gamma$ is.

 

[Terrell]

So is your function holomorphic?

$\frac{\cos^2(\frac{\pi}{6}+2z)}{iz}$ would not be holomorphic because it has an isolated singularity at the origin...? Also, this singularity is inside the curve we evaluate the integrand so this is where I use Cauchy's Integral Theorem, correct?

 

[Orodruin]

Exactly. Does your curve enclose the origin?

 

[Terrell]

It is better to write ∮Γ∮Γ\oint_\Gamma and specify what ΓΓ\Gamma is.

Now, I see why this is important. Thanks!

 

[Terrell]

Exactly. Does your curve enclose the origin?

Yes!

 

[Orodruin]

So can you apply Cauchy’s integral theorem (which only deals with holomorphic functions) or do you need to use a bigger gun (more general theorem)?

 

[Terrell]

So can you apply Cauchy’s integral theorem (which only deals with holomorphic functions) or do you need to use a bigger gun (more general theorem)?

Is this bigger gun the "Cauchy Integral Theorem for Derivatives"? I think the regular gun would suffice?

 

[Orodruin]

Is this bigger gun the "Cauchy Integral Theorem for Derivatives"? I think the regular gun would suffice?

No. The regular gun deals with integrals of holonomic functions and your function is not holonomic (you just confirmed that it has a pole at z=0). The improved gun is a generalisation of the integral theorem and starts with an R.

However, it strikes me now that you may be referring to the Cauchy integral formula rather than the Cauchy integral theorem. You can apply the formula, not the theorem.

 

[Terrell]

integrals of holonomic functions and your function is not holonomic

What is holonomic? Did you meant holomorphic? I googled holonomic and it exists and it means something else.

However, it strikes me now that you may be referring to the Cauchy integral formula rather than the Cauchy integral theorem. You can apply the formula, not the theorem.

Yes, I meant formula - sorry. But, the formula is applicable right?

 

[Orodruin]

What is holonomic? Did you meant holomorphic?

Yes, it is late here (4 am) and I am suffering from insomnia so not completely alert ...

Yes, you can apply the Cauchy integral formula with $z_0 = 0$. The more general theorem is the residue theorem.

 

[Terrell]

Yes, it is late here (4 am) and I am suffering from insomnia so not completely alert ...

Yes, you can apply the Cauchy integral formula with z0=0z0=0z_0 = 0. The more general theorem is the residue theorem.

Got it! Thank you, I think I can handle this from here. Sorry if I am taking a while, I am multi-problem solving right now . Exam is tomorrow.

 

[Terrell]

Just to have closure. I'll put the entire solution here. By letting $z=e^{i\theta}$ and $dz=ie^{i\theta}d\theta$. we have
\begin{align}
\frac{1}{ie^{i\theta}}\int_{c}\cos^2(\frac{\pi}{6}+2e^{i\theta})ie^{i\theta}d\theta = \frac{1}{i}\int_{c}\frac{\cos^2(\frac{\pi}{6}+2z)}{z}dz
\end{align}
Since $\frac{\cos^2(\frac{\pi}{6}+2z)}{z}$ is analytic everywhere except at the origin - which is encapsulated by the given contour/curve we are integrating on, then by the Cauchy Integral Formula, letting $f(z)=\cos^2(\frac{\pi}{6}+2z)$ so $f(0)=3/4$ we have
\begin{align}
\int_c\frac{f(z)}{z}dz=2\pi i\frac{3}{4}=\frac{3\pi i}{2}
\end{align}

 

[Ray Vickson]

Just to have closure. I'll put the entire solution here. By letting $z=e^{i\theta}$ and $dz=ie^{i\theta}d\theta$. we have
\begin{align}
\frac{1}{ie^{i\theta}}\int_{c}\cos^2(\frac{\pi}{6}+2e^{i\theta})ie^{i\theta}d\theta = \frac{1}{i}\int_{c}\frac{\cos^2(\frac{\pi}{6}+2z)}{z}dz
\end{align}
Since $\frac{\cos^2(\frac{\pi}{6}+2z)}{z}$ is analytic everywhere except at the origin - which is encapsulated by the given contour/curve we are integrating on, then by the Cauchy Integral Formula, letting $f(z)=\cos^2(\frac{\pi}{6}+2z)$ so $f(0)=3/4$ we have
\begin{align}
\int_c\frac{f(z)}{z}dz=2\pi i\frac{3}{4}=\frac{3\pi i}{2}
\end{align}

Good, so you have used that "famous" integral I mentioned. Every single textbook about complex analysis will have lots of material using that integral formula, since it is possibly one of the most important formulas in the whole field. In applications it is used over and over again, roughly in the manner that you just now used it.