I am studying for a Complex analysis exam that I have coming up in the next few weeks and I am working through some practice problems. I happen to have gotten stuck on the following integral;

$$ \int_{\gamma} \exp(z)\log(z) dz $$

Where we take the logarithm to be $$ \log(z) = \log|z|+i\arg(z), \ \ \ -\pi<\arg(z)<\pi $$ and $\gamma$ is the parabola $$ \gamma(t) = 1-t^2+it, \ \ \ -\infty<t<\infty $$ I have tried computing this directly by the line integral definition but that was fruitless and I am wondering if that was just because I couldn't see how to do it or if there is a different approach that needs to be taken.


You should be able to calculate this integral with the following contour:

$\gamma_1:[-s,s]\to\mathbb{C}$, $\gamma_1(t):=1-t^2+it$.

Then $\gamma_2$ is a straight line from $\gamma_1(s)$ to $\varepsilon e^{i\arg(\gamma_1(s))}$.

$\gamma_3:[\arg(\gamma_1(s)),\arg(\gamma_1(-s))]\to\mathbb{C}$, $\gamma_3(\theta):=\varepsilon e^{i\theta}$,

Finally, $\gamma_4$ will be a straight line from $\varepsilon e^{i\arg(\gamma_1(-s))}$ to $\gamma_1(-s)$.

If we let $s$ go to infinity and $\varepsilon$ to $0$, then $\gamma_2$, $\gamma_3$ and $\gamma_4$ can be calculated and $\gamma_1$ will equal the requested integral. Because $\exp(z)\log(z)$ has no singularities in the given domain, we find that the sum of the integrals over $\gamma_i$ will be $0$. This should help you find the requested integral.


So you want help actually calculating these integrals. You already deduced that the integral over $\gamma_3$ equals $0$.

Notice that $-\gamma_2$ and $\gamma_4$ are of the form

$\gamma:[\varepsilon,R]\to\mathbb{C}$, $\gamma(t)=te^{i\theta}$.

In both cases we let $\varepsilon\to0$ and $R\to\infty$. For $-\gamma_2$ we let $\theta\to\pi$ and for $\gamma_4$ we let $\theta\to-\pi$. We calculate

$$ \int_{\gamma_2}f(z)\mbox{d}z \\=-\int_{-\gamma_2}f(z)\mbox{d}z \\=-\int_\varepsilon^R(-\gamma_2)'(t)f((-\gamma_2)(t))\mbox{d}t \\=-\int_\varepsilon^Re^{i\theta}f(te^{i\theta})\mbox{d}t \\=-e^{i\theta}\int_\varepsilon^Re^{te^{i\theta}}\log(te^{i\theta})\mbox{d}t \\=-e^{i\theta}\int_\varepsilon^Re^{te^{i\theta}}(\log(t)+i\theta)\mbox{d}t \\\to\int_0^\infty e^{-t}(\log(t)+i\pi)\mbox{d}t $$

Similarly, we get

$$ \int_{\gamma_4}f(z)\mbox{d}z \\\to-\int_0^\infty e^{-t}(\log(t)-i\pi)\mbox{d}t $$

Therefore, we get

$$ \int_{\gamma_2}f(z)\mbox{d}z+\int_{\gamma_4}f(z)\mbox{d}z \\\to\int_0^\infty e^{-t}2i\pi\mbox{d}t \\=2i\pi $$

So we can conclude that

$$ \int_{\gamma_1}f(z)\mbox{d}z\to-2i\pi $$

Remark: When summing the integrals over $\gamma_2$ and $\gamma_4$, we used the fact that the integral

$$ \int_0^\infty e^{-t}\log(t)\mbox{d}t $$

converges. See if you can verify why this is true.

  • $\begingroup$ Okay, contour integrals are great, but now that we have these contours, this is where I get stuck. How do we actually compute the integral along gamma2 and gamma4, gamma 3 obviously goes to zero. $\endgroup$ – Jandré Snyman Oct 18 '18 at 0:33
  • $\begingroup$ I made an edit. I hope this helps. $\endgroup$ – SmileyCraft Oct 18 '18 at 1:11
  • $\begingroup$ thank you that helps a lot. $\endgroup$ – Jandré Snyman Oct 18 '18 at 1:18

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