# Calculating an integral.

1. Jan 19, 2008

### MathematicalPhysicist

I probably need to calculate this integral with complex residues calculus, but i'm having difficulty with it:
$$\int_{0}^{\infty}\frac{dx}{x^a(x-4)}$$
where 0<a<1 is real number.
I mean trying to find a closed contour here seems to be a pain in the neck, I mean how do i calculate the residue of 1/(x^a), assuming my contour runs with a circle around it (not from inside the curve but outside it, i hope it's clear.

2. Jan 19, 2008

### malawi_glenn

if a where an integer I would know how to do it, not by residues.

If you want to solve this by residues, I have no idea.

It have to bee solved analytically?

And the pole in x= 0 is of the order a, which is a real number, not an integer. No ideas.

3. Jan 19, 2008

### MathematicalPhysicist

yes, analytically.

4. Jan 19, 2008

### weburbia

You need to put a cut from the singularity at the origin to infinity. Have it go along the real axis but skirting round the pole at x=4. Now look for a closed contour that does not cross the cut. When you go round the singularity at the origin in a small circle you pick up a factor....

5. Jan 19, 2008

### mjsd

you will need a so-called "keyhole contour" (i guess that's what the previous poster is suggesting). I've got a feeling that it may also work with a half-disk "indented contour" with the help of fractional residue theorem (but not tried that before).

let me give you a hint by giving you a standard result:
$$\int_0^{\infty}\frac{1}{z^a (z+1)} dz = \frac{\pi}{\sin(\pi a)},\quad 0<a<1$$

6. Jan 19, 2008

### mjsd

After thought: you may need to check the convergence of the integral in $$\left[0,\infty\right)$$ since my example involves "+1" while yours is "-4". the pole location is different and i've go the feeling that it may have problems. if you have have
$$\left[0,-\infty\right)$$ instead, it would be just as my example.

EDIT: now I understand you difficulty. you can't have a branch cut from 0 to +ve infinity.

Last edited: Jan 19, 2008
7. Jan 19, 2008

### Rainbow Child

Try the following contour, and the integral

$$I=\oint_C\frac{d\,z}{z^\alpha(z-4)}=\oint_C f(z)\,d\,z$$

Since there are no ploles inside $$C$$, we have $$I=0$$.

It is easy to prove that
$$\int_{C_1}f(z)\,d\,z=\int_{C_2}f(z)\,d\,z=0$$
and
$$\lim_{\epsilon \rightarrow 0}\int_{C_3}f(z)\,d\,z=-4^{-\alpha}\,i\,\pi$$

Now write down the rest of the integrals, and equate real and imaginary parts. You should arrive to

$$\int_{0}^{\infty}\frac{dx}{x^a(x-4)}=4^{-\alpha}\,\pi\,\cot(\pi\,\alpha)$$

Be careful for the AB integral. There $$z=x\,e^{i\,\pi}$$

EDIT Change sign in $$C_3$$ integral.

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Last edited: Jan 20, 2008
8. Jan 20, 2008

### MathematicalPhysicist

shouldn't the integral of C_3 when epsilon approaches zero equal minus what youv'e wrriten, we learned it should be this way.

besides, this thanks, i did it before i asked, i just wasn't sure what is a_-1 of 1/z^a, but from writing i guess it should be zero, correct? (when a is between 0 and 1).

9. Jan 20, 2008

### MathematicalPhysicist

how do i calculate the integral of AB?
it goes from -R to -epsilon

10. Jan 20, 2008

### Rainbow Child

You are right for the integral of $$C_3$$. I changed it in my previous post. Calculations are not my best thing!

Now let's denote $$I$$ your integral. The integral on AB is

$$\int_{-R}^{-\epsilon}\frac{e^{-i\,\pi\,\alpha}}{x^\alpha(x+4)}\,d\,x \xrightarrow[\substack{\epsilon \to 0\\ R \to \infty}]{} e^{-i\,\pi\,\alpha}\,\int_{-\infty}^0 \frac{d\,x}{x^\alpha(x+4)}=e^{-i\,\pi\,\alpha}\,J$$

Thus we have

$$\oint_C f(z)\,d\,z=0\Rightarrow I+e^{-i\,\pi\,\alpha}\,J-i\,4^{-\alpha}\,\pi=0 \quad (I,J) \in \mathbb{R}$$

In this equation equate real and imaginary parts, in order to arrive to the wanted result.

Last edited: Jan 21, 2008
11. Jan 21, 2008

### mjsd

I am puzzled as to how you get this result... either there is a typo here somewhere or you have done something wrong.

but regardless of that, how did you manage to pick out a piece that corresponds to integration from 0 to infinity from your "half-disk indented contour"?

I've got a feeling that you have either done the following
$$\int_{0}^{\infty} \frac{dx}{x^a (x+4)} = \frac{4^{-a}\pi}{\sin(a\pi)}, \quad 0<\text{Re}(a)<1$$
or
$$\int_{-\infty}^{0} \frac{dx}{x^a (x-4)}=-\frac{(-4)^{-a}\pi}{\sin(a\pi)}, \quad 0<\text{Re}(a)<1$$

12. Jan 21, 2008

### Rainbow Child

Do we agree that in contour at hand

$$\oint_C f(z)d\,z=0 \, ?$$

Then

$$\oint_C=\int_{C_1}+\int_{AB}+\int_{C_2}+\int_{CD}+\int_{C_3}+\int_{EF}=0$$

At the limit $$\epsilon\to 0,R \to \infty$$
• The 1st and the 3rd integral vanishes.
• The 4th and the 6th integral combine to
$$I=\int_{0}^{\infty}\frac{dx}{x^a(x-4)}$$
• The 5th integral gives
$$\int_{C_3}f(z)\,d\,z=-4^{-\alpha}\,i\,\pi$$
• The 2nd integral
$$e^{-i\,\pi\,\alpha}\,\int_{-\infty}^0 \frac{d\,x}{x^\alpha(x+4)}=e^{-i\,\pi\,\alpha}\,J$$

Now apply what is said in my previous post and you will calulate both

$$I=\int_{0}^{\infty}\frac{dx}{x^a(x-4)},\, J=\,\int_{-\infty}^0 \frac{d\,x}{x^\alpha(x+4)}$$

This is a general trick, applied when you have a pole on the real line and the integral "passes" through it, i.e. Cauchy principal value.

Does this clears up things?

13. Jan 21, 2008

### mjsd

I think I know why I had a concern that you didn't have. just a matter of how we interpreted the question:
you took it as "calculating the given integral, if it is not absolutely convergent then just find the PV of it"
while
I took it as "calculating the given integral as it is, if it is not absolutely convergent then...say it is not doable..."

ok, fair enough, i guess the PV would be useful anyway. But I did thought that since the singularity at +4 is non-removable, you probably can't evaluate the original integral in the same sense as you would evaluate, say

$$\int_{0}^{\infty} \frac{1}{x^a(x+1)}dx = \frac{\pi}{\sin(\pi a)}$$

which can be done (by using a keyhole contour which picks up the residue at -1) without resorting to just finding the PV.

Perhaps you see where I was coming from before, when I commented divergence etc. And since in the original post there is no mention of PV of the integral... I thought you are not meant to do that.. but in hindsight it is probably what the intention was.