# values of integral

• Mar 11th 2009, 08:11 AM
manjohn12
values of integral
Compute all possible values of $\displaystyle \int_{C} \frac{z+1}{z(z-1)(z-2)} \ dz$. Assume $\displaystyle C$ is positively oriented and none of the points $\displaystyle 0,1,2$ lie on the simple closed curve $\displaystyle C$.

So deform $\displaystyle C$ into $\displaystyle C_1, C_2, C_3$. Then we have $\displaystyle \int_{C} \frac{z+1}{z(z-1)(z-2)} \ dz = \int_{C_1} \frac{z+1}{(z-1)(z-2)} \ dz + \int_{C_2} \frac{z+1}{z(z-2)} \ dz + \int_{C_3} \frac{z+1}{z(z-1)} \ dz$.

This is equaled to: $\displaystyle 2 \pi i(1/2)+ 2 \pi i (-2) + 2 \pi i (3/2)$.

Is this correct?
• Mar 14th 2009, 02:56 PM
sitho
Yes, with the tiny correction:

$\displaystyle \int_C {\frac{z + 1}{z(z - 1)(z - 2)} dz} = 2\pi i \cdot \left(\lim_{z \rightarrow 0} \frac{z + 1}{(z - 1)(z - 2)} + \lim_{z \rightarrow 1} \frac{z + 1}{z(z - 2)} + \lim_{z \rightarrow 2} \frac{z + 1}{z(z - 1)}\right) =$
$\displaystyle = 2\pi i \left(\frac{1}{2} - 2 + \frac{3}{2} \right) = 0$

using Cauchy's integral formula and calculating the residues for the separate singularities of the integrand. The integral has this value assuming $\displaystyle C$ encircles both 0, 1 and 2. Assuming any closed curve in the complex plane is possible, we keep all different combinations of the terms in the final sum, giving the possible values of the integral to be $\displaystyle \pi i \{-4, -3, -1, 0, 1, 3, 4\}$.
• Mar 14th 2009, 03:07 PM
mr fantastic
Quote:

Originally Posted by manjohn12
Compute all possible values of $\displaystyle \int_{C} \frac{z+1}{z(z-1)(z-2)} \ dz$. Assume $\displaystyle C$ is positively oriented and none of the points $\displaystyle 0,1,2$ lie on the simple closed curve $\displaystyle C$.

So deform $\displaystyle C$ into $\displaystyle C_1, C_2, C_3$. Then we have $\displaystyle \int_{C} \frac{z+1}{z(z-1)(z-2)} \ dz = \int_{C_1} \frac{z+1}{(z-1)(z-2)} \ dz + \int_{C_2} \frac{z+1}{z(z-2)} \ dz + \int_{C_3} \frac{z+1}{z(z-1)} \ dz$.

This is equaled to: $\displaystyle 2 \pi i(1/2)+ 2 \pi i (-2) + 2 \pi i (3/2)$.

Is this correct?

Since for a given C there can be only one value for the integral, I'd suggest that the question wants you to consider the different values arising from the different paths.

eg. If C encloses none of the singularites of f(z) then the integral is equal to zero. So one possible value of the inetrgal is zero.

Clearly there are several other possible values, depending on which and how many of the singularities are enclosed.