# Thread: Help with integration (complex analysis)

1. ## Help with integration (complex analysis)

I'd like to know what is wrong in the following steps:

This will be 0, and I know for sure that this should be 2 PI i. What am I doing wrong?

PS: I know I can use other techniques such as residues and such, but I want to understand what I'm doing wrong when trying to integrate by hand.

2. Originally Posted by devouredelysium
I'd like to know what is wrong in the following steps:

This will be 0, and I know for sure that this should be 2 PI i. What am I doing wrong?

PS: I know I can use other techniques such as residues and such, but I want to understand what I'm doing wrong when trying to integrate by hand.
It would only be 0 if the limits were equal, and that's not very likely, is it?

3. Originally Posted by Mush
It would only be 0 if the limits were equal, and that's not very likely, is it?
I mean, from 0 to 2PI.

4. Originally Posted by devouredelysium
I mean, from 0 to 2PI.

What is the COMMON value of $e^{2\pi i}=e^0$?

Tonio

$\textit{Addition to the above}$: Ok, so by the above you get 0, but in may all depend on what value of Log(z) you're choosing: remember that Log(z) is multivalued and the argument of its value is determined only up to a multiple of $2\pi i$, so it may be this the reason why you think (or you were given or you read in the solutions sheet) that the value is $2\pi i$. It can as well be any multiple of this, too.

Tonio

5. Originally Posted by tonio
What is the COMMON value of $e^{2\pi i}=e^0$?

Tonio
From my understanding, e^i 2PI = e^0, or not?

6. Originally Posted by devouredelysium
From my understanding, e^i 2PI = e^0, or not?
Yes, and they're both equal to 1, and hence you're right that the integral is equal to 0. What makes you 'know for sure' that it's supposed to be 2 i pi?

7. Originally Posted by Mush
Yes, and they're both equal to 1, and hence you're right that the integral is equal to 0. What makes you 'know for sure' that it's supposed to be 2 i pi?
Well, it is known in complex analysis that functions of that kind have 2 PI i as integral. If I try to directly calculate it in mathematica, it will give me 2 PI i, too.

8. Originally Posted by devouredelysium
Well, it is known in complex analysis that functions of that kind have 2 PI i as integral. If I try to directly calculate it in mathematica, it will give me 2 PI i, too.

It is "known"? What are "those" functions that have that value "as integral"? And what "integral", exactly?

Can you be more specific and, perhaps, give some references to these claims?

Tonio

9. Originally Posted by tonio
It is "known"? What are "those" functions that have that value "as integral"? And what "integral", exactly?

Can you be more specific and, perhaps, give some references to these claims?

Tonio
Ok, that was it. Thanks!

10. Originally Posted by devouredelysium
What I am trying to do is calculate the integral over a closed curve with radius 2, centered at the origin. The function which I am trying to integrate over there is 1/(z+1). So I defined g(t) = 2exp(it) from 0 to 2PI as the integration path and that gave the integrals I have shown before.
Oh, I think I see now...sometimes posters change a question they're given and they don't add information relevant: of course, the complex integral you want indeed equals $2\pi i$ and this is due to the fact I added as $\textit{addition}$ to my second post: after revolving around a circle around a zero of $z+1$, indeed the integral $\oint\limits_{|z|=2}\frac{dz}{z+1}=2\pi i$ since the value of the primitive function $Log(z+1)$ (complex logarithm, of course) adds $2\pi i$ to its argument!

Tonio

11. Originally Posted by tonio
Oh, I think I see now...sometimes posters change a question they're given and they don't add information relevant: of course, the complex integral you want indeed equals $2\pi i$ and this is due to the fact I added as $\textit{addition}$ to my second post: after revolving around a circle around a zero of $z+1$, indeed the integral $\oint\limits_{|z|=2}\frac{dz}{z+1}=2\pi i$ since the value of the primitive function $Log(z+1)$ (complex logarithm, of course) adds $2\pi i$ to its argument!

Tonio
That log argument thing was it. Thanks!