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

  • Thread starter Thread starter jjangub
  • Start date Start date
  • Tags Tags
    Integral
jjangub
Messages
20
Reaction score
0

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...
 
Physics news on Phys.org
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.
 
sorry, but can you explain with other formula?
i didn't learn this formula, i instead learned an and bn to find it.
but this function is different...
 
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.
 
well...I tried... but how do I get 3sqrt(3)?
I can't even get any roots...
some hints please...
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top