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Complex Analysis Review Question

thatonekid

New member
Nov 27, 2012
2
Use residues toe evaluate the improper integral

Use residues toe evaluate the improper integral
[FONT=MathJax_Main][FONT=MathJax_Main] [/FONT][/FONT]

$$\int_{0}^{\infty} \dfrac{dx}{(x^2 +9)^3}.$$

Explain all steps including convergence. No need to simplify the final answer.

I took this off a old mid-term that I was looking at, no solution is provided, wonder if anyone had any idea how to start this problem off. Thanks!
 
Last edited by a moderator:

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
My complex analysis skills are a bit rusty, but I'll try to see what I can do. As I remember, by residue methods you'll calculate the value of

$$\int_{- \infty}^{+ \infty} \frac{dx}{(x^2 +9)^3}.$$

You will find the desired integral by noting that

$$\int_{0}^{+ \infty} \frac{dx}{(x^2 +9)^3} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dx}{(x^2 +9)^3}.$$

Making the passage to the complex plane, the denominator becomes

$$\frac{1}{(z^2 +9)^3} = \frac{1}{(z-3i)^3 (z+3i)^3}.$$

This means we have two poles of order three: one at $z=3i$ and the other at $z=-3i$. To find the residue you must evaluate only the pole at $z=3i$.

$$\lim_{z \to 3i} \frac{1}{2!} \frac{d^2}{dz^2} (z-3i)^3 f(z),$$

where

$$f(z) = \frac{1}{(z-3i)^3 (z+3i)^3}.$$

I've found that

$$\lim_{z \to 3i} \frac{1}{2!} \frac{d^2}{dz^2} (z-3i)^3 f(z) = \lim_{z \to 3i} \frac{6}{(z+3i)^5} = \frac{6}{7776i} = - \frac{i}{1296}.$$

Therefore

$$\int \frac{dz}{(z^2 +9)^3} = 2 \pi i \left[ \sum \mathcal{Res} \right] = 2 \pi i \left[ - \frac{i}{1296} \right] = \frac{\pi}{648}$$

and finally

$$\int_{0}^{+ \infty} \frac{dx}{(x^2 +9)^3} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dx}{(x^2 +9)^3} = \frac{1}{2} \cdot \frac{\pi}{648} = \frac{\pi}{1296}.$$

This gave me some trouble! (Speechless) Hope you get the general idea. Also, try to write out all your formulas in LaTeX. You can find the codes I used by right-clicking and choosing "Show Math As $\to$ TeX Commands". Do it in your other thread as well.
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
My complex analysis skills are a bit rusty, but I'll try to see what I can do. As I remember, by residue methods you'll calculate the value of

$$\int_{- \infty}^{+ \infty} \frac{dx}{(x^2 +9)^3}.$$

You will find the desired integral by noting that

$$\int_{0}^{+ \infty} \frac{dx}{(x^2 +9)^3} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dx}{(x^2 +9)^3}.$$

Making the passage to the complex plane, the denominator becomes

$$\frac{1}{(z^2 +9)^3} = \frac{1}{(z-3i)^3 (z+3i)^3}.$$

This means we have two poles of order three: one at $z=3i$ and the other at $z=-3i$. To find the residue you must evaluate only the pole at $z=3i$.

$$\lim_{z \to 3i} \frac{1}{2!} \frac{d^2}{dz^2} (z-3i)^3 f(z),$$

where

$$f(z) = \frac{1}{(z-3i)^3 (z+3i)^3}.$$

I've found that

$$\lim_{z \to 3i} \frac{1}{2!} \frac{d^2}{dz^2} (z-3i)^3 f(z) = \lim_{z \to 3i} \frac{6}{(z+3i)^5} = \frac{6}{7776i} = - \frac{i}{1296}.$$

Therefore

$$\int \frac{dz}{(z^2 +9)^3} = 2 \pi i \left[ \sum \mathcal{Res} \right] = 2 \pi i \left[ - \frac{i}{1296} \right] = \frac{\pi}{648}$$

and finally

$$\int_{0}^{+ \infty} \frac{dx}{(x^2 +9)^3} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dx}{(x^2 +9)^3} = \frac{1}{2} \cdot \frac{\pi}{648} = \frac{\pi}{1296}.$$

