[SOLVED]Complex integral oriented counterclockwise

dwsmith

Well-known member
$\gamma$ is the unit circle oriented counterclockwise.

$\displaystyle\int_{\gamma}\dfrac{e^z}{z}dz$

$\gamma(t) = e^{it}$ for $0\leq t\leq 2\pi$

$\gamma'(t) = ie^{it}$

Using $\int_{\gamma} f(\gamma(t))\gamma'(t)dt$, I obtain

$\displaystyle i\int_0^{2\pi}e^{e^{it}}dt$

Not quite sure how to integrate this one.

Fernando Revilla

Well-known member
MHB Math Helper
Better use the Cauchy Integral Formula, then $\displaystyle\int_{\gamma}\dfrac{e^z}{z}dz=2\pi i e^0=2\pi i$

dwsmith

Well-known member
Better use the Cauchy Integral Formula, then $\displaystyle\int_{\gamma}\dfrac{e^z}{z}dz=2\pi i e^0=2\pi i$
Evaluating $\frac{\cos(z)}{z}$ and $\frac{\cos(z^2)}{z}$ will be $2\pi i$ as well and then $\frac{\sin(z)}{z}$ is 0, correct?

Random Variable

Well-known member
MHB Math Helper
Evaluating $\frac{\cos(z)}{z}$ and $\frac{\cos(z^2)}{z}$ will be $2\pi i$ as well and then $\frac{\sin(z)}{z}$ is 0, correct?
Yes, that''s correct.

In you expand each function in a Laurent series about zero (since that's where a singularity exists for each function), the only term that will contribute anything is the $\displaystyle \frac{1}{z}$ term. Specifically it will contribute $2 \pi i$ times its coefficient.

$\displaystyle \frac{\cos \ z}{z} = \frac{1}{z} \left( 1 - \frac{z^{2}}{2!} + \frac{z^{4}}{4!} + \ldots \right) = {\bf \frac{1}{z}} - \frac{z}{2!} + \frac{z^{3}}{4!} + \ldots$

$\displaystyle \frac{\cos \ z^{2}}{z} = \frac{1}{z} \left( 1 - \frac{z^{4}}{2!} + \frac{z^{8}}{4!} + \ldots \right) = {\bf \frac{1}{z}} - \frac{z^{3}}{2!} + \frac{z^{7}}{4!} + \ldots$

$\displaystyle \frac{\sin z}{z} = \frac{1}{z} \left( z - \frac{z^{3}}{3!} + \frac{z^{5}}{5!} + \ldots \right) = 1 - \frac{z^{2}}{3!} + \frac{z^{4}}{5!} + \ldots$

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dwsmith

Well-known member
$\gamma$ is the unit circle oriented counterclockwise.

$\displaystyle\int_{\gamma}\dfrac{e^z}{z}dz$

$\gamma(t) = e^{it}$ for $0\leq t\leq 2\pi$

$\gamma'(t) = ie^{it}$

Using $\int_{\gamma} f(\gamma(t))\gamma'(t)dt$, I obtain

$\displaystyle i\int_0^{2\pi}e^{e^{it}}dt$

Not quite sure how to integrate this one.
How can this be integrate without using the Cauchy Integral Formula?

$\displaystyle\int_0^{2\pi}\left(\frac{1}{z}+1+ \frac{z}{2!} +\frac{z^2}{3!}+\cdots\right)dz$

Now, what should be done?

Never mind I found a Theorem to use in my book that applies to the expansion.

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Random Variable

Well-known member
MHB Math Helper
How can this be integrate without using the Cauchy Integral Formula?

$\displaystyle\int_0^{2\pi}\left(\frac{1}{z}+1+ \frac{z}{2!} +\frac{z^2}{3!}+\cdots\right)dz$

Now, what should be done?

Never mind I found a Theorem to use in my book that applies to the expansion.

$\displaystyle \int_{\gamma} \frac{e^z}{z}dz = \int_{\gamma}\left(\frac{1}{z}+1+ \frac{z}{2!} +\frac{z^2}{3!}+\cdots\right)dz$

$\displaystyle = \int_{0}^{2 \pi} \left( e^{-it} + 1 + \frac{e^{it}}{2!} + \frac{e^{2it}}{3!} + \ldots \right) \ i e^{it} \ dt$

$\displaystyle = i \int_{0}^{2 \pi} \left(1 + e^{it} + \frac{e^{2it}}{2!} + \frac{e^{3it}}{3!} \ldots \right) = i \int_{0}^{2 \pi} \left(1 + \cos(t) + i \sin(t) + \frac{\cos (2t)}{2!} + i\frac{\sin (2t)}{2!} + \ldots \right)$

$\displaystyle = i \left( 2 \pi + 0 + 0 + 0 + 0 + \ldots \right)= 2 \pi i$