# Complex analysis

1. Jan 25, 2009

### asi123

Hey guys.
I need to calculate this integral so I was thinking about using the residue theorem.
The thing is that the point 0 is not enclosed within the curve that I'm about to build, it's on it.
Can I still use the theorem?

Thanks a lot.

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2. Jan 25, 2009

### MathematicalPhysicist

So use Cauchy Principal value to bypass the singularity at 0.

3. Jan 25, 2009

### lurflurf

You have a bigger problem then that, the residue at x=0 is zero (removeable singularity). This is a standard example and is often done as follows
consider
To fix the residue=0 problem add in an odd function the usual one is to use
exp(x i)/x=cos(x)/x+i sin(x)/x
the cos(x)/x diverges as an improper integral, but as lqg states cauch principle sense may be used
now consider ther contour formed by theese pieces
(-R,r) straight line
(r, R) straight line

4. Jan 25, 2009

### asi123

I didn't quite understand the contour you described there.
What's R and What's r?

Thanks a lot.

5. Jan 25, 2009

### lurflurf

Take limits R-> +infinity,r->+0
If you draw a pictuis an upper semicircle that in the limit is large with a tiny semicircle at the origin then you get for the various integrals

(-R,r) straight line
-infinity+(Integral you want)/2
{+,-}[+ if it was upper - if it was lower] pi*i (residue theorem)
(r, R) straight line
-infinity+(integral you want)/2
0

6. Jan 25, 2009

### asi123

Is it something like that?

Thanks again.

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7. Jan 26, 2009

### lurflurf

That is it. I guess that is one reasone why some people do not like old math books with no pictures. Were you able to finish?

8. Jan 26, 2009

### asi123

I'll try, if I'll have some troubles, I'll be back .
Thanks a lot.