Complex Integral help?

[Darya]

Hi, I have to solve the following integral:
My understanding is quite chaotic. But I have so far translated it into integral in terms of z only, using a theorem for complex integration.

My question is, can I now say that the primitive function of that is some ln(2z²+1)/4 and just plug in -1 and 1? I have some serious confusion about this because of discontinuity of ln(z). Do I have to state the interval for the argument? The singularities of the following function are ±√2i, does it matter?

 

[topsquark]

Hi, I have to solve the following integral:
My understanding is quite chaotic. But I have so far translated it into integral in terms of z only, using a theorem for complex integration.

My question is, can I now say that the primitive function of that is some ln(2z²+1)/4 and just plug in -1 and 1? I have some serious confusion about this because of discontinuity of ln(z). Do I have to state the interval for the argument? The singularities of the following function are ±√2i, does it matter?View attachment 32655

You can't really take it back to z because you then have to go back and integrate over your contour and you've gained nothing. You've got an integral over a real variable so stick with it!

What I'd do is this: There are far too many i's in your t integration for my comfort. So I'd expand [imath]e^{2it} = cos(2t) + i ~ sin(2t)[/imath] and separate the integral into a real and imaginary part. That leaves you with two (slightly ugly) integrals that contain no imaginary numbers.

-Dan

 

[Darya]

You can't really take it back to z because you then have to go back and integrate over your contour and you've gained nothing. You've got an integral over a real variable so stick with it!

What I'd do is this: There are far too many i's in your t integration for my comfort. So I'd expand [imath]e^{2it} = cos(2t) + i ~ sin(2t)[/imath] and separate the integral into a real and imaginary part. That leaves you with two (slightly ugly) integrals that contain no imaginary numbers.

-Dan

Thanks! I'll try to solve it this way. Our lector used to do this substitution z=e^it to probably get rid of |z| and z conjugate. To then express this function in terms of z only. But it was an example where the contour was the whole unit circle.

 

[topsquark]

Thanks! I'll try to solve it this way. Our lector used to do this substitution z=e^it to probably get rid of |z| and z conjugate. To then express this function in terms of z only. But it was an example where the contour was the whole unit circle.

Yes, that method would work much better over a closed contour.

-Dan