1. ## Evaluating Improper Integrals

Hi Again

I have another question for you that I'm stuck on. I get an answer on ipi*e^(ipi/6) can anyone confirm this?

Heres the question:

Note that theorem 1.1 is as follows:

Let p and q be polynomial functions such that:
1. the degree of q exceeds the degree of p by at least 2
2. any poles of p/q on the non-negative real axis are simple

Then for 0<a<1,

Integral (p(t)/q(t))t^a dt = -(pie^-piai cosec pia)S - (picot(pia)T

Where S is the sum of the residues of the function
f1(z) = p(z)/q(z) exp(alog2pi(z))

in C2pi and T is the residues of the function

f2(z)=p(z)/q(z) exp (alogz)

on the positive real axis

Thanks again

2. I am not familar with this theorem.

Let $\displaystyle x=t^{1/3}$ then,
Thus,
$\displaystyle \int_0^{\infty} \frac{3x^3}{x^6+1}dx$

Consider a contour $\displaystyle -R,+R$ in a semicircular fashion with $\displaystyle f(z) = \frac{3z^3}{z^6+1}$. So we get,
$\displaystyle \int_{-R}^R \frac{3x^2}{x^6+1} dx + \int_0^{\pi} \frac{3R^4 e^{3\theta i} i}{R^6 e^{6\theta i}+1} d\theta = 2\pi i \mbox{ RESIDUE }$

Now,
$\displaystyle \left| \frac{3R^4e^{3\theta i}i}{R^6e^{6\theta i}+1} \right| \leq \frac{3R^4|e^{3\theta i}i|}{|R^6e^{6\theta i}| - |1|} \leq \frac{3R^4}{R^6-1}\to 0 \mbox{ as }R\to \infty$

Now you need to find the poles and their residues and add them up to get your result.

3. Originally Posted by moolimanj
Hi Again

I have another question for you that I'm stuck on. I get an answer on ipi*e^(ipi/6) can anyone confirm this?
My only concern here (I'm not competent to help you with your theorem) is that you are apparently stating that the result of the integral is imaginary? This is (apparently) an integral over a real variable!

-Dan

4. Originally Posted by topsquark
This is (apparently) an integral over a real variable!
I am sure the poster means the the real part of the number.