# 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 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.

## 1 Answer

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.

• 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. – Jandré Snyman Oct 18 '18 at 0:33
• I made an edit. I hope this helps. – SmileyCraft Oct 18 '18 at 1:11
• thank you that helps a lot. – Jandré Snyman Oct 18 '18 at 1:18