# Contour Integral with Branch Cut

1. Aug 3, 2005

### cjellison

Hi, I've typed up my work. Please see the attached pdf.

Basically, I am trying to sovle this problem.

$$\int_0^\infty \frac{x^\alpha}{x^2+b^2} \mathrm{d}x$$

for $0 <\alpha < 1$. I follow the procedure given in Boas pg 608 (2nd edition)...and everything seems to work. However, when I check my results, with mathematica for various values of $\alpha$ and $b$, I get incorrect results.

Something I noticed: If I take the absolute value of my generally complex answer, I always get the correct (Mathematica) answer.

So, I would like someone to explain why this happens. What mistake have I made in my solution, and why doesn't this mistake effect Boas in her solution to:

$$\int_0^\infty \frac{r^{p-1}}{1+r} \mathrm{d}r$$

Both problems are solved in the same way...yet her solution $\pi / \sin(\pi p)$ is valid and does not require you to take the absolute value of it.

Thanks so much.

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2. Aug 3, 2005

### lurflurf

The most likely error is that somewhere you use a wrong branch.
The log you define has
0<=Im(log(z))<=2pi
the principle value has
-pi<=Im(log(z))<=pi
Mathematica uses principle values
remeber for your log
log(-i)=3pi*i/2

3. Aug 3, 2005

### lurflurf

So all shown work was right up to this the error comes after
$$I=i\pi\frac{[(ib)^{\alpha-1}+(-ib)^{\alpha-1}]}{1-e^{2\pi i \alpha}}$$
Just recall that the powers are not principle values
$$I=\pi b^{\alpha-1}\frac{[(i)^{\alpha}-(-i)^{\alpha}]}{1-e^{2\pi i \alpha}}$$
using -i=exp(3pi i/2) per nonprinciple roots
$$I=\pi b^{\alpha-1}\frac{exp(\frac{\pi\alpha i}{2})-exp(\frac{3\pi\alpha i}{2})}{1-exp(2\pi i \alpha)}=\frac{\pi}{2}b^{\alpha-1}\sec(\frac{\pi\alpha}{2})$$
thus your answer agrees with mine which is
(pi/2)(b^(a-1))sec(pi a/2)
also a substitution u=(x/b)^2
reduces this integral to your other example.

Last edited: Aug 3, 2005
4. Aug 3, 2005

### cjellison

Thanks.

So the error was in the way I instructed Mathematica to obtain a numerical answer...yet another lesson that Mathematica must be used carefully.

Can you explain a little more about your comments with the log? I don't understand how that comes into play...since I wasn't using logs at all (at least, I was not aware of it). Does it come from:

$$z^\alpha := e^{\alpha \log z} [/itex] Even so, why do we say that we are restricting $\log z$ to (0,2pi) rather than restricting z? 5. Aug 3, 2005 ### cjellison I think I got it....we are trying to restrict our numbers between 0 and 2 pi...so the log z must be restricted to 0 and 2 pi (since alpha is less than 1). 6. Aug 3, 2005 ### lurflurf yep thats it mathematica uses the principle value of log so log(-i)=-pi/2 your log has log(-i)=3pi/2 the log comes in indirectly as you showed through the raising to a power. the restriction is really [tex]\Im(\log(z))$$ to (0,2pi)
it just has to do with which branch of log is used
remeber that since mathematica used the principle branch you must convert before the calculation

7. Aug 3, 2005

### cjellison

Thanks again. Sorry for so many questions...I'm still trying to get a handle on this...in my classes (physics) we learned contour integration without really discussing branch points.

So, is it still true that
$$\cos \left( \frac{\pi a}{2}\right) = \frac{1}{2} \left( \mathrm{e}^{\mathrm{i} \pi a/2} + \mathrm{e}^{-\mathrm{i} \pi a/2} \right)$$

with my branch cut....it seems like the answer is no (since the minus sign puts the angle out of the range). It seems like it should become:

$$\cos \left( \theta\right) = \frac{1}{2} \left( \mathrm{e}^{\mathrm{i} \theta} + \mathrm{e}^{\mathrm{i} (2\pi - \theta)}\right)$$

If so, then I am having "trouble" converting

$$\frac{\exp(\frac{\pi\alpha i}{2})-\exp(\frac{3\pi\alpha i}{2})}{1-\exp(2\pi i \alpha)} = \frac{1}{2} \sec\left(\frac{\pi\alpha}{2}\right)$$

When I factor, I get an exponent which is negative...but this is not allowed (right?) since I must keep angles within (0,2pi). Can you comment on this and spell it out for me (a hint is fine too)?

I hope I don't slap myself after I see the responses to this.

8. Aug 4, 2005

### lurflurf

The answer is yes. Here you have cos in terms of exp. It is usual to retain the usual exp and let all the multibranch matters be delt with in the log. In any case in the above the correction was made in passing from
(-i)^a=exp(3pi*i*a/2) and the branch cut matters are now resolved

I tried to make clear that by this step we are back to principle values, and even if we were not we take exp to be single valued. Thus from this point we need not worry about branch cuts. A hint is to multiply the numerator and denominator of the left hand side by exp(-pi*alpha). Negative exponents are not a problem. This will give familar looking results since
2i*sin(z)=exp(i z)-exp(-i z)
some simple trig will then finish things

Last edited: Aug 4, 2005