Calculate:
$\int_{\gamma} |z|dz$ with $\gamma: [0,1]\rightarrow\mathbb{C},t\mapsto i +\exp(i\pi t)$
I tried calculating it and actually made some progress, where one term vanished when splitting the integral in real and imaginary part, but the substitutions needed to arrive at the value afterwards just seem to complicated (I checked with Wolfram Alpha). Are there any tricks to bring to make this integral more managable ?
Well, writing down using the definitions: \begin{align}\int_\gamma |z|dz & = i\pi\int_0^1|i+e^{i\pi t}|e^{i\pi t}dt \\& = i\pi\int_0^1 \left(cos(\pi t)^2+(1+\sin(\pi t))^2\right)^{1/2}(\cos(\pi t)+i\sin(\pi t))dt .\end{align}
Now, we compute the real and imaginary part of that integral in two separated integrals:
\begin{align} \int_0^1\left(cos(\pi t)^2+(1+\sin(\pi t))^2\right)^{1/2}\cos(\pi t)dt & =\int_0^1 \left(1-\sin^2(\pi t)+(1+\sin(\pi t))^2\right)^{1/2}\cos(\pi t)dt \\ & = \sqrt{2}\int_0^1 (1+\sin(\pi t))^{1/2}\cos(\pi t)dt \\ & = \frac{2\sqrt{2}}{3\pi}\left(1+\sin(\pi t)\right)^{3/2}\Bigr|_0^1 \\ & = 0\end{align} since $\int (1+\sin(\pi t))^{1/2}\cos(\pi t)=\frac{1}{\pi}\int u^{1/2} du$ using $u=1+\sin(\pi t)$.
Now, the imaginary part: It's harder to calculate, but using several times chain rule, you will get: $$\int (1+\sin(\pi t))^{1/2}\sin(\pi t)=-\frac{\sqrt{\sin(\pi x)+1}(-4\sin^3(\frac{\pi x}{2})+3\cos(\frac{\pi x}{2})+\cos(\frac{3\pi x}{2}))}{3\pi(\sin(\frac{\pi x}{2})+\cos(\frac{\pi x}{2}))},$$ and replacing: $$\int_0^1 (1+\sin(\pi t))^{1/2}\sin(\pi t)=\frac{8}{3\pi}.$$
So $$\int_\gamma |z|dz= i\frac{8}{3\pi}.$$