Finding the Integral of 1/((x^2+x+1))^2

[jjangub]

Homework Statement

Show $$\int_{-\infty}^{\infty} 1/((x^2+x+1))^2 dx$$ = 4$$\pi$$/(3sqrt(3))

Homework Equations

The Attempt at a Solution

First, I tried to find the singularities, I used (-b$$\pm$$$$\sqrt{b^2-4ac}$$)/2a
so I got two singularities, but only (-b+$$\sqrt{b^2-4ac}$$)/2a works.
Now, I got stuck...for 1/((x^2+x+1))^2
I tried to use the q(z)/p(z) when p(z) is two or higher degree then q(z),
we can do q(z)/p'(z). But it didn't work.
How should I do this...
I need some help...
Thank you...

 

[hgfalling]

When you have poles of order 2, the q(z)/p'(z) formula doesn't work; you need to go back to the main expression for residues:

$$res_{z_0} = \lim_{z\rightarrow z_0} \left(\frac{1}{(n-1)!}\right) \frac{d^{n-1}}{dx^{n-1}}\left[(z-z_0)^n f(z) \right]$$

Then you should be able to get your residues, etc.

 

[jjangub]

sorry, but can you explain with other formula?
i didn't learn this formula, i instead learned a_n and b_n to find it.
but this function is different...

 

[hgfalling]

Suppose you have a function $f(z)$ which has a pole of order n at $z=z_0$, and you want to find the residue. Recall that the residue is the coefficient of the $(z-z_0)^{-1}$ term of the series expansion of $f(z)$.

Now let $g(z)=(z-z_0)^n f(z)$. Then g is holomorphic at $z_0$. (Do you see why?)

But since g is holomorphic, we can write it as a power series, like so:
$$g(z)=\sum_{k=0}^\infty a_k (z-z_0)^k=\sum_{k=0}^\infty \frac{g^{(k)}(z_0)(z-z_0)^k}{k!}$$

Now we can't just go dividing by $z-z_0$ because it's zero, but if we take a limit, we have:

$$f(z) = \lim_{z\rightarrow z_0} \frac{g(z)}{(z-z_0)^n}$$

And since what we want is the coefficient associated with the $(z-z_0)^{-1}$ term, we see that we have:

$$res_{z_0} = \lim_{z\rightarrow z_0} \frac{1}{(n-1)!} g^{(k)}$$

which matches the formula i posted before. But the idea is just that you expand f in a series and pull the coefficient right off the appropriate term.

 

[jjangub]

well...I tried... but how do I get 3sqrt(3)?
I can't even get any roots...
some hints please...

 

[hgfalling]

OK, here are the steps to doing this problem (or some other problems like it). If you want more help than this, you'll need to show some work.

Step 0: Draw a big half-circle connected to the real axis with radius R.
Step 1: Show that the big half-circle part goes to zero as R goes to infinity.
Step 2: Show that the integral over the entire contour equals the quantity you are trying to find as R goes to infinity.
Step 3: Find the poles that are inside the contour.
Step 4: Find the residue at the poles, using the formula above, or whatever method you like.
Step 5: Use the residue formula to find the value of the integral over the total contour.
Step 6: Win!

I will give you this one hint: the $\sqrt{3}$ comes from the residue calculations.