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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.

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  • $\begingroup$ What does $S(0,3)$ denote? $\endgroup$
    – user170231
    Commented Nov 8, 2022 at 15:28
  • $\begingroup$ Circle with center at $0$ and radius $3$. I changed that to $C(0,3)$ to be more intuitive. $\endgroup$
    – thefool
    Commented Nov 8, 2022 at 15:30
  • $\begingroup$ Your answer is correct. $\endgroup$
    – user170231
    Commented Nov 8, 2022 at 17:01

1 Answer 1

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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.

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