Complex Analysis and Change of Variables in Line Integrals

[Edwin]

Consider the function:

g(z(t)) = i*f '(c+it)/(f(c+it) - a)

Where {-d <= t < d}

If we let z = c+it

By change of variables don't we get:

Line integral of g(z(t)) = i ln[f(c+it) - a]

evaluated from t = - d to t = d?

note: ln is the natural log.

Inquisitively,

Edwin G. Schasteen

 

[leach]

You may consider the line integral as the complex path integral

\int_\gamma\, \frac{f&#039;(z)}{f(z)-a}\,dz

where \gamma(t) = c + it, for -d\le t\le d.

Since the integrand has the trivial primitive G(z) = \ln(f(z)-a), you may indeed consider that:

\int_\gamma\, \frac{f&#039;(z)}{f(z)-a}\,dz \quad = \quad G(c+id) - G(c-id)

You should be cautious anyway, since for this integral to be right, it must be f(z) \ne a along \gamma. Assuming that g(z) has no singularities in a domain containing the path \gamma, I think you can safely consider the primitive G(z) as correct to compute the integral.

 

[Edwin]

That makes sense.

Thanks! I appreciate your help!

Best Regards,

Edwin