MHB Reverse direction for complex functions

[ognik]

Hi

An exercise asks to show $ \int_{a}^{b}f(z) \,dz = -\int_{b}^{a}f(z) \,dz $
I can remember this for real functions, something like $ G(x) = \int_{a}^{b}f(x) \,dx = G(b) - G(a), \therefore \int_{b}^{a}f(x) \,dx = G(a) - G(b) = -\int_{a}^{b}f(x) \,dx $

I have seen 2 approaches, either parametizing w.r.t t or expanding z=u+iv, what I don't understand is why I need to paramatise/expand f(z)? Both of those approaches also have complex intergrands, so why does the paramatise/expand step make a difference?

 

[chisigma]

Hi

An exercise asks to show $ \int_{a}^{b}f(z) \,dz = -\int_{b}^{a}f(z) \,dz $
I can remember this for real functions, something like $ G(x) = \int_{a}^{b}f(x) \,dx = G(b) - G(a), \therefore \int_{b}^{a}f(x) \,dx = G(a) - G(b) = -\int_{a}^{b}f(x) \,dx $

I have seen 2 approaches, either parametizing w.r.t t or expanding z=u+iv, what I don't understand is why I need to paramatise/expand f(z)? Both of those approaches also have complex intergrands, so why does the paramatise/expand step make a difference?

If the integral $\displaystyle \int_{a}^{b} f(z)\ d z$ does not depend on the path connecting a and b in the complex plane, then f(z) is analytic in the whole complex plane and effectively is $\displaystyle \int_{a}^{b} f(z)\ d z = - \int_{b}^{a} f(z)\ d z$...

... otherwise, i.e. if f(z) has some singularities in the complex plane, that may be not true...

Kind regards

$\chi$ $\sigma$

 

[ognik]

Thanks χ σ , but the exercise doesn't specify that f(z) is analytic.

Maybe more context would help - it appears just after a section on Cauchys Integral theorem.

That section also says: "A consequence of the Cauchy integral theorem is that for analytic functions the line integral is a function only of its end points, independent of the path of integration," - which would surely mean I could directly use the same argument I used for real functions above, WITHOUT paramatising or expanding, wouldn't it?

 

[chisigma]

Thanks χ σ , but the exercise doesn't specify that f(z) is analytic.

Maybe more context would help - it appears just after a section on Cauchys Integral theorem.

That section also says: "A consequence of the Cauchy integral theorem is that for analytic functions the line integral is a function only of its end points, independent of the path of integration," - which would surely mean I could directly use the same argument I used for real functions above, WITHOUT paramatising or expanding, wouldn't it?

The difference between integrating in $\mathbb {R}$ and integrate in $\mathbb {C}$ is that in the second case it must be generally specify not only the limits of integration a and b, but also the path that connects a and b...

Let's consider the following illustrative example taking $f(z) = \frac{1}{z}, a=1, b=-1$ and suppose to choose as path of integration the upper half cirle of the figure...
http://cdn.simplesite.com/i/c2/c2/286260055773790914/i286260064307432655._szw1280h1280_.jpg

Setting $\displaystyle z=e^{i\ \theta}$ we have...

$\displaystyle \int_{1}^{-1} \frac{d z}{z} = i\ \int_{0}^{\pi} d \theta = i\ \pi\ (1)$

Now we have to integrate from -1 to 1 and, among other, we have the opportunity to choose two different paths...

a) the upper half circle in the figure, i.e. the reverse path than the previous case, and we have...

$\displaystyle \int_{-1}^{1} \frac{d z}{z} = i\ \int_{\pi}^{0} d \theta = - i\ \pi\ (2)$

... so that Your relation is verified...

b) the bottom half circle of the figure and we have...

$\displaystyle \int_{- 1}^{1} \frac{d z}{z} = i\ \int_{\pi}^{2\ \pi} d \theta = i\ \pi\ (3)$

... and Your relation isn't verified...

The reason of that is that the function $f(z)= \frac{1}{z}$ is not analytic in the whole complex plan, having a singularity in z=0...

Kind regards

$\chi$ $\sigma$

 

[ognik]

Thanks χ σ ,
I know that if a function is complex analytic, it is independent of the path - so it seems to me I have no other option but to assume F(z) is analytic?

My concern is that the solutions I have seen either expand the function (u,v) or parametise it (t) - and then conclude that the reverse integral does change sign. But my logic says - if the function is not analytic, then neither expanding nor parametising will suddenly make it analytic?