Evaluate $\int_{\gamma} \frac{2 \sin(z) +e^z}{z^2 - 2z} dz$ where $\gamma = C(0,3)$ with positive orientation.

Question

I need to evaluate $\int_{\gamma} \frac{2 \sin(z) +e^z}{z^2 - 2z} dz$ where $\gamma = C(0,3)$ with positive orientation.

We know that: $$\int_{\gamma} \frac{2 \sin(z) +e^z}{z^2 - 2z} dz = 2 \int_{\gamma} \frac{ \sin(z)}{z^2 - 2z} dz + \int_{\gamma} \frac{e^z}{z^2 - 2z} dz = 2 \int_{\gamma} \frac{ \sin(z)}{z(z - 2)} dz + \int_{\gamma} \frac{e^z}{z(z - 2)} dz$$

To evaluate $2\int_{\gamma} \frac{ \sin(z)}{z(z-2)} dz$ we take:

$$2\int_{\gamma} \frac{ \sin(z)}{z(z-2)} dz = 2 \int_{\gamma_1} \dfrac{ \sin(z)/z}{(z-2)}dz + 2\int_{\gamma_2} \dfrac{ \sin(z)/\{z-2\}}{z}dz $$ Where $\gamma_1$ is a circle around pole $z = 2$ and $\gamma_2$ is a circle around pole $z = 0$. Using Cauchy's Integral Formula: $$2 \int_{\gamma_1} \dfrac{ \sin(z)/z}{(z-2)}dz = 2 * 2 \pi i \frac{ \sin(2)}{2} = 2 \pi i \sin(2)$$ $$2\int_{\gamma_2} \dfrac{ \sin(z)/\{z-2\}}{z}dz = 2 * 2 \pi i \frac{ \sin(0)}{-2} = 0$$

To evaluate $\int_{\gamma} \frac{e^z}{z(z - 2)} dz$ we take:

$$\int_{\gamma} \frac{e^z}{z(z - 2)} dz = \int_{\gamma_1} \dfrac{e^z/z}{(z-2)}dz + \int_{\gamma_2} \dfrac{e^z/\{z-2\}}{z}dz $$ Where $\gamma_1$ is a circle around pole $z = 2$ and $\gamma_2$ is a circle around pole $z = 0$.

Using Cauchy's Integral Formula: $$\int_{\gamma_1} \dfrac{e^z/z}{(z-2)}dz = 2 \pi i \frac{e^2}{2} = \pi i e^2$$ $$\int_{\gamma_2} \dfrac{e^z/\{z-2\}}{z}dz = 2 \pi i \frac{e^0}{-2} = -\pi i$$

Taking everything together we get: $$\int_{\gamma} \frac{2 \sin(z) +e^z}{z^2 - 2z} dz = 2 \pi i \sin(2) + \pi i e^2 -\pi i$$

Is that correct? I did it just by looking in the textbook and examples so it may be all wrong.

Answer 1

While evaluating $\int_{\gamma_1} \frac{\sin z/z}{z-2}dz$ using Cauchy integral formula, the assumption $\frac{\sin z}{z}$ must be analytic within and on $\gamma_1$ must hold. You can see that $z=0$ is a removable singularity of $\frac{\sin z}{z}$ and defining $f(0)=1$ is required there; so the application of formula is not as trivial and needs some background assumptions and theorems like analytic continuation. Though Cauchy's Residue Theorem gives you the correct result without using deformation principle.