Evaluate:
∫C e^z (z-1)(z-2)^2 / sin^2 pi z
where C is the circle │z│= 2.5
Are we talking about:
$\displaystyle
\int\limits_C {\frac{{e^z \left( {z - 1} \right)\left( {z - 2} \right)^2 }}{{\sin ^2 \left( {\pi z} \right)}}} dz
$
Because then I find non-zero residues for all the poles (-2,-1,0,1).
Also, from the residue theorem, aren't we forgetting a factor 2pi.i ?
The factor of 2 pi i helps, and I would be happry to be convinced about the
residues at the other points as I still don't have agreement with my numerical
integral (he said giving away the secret of why he was not happy with his previous
post).
But this is the problem I have. At 0 for instance:
$\displaystyle z^2{\frac{{e^z \left( {z - 1} \right)\left( {z - 2} \right)^2 }}{{\sin ^2 \left( {\pi z} \right)}}} $
is analytic so the residue is (2 pi i times) the coefficient of x in the taylor series
of this about $\displaystyle z=0$, which of course now that I compute it I find it non zero
(where before it seemed to be 0 )
So now I am with you.
RonL
Now that agrees with my numerical integration
RonLCode:This is EULER, Version 2.3 RL-06. Type help(Return) for help. Enter command: (20971520 Bytes free.) Processing configuration file. Done. > > >dthe=0.001*pi; > >the=dthe/2:dthe:2*pi; >rc=2.5; >z=rc*exp(I*the); > >fz=exp(z)*(z-1)*(z-2)^2/(sin(pi*z))^2; > > >..numerical integral: > >II=sum(fz*rc*I*exp(I*the))*dthe 7.26249e-014+4.29033i > > >e=exp(1) 2.71828 > >..TD!'s residues > >(-8/e^2/pi^2+3/e/pi^2+4/pi^2+e/pi^2)*2*pi*I 0+4.29033i >
Why. They are complementary, a mistake can occur either way and
their agreement is a sanity check.
It is always important to be able to have two or more independent
techniques available. The ability of the mind to see what it wants is
always with us. An idependent check helps prevent us being carried away
(resolution disagreements also helps develop technique and understanding
of the limitations of our methods).
Finally, it is also instructive to set about computing the integral numericaly
a few things becom clear that otherwise would remain hidden.
RonL