integral of $e^z \log z$ over parabola in complex plane 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.
 A: 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.
EDIT:
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.