This gave me some trouble! (Speechless) Hope you get the general idea. Also, try to write out all your formulas in LaTeX. You can find the codes I used by right-clicking and choosing "Show Math As $\to$ TeX Commands". Do it in your other thread as well.
This is an ok response (incomplete), but it can be improved upon. There are a couple additional things that need to be addressed:

1. The contour the integration is being done over. In this case, I'd suggest going with the upper half circle of radius $R$, i.e. $\Gamma = [-R,R]\cup C_R$, where $C_R=\{Re^{i\theta}:\theta\in[0,\pi]\}$ is the arc of the upper half circle of radius $R$.

2. You need to show that $\left|\int_{C_R}f(z)\,dz\right|\rightarrow 0$ as $R\rightarrow \infty$.

These are important because after making the change from real variables to complex variables and applying the residue theorem, we have

\[2\pi i\sum\text{res}(f(z)) = \int_{\Gamma} f(z)\,dz = \int_{-R}^R\frac{\,dx}{(x^2+9)^3} + \int_{C_R}\frac{\,dz}{(z^2+9)^3}\]

Let us focus on

\[\left|\int_{C_R}\frac{\,dz}{(z^2+9)^3}\right|\]

We know that

\[\left|\int_C f(z)\,dz\right|\leq \ell(C)\max_{z\in C}|f(z)|\]

where $\ell(C)$ is the length of the contour. So in our case, we have that

\[\left|\int_{C_R}\frac{\,dz}{(z^2+9)^3}\right|\leq \pi R\max_{z\in C_R}\left|\frac{1}{(z^2+9)^3}\right|\]

Now, we note that

\[\left|\frac{1}{(z^2+9)^3}\right|=\left|\frac{1}{z^2+9}\right|^3\leq\frac{1}{\left||z|^2-9\right|^3}\]

Therefore,

\[\max_{z\in C_R}\left|\frac{1}{(z^2+9)^3}\right|\leq\max_{z\in C_R}\frac{1}{||z|^2-9|^3} = \frac{1}{(R^2-9)^3}\]

and we now see that as $R\rightarrow \infty$,

\[\pi R\max_{z\in C_R}\left|\frac{1}{(z^2+9)^3}\right|\leq \frac{\pi R}{(R^2-9)^3}\rightarrow 0.\]

Thus, as $R\rightarrow\infty$,

\[2\pi i\sum\text{res}(f(z)) = \int_{-R}^R\frac{\,dx}{(x^2+9)^3} + \int_{C_R}\frac{\,dz}{(z^2+9)^3}\rightarrow 2\pi i \sum\text{res}(f(z)) = \int_{-\infty}^{\infty}\frac{\,dx}{(x^2+9)^3}\]

which is what we were after (modify this result to get the answer we really want).

I hope this helps!
 
Last edited:

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Thank you Chris. If I can excuse myself for not including all those details, that is because I didn't learn how to do them. When I took my complex variable course the teacher didn't quite teach us how to do this contour integration, but rather that we should take the poles on the positive axis and that would be it. What followed is what you see in my post: doing the calculations. I'll keep this in mind in next complex analysis problems! (Nod)
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Thank you Chris. If I can excuse myself for not including all those details, that is because I didn't learn how to do them. When I took my complex variable course the teacher didn't quite teach us how to do this contour integration, but rather that we should take the poles on the positive axis and that would be it. What followed is what you see in my post: doing the calculations. I'll keep this in mind in next complex analysis problems! (Nod)
I didn't mean to call you out!

What you did there was correct; I was just filling in the missing details. And no worries about that; I didn't learn how to fill in the details for things like this until I took a graduate level course in complex analysis.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
I didn't mean to call you out!

What you did there was correct; I was just filling in the missing details. And no worries about that; I didn't learn how to fill in the details for things like this until I took a graduate level course in complex analysis.
I didn't feel called out! I appreciate when someone fills in missing details / correct me. That way I learn more than if I had gotten the question "right" the first time. (Clapping) I hope I can do a graduate level course in complex analysis soon enough. I believe I will greatly enjoy it!