Fantini said:
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 http://www.mathhelpboards.com/f13/need-some-help-these-review-questions-complex-analysis-please-2590/ 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!
