Branch cut integration

  • Thread starter gonzo
  • Start date
  • #1
277
0
I need help with a branch cut intgration. The problem is to show the following for [itex]0< \alpha <1[/itex]:

[tex]
\int_{0}^{\infty}{x^{\alpha - 1} \over x+1}={\pi \over sin\alpha\pi}
[/tex]

I used the standard keyhole contour around the real axis (taking that as the branch cut), but using the residue theorem I end up with:

[tex]
-\pi i e^{i\alpha\pi}
[/tex]

Which obviously doesn't match. Although this does match up for alpha equals one half.

Some help would be appreciated.
 

Answers and Replies

  • #2
shmoe
Science Advisor
Homework Helper
1,992
1
It would be easier to see what went wrong if you showed more work!

What was your residue? (I think you were Ok here)

How did you deal with the part of the keyhole contour that lies just below the x-axis? I think this is what went wrong. On this part you will be working with a different branch of the logarithm, so z^{alpha} will take on different values here than the bit above the x-axis.
 
  • #3
277
0
Thanks, you actually showed me my stupid mistake. I had just worked through the problem with alpha as one half, in which case the lower and upper part of the branch cuts are the same (well, negatives, but going in different directions, so you get a divisor of 2).

I had naively assumed that it would be the same for this problem. Your comment inspired me to take a closer look and I realized it wasn't the case, and then the answer was trivial.

Thanks!
 

Related Threads on Branch cut integration

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
9
Views
11K
  • Last Post
Replies
7
Views
10K
Replies
1
Views
4K
Replies
1
Views
2K
  • Last Post
Replies
3
Views
5K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
1K
Top