Complex analysis - path integrals

Question

I need to evaluate the following. $$\int_\gamma f(z) \: \text{d}z = \int_\gamma z^3+\cosh z \: \text{d}z$$ where $\gamma(t)=t^2+2it$ for $0\leq t \leq 1$. At first, I used the standard approach: $$\int_\gamma f(z) \: \text{d}z =\int_\gamma f(\gamma(t))\gamma '(t) \: \text{d}t=\left( (t^2+2it)^3 +\cosh(t^2+2it)\right)(2t+2i) \: \text{d}t.$$ But this seems ridiculous. Is there a better approach to this? Any tips would be great!

Answer 1

One can proceed as in the OP with

$$\begin{align} \int_\gamma f(z)\,dz&=\int_\gamma (z^3+\cosh(z))\,dz\\\\ &=\int_0^1 (t^2+2it)^3\,(2t+i2)\,dt+\int_0^1 \cosh(t^2+2it)\,(2t+i2)\,dt\\\\ &=\left.\left(\frac14(t^2+2it)^4+\sinh(t^2+2it)\right)\right|_{0}^{1}\\\\ &=\frac14 (1+i2)^4+\sinh(1+i2) \end{align}$$

But if one understands that since $f(z)$ is analytic everywhere, the value of the integral over any path from $(0,0)$ to $(1,2)$ is path independent, then one can write

$$\begin{align} \int_\gamma f(z)\,dz&=\int_\gamma (z^3+\cosh(z))\,dz\\\\ &=\left.\left(\frac14 z^4+\sinh(z)\right)\right|_{z=0}^{z=1+i2}\\\\ &=\frac14 (1+i2)^4+\sinh(1+i2) \end{align}$$

and we are done!