For example, how can I calculate
$$\int_{\left | z \right |=1}\frac{dz}{z}$$
I know that the answer is $2\pi i$ but how do I do it using Mathematica?
$\begingroup$
$\endgroup$
1
-
2$\begingroup$ Parametrize the range of integration. Then make sure you properly rewrite dz using the chain rule. (@Nasser gave a response with explicit detail in case this is not clear.) $\endgroup$– Daniel LichtblauCommented Aug 23, 2013 at 23:08
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
0
If you want to calculate this integral with Mathematica use beautiful (and very powerful!)$\;$ Cauchy Integral Formula implying an adequate theorem of Complex Residue. Thus we have $$\int_{\left | z \right |=1}\frac{dz}{z}= 2\pi i\; Res_{z_{0}=0} f$$ where $f(z)=\frac{1}{z}$.
We can find the residue at $z_0=0$ of $f(z)$ in Mathematica with Residue:
Residue[1/z, {z, 0}]
1
therefore:
$$\int_{\left | z \right |=1}\frac{dz}{z}= 2\pi i$$
$\begingroup$
$\endgroup$
4
z = Exp[I t];
dz = D[z, t];
Integrate[(1/z) dz, {t, 0, 2 Pi}]
The idea is to convert it to complex contour integration.
-
$\begingroup$ This is incorrect,
dtis "infinitesimal" whileD[z[t],t]is not (indt = D[z[t], t]), etc... $\endgroup$– ArtesCommented Aug 23, 2013 at 23:08 -
$\begingroup$ @Artes, so I assume you do not want the first way of expression it? i.e. use explicit function? Will remove the function, and keep the expression. no problem. $\endgroup$– NasserCommented Aug 23, 2013 at 23:14
-
$\begingroup$ You could write
dz = D[z, t] dtbut you musn't writedz = D[z, t]. NextIntegrate[(1/z) dz, {t, 0, 2 Pi}]doesn't make sense. $\endgroup$– ArtesCommented Aug 23, 2013 at 23:17 -
1$\begingroup$ Now, your approach is more appropriate, unfortunately using this symbol
dzit seems to be a differential while it seems you are using it in a different context. You haven't demonstrated seamlessly why you change $\frac{dz}{z}$ to another function using the same symbols. $\endgroup$– ArtesCommented Aug 24, 2013 at 10:11