Integration a long closed curve is 0

[alyafey22]

\int_{\gamma (t) }\, f(z) dz

\int_{\alpha}^{\beta} \, f(\gamma (t))\, \gamma '(t) \, dt

\text{Use the substitution : } \gamma (t) = \xi

\int_{\gamma (\alpha) }^{\gamma (\beta)} \, f(\xi )\, d \xi

\text{If we integrate around a closed loope : }\gamma (\alpha) = \gamma(\beta)

\int_{\gamma (\alpha) }^{\gamma (\alpha)} \, f( \xi )\, d \xi =0

\text{This is only true if the function is analytic }

Feel free to leave any comments .

 

[topsquark]

\int_{\gamma (t) }\, f(z) dz

\int_{\alpha}^{\beta} \, f(\gamma (t))\, \gamma '(t) \, dt

\text{Use the substitution : } \gamma (t) = \xi

\int_{\gamma (\alpha) }^{\gamma (\beta)} \, f(\xi )\, d \xi

\text{If we integrate around a closed loope : }\gamma (\alpha) = \gamma(\beta)

\int_{\gamma (\alpha) }^{\gamma (\alpha)} \, f( \xi )\, d \xi =0

\text{This is only true if the function is analytic }

Feel free to leave any comments .

Comment, not a full answer.

This will work if your function f is conservative. I don't how that relates to the analyticity of f.

-Dan

 

[alyafey22]

Comment, not a full answer.

This will work if your function f is conservative. I don't how that relates to the analyticity of f.

-Dan

But how to define conservative functions mathematically ?
We define analytic functions as those which satisfy the Cauchy-Reimann equations and the partial derivatives exist and are continuous so if they have a pole then we can use the Cauchy-integral formula to find the integral along the loop this is illustrated by the deformation hypothesis .

 

[topsquark]

But how to define conservative functions mathematically ?
We define analytic functions as those which satisfy the Cauchy-Reimann equations and the partial derivatives exist and are continuous so if they have a pole then we can use the Cauchy-integral formula to find the integral along the loop this is illustrated by the deformation hypothesis .

Ah! It's a complex integration. You didn't tell us that. (Tmi)

Then as far as I know, so long as you have a closed path (that doesn't contain any nasty singularities) then the answer is 0.

-Dan

Come to think about it, if it's analytic I think that means no singularities. I'm too lazy to check that. Time for a nap! (Yawn)

-Dan

 

[Ackbach]

If $f$ is analytic inside the region enclosed by $\gamma$, the integral in question will be zero. Are you putting forth a proof of that? I'm a little unclear what it is you're after.

 

[alyafey22]

If $f$ is analytic inside the region enclosed by $\gamma$, the integral in question will be zero. Are you putting forth a proof of that? I'm a little unclear what it is you're after.

Yes, indeed.

 

[Ackbach]

What are your assumptions? What theorems are you allowed to use?

 

[alyafey22]

I am using the contour integration formual :

\int_{\gamma (t) }\, f(z) dz =\int_{\alpha}^{\beta} \, f(\gamma (t))\, \gamma '(t) \, dt