I cannot figure out why I get differing results for $\int _\gamma z\,dz$, $\gamma (t)=te^{it}$ when using integration by parts and primitive of $z$

Question

Let $\gamma(t) =te^{it}$, $t\in [0,\pi ]$ and consider $\displaystyle \int \limits _\gamma z\,dz$. Since the primitive of $z$ is $\frac{z^2}{2}$, we can evaluate $\displaystyle \int \limits _\gamma z\,dz$ directly as $\displaystyle \int \limits _\gamma z\,dz=\frac{1}{2}\left (\gamma (\pi )^2-\gamma (0)^2\right )=\frac{\pi ^2}{2}$.

An alternative, yet more cumbersome, method is to write\begin{align*}\int \limits _\gamma z\,dz & =\int \limits _0^\pi te^{it}\cdot e^{it}(1+it)\,dt \\ & =\int \limits _0^\pi te^{2it}(1+it)\,dt \\ & =\frac{1}{2i}\int \limits _0^\pi 2ite^{2it}(1+it)\,dt, \end{align*}so that integration by parts yields\begin{align*}\int \limits _\gamma z\,dz & =\frac{1}{2i}\left (\left .(1+it)e^{2it}\right |_0^\pi -\int \limits _0^\pi ie^{2it}\,dt\right ) \\ & =\frac{1}{2i}\left ((1+i\pi )e^{2i\pi}-(1+0)e^0-\frac{i}{2}\left .e^{2it}\right |_0^\pi \right) \\ & =\frac{1}{2i}\left (i\pi -\frac{i}{2}\cdot 0\right ) \\ & =\frac{\pi}{2}, \end{align*}which is missing a factor of $\pi$. So what is my error?

Answer 1

You are missing a factor of $t$. The antiderivative of $2i\color{red}{t}e^{it}$ is not simply $e^{it}$. You get:

$$\begin{align}\frac{1}{2i}\left((1+it)\color{red}{t}e^{2it}\Big |_0^\pi-\int_0^\pi(i\color{red}{t}+1+it)e^{2it}\,\mathrm{d}t\right)&=\frac{1}{2i}\left(\pi(1+i\pi)-\frac{1}{2i}((2it+1)e^{2it})_0^\pi+\frac{2i}{2i}\int_0^\pi e^{2it}\,\mathrm{d}t\right)\\&=\frac{1}{2i}\left(\pi+i\pi^2-\frac{1}{2i}[2i\pi+1-1]+0\right)\\&=\frac{\pi^2}{2}\end{align}$$