Cauchy Principal Value integral

[eXorikos]

Homework Statement

Calculate I=P\int^{\infty}_{- \infty} \frac{e^{ikx}}{x} dx

Homework Equations

I=P\int^{\infty}_{- \infty} \frac{f(x)}{x-x_0} dx = i \pi f(x_0) + 2 \pi i \sum a_{-1}(z_+)

The Attempt at a Solution

According to Maple the solution is 2i\pi. Now if I try to calculate it using the above formula, I find f(x_0)=e^0=1
and since f(x)/x doesn't have any poles in the upper halfplane the sum of the residues is zero. This leads to
I=P\int^{\infty}_{- \infty} \frac{f(x)}{x-x_0} dx = i \pi

Where did I go wrong?

Also, to use this formula for the integral the line integral over the upper halfplane must be zero. So to prove this you have to calculate: \lim_{R\rightarrow \infty} R f(R) This is not zero. I'm confused now...

 

[jackmell]

Try to analyze it more slowly and carefully. What, exactly, is the contour you're using? Suppose for now, it's the half-washer in the upper half-plane, the one with an indentation around the origin. Ok, so that looks likes:

\int_P+\int_O+\int_R=0

where P is principal-value, O is the indentation around the origin, and R is the half-circle contour. Just look at the integral over R for now. Wouldn't that be:

\int_0^{\pi} \frac{e^{ikRe^{it}}}{Re^{it}} Rie^{it} dt

Can you simplify that, convert to sines and cosines, and show what k has to be in order for that to converge to zero? Ok, now look at closing the contour in the lower half-plane and do the same analysis. What must k be in order for that one to converge? Get that straight then look up that theorem that discusses what happens when you integrate over an indentation around a simple pole when the radius of the indentation goes to zero. Find that theorem and study it. It's something like it's equal to i\theta r_0 where theta is the (signed) radian measure of the contour and r_0 is the residue. That should do it huh?

 

[eXorikos]

It was stated explicitly that k is a real positive number.

So I get i \int^\pi_0 e^{ikRcos(t)-kRsin(t)} dt but where do I go from there? If I take take the limit of tje exp(-kRsin(t)) for R going to infinity it is clear that this is zero. But if the exponential including the cosine is still infinite, this gets me nowhere?

I'm really fuzy on my calculus, since it's been 4 years... :/

 

[jackmell]

You have:

i\int_0^{\pi} e^{ikR e^{it}}dt

now convert that to trigs:

i\int_0^{\pi} e^{-kR\sin(t)+ikR\cos(t)}dt

Now, we're only interested in the bounds of that integral as R goes to infinity, so we could take the absolute value of the integrand and write that the absolute value of the bound will be no more than:

\int_0^{\pi}e^{-kR\sin(t)}dt

that's bingo-bango right?

 

[eXorikos]

That's indeed correct. Thanks for that.

Now what about my first question?

 

[jackmell]

Tell you what, it's up to you what you want to do. Now, we've established:

\int_P=-\int_O

Ok, there you go. Unless you are required to do all that other stuff you mentioned, then I would think the easiest thing to do now is compute that integral over O, negate it, and you have the answer. But I think you really should look up the theorem on that first, get it straight, then you'll see the answer is easy.

Oh, I see what you're asking. Maple is wrong or rather I think it's more likely you didn't state the problem correctly for Maple. The answer is pi i.

 

[eXorikos]

Ok, thanks for your help!

I used the commando:
int(f(x),x=-infinity..infinity,CauchyPrincipalValue) assuming positive;