# $\int \pi e^{\pi\overline{z}}\ dz$ over a square

I need to calculate

$$\int \pi e^{\pi\overline{z}}\ dz$$

over a square of vertices $z=0, z=1, z=1+i, z=i$. These are my parametrizations for the square:

$$z_1 = t, 0\le t\le 1\\z_2 = 1+ti, 0\le t\le 1\\z_3 = t+i, 0\le t\le 1\\z_4 = ti, 0\le t\le 1$$

Therefore, the integral can be brokwn in $4$ cases:

$$\int_{z_1}\pi e^{\pi \overline{z_1}}\ dz_1+\int_{z_2}\pi e^{\pi \overline{z_2}}\ dz_2+\int_{z_3}\pi e^{\pi \overline{z_3}}\ dz_3+\int_{z_4}\pi e^{\pi \overline{z_4}}\ dz_4$$

$$\int_{z_1}\pi e^{\pi \overline{z_1}}\ dz_1 = \int_0^1 \pi e^{\pi t}\ dt = e^{\pi}-1$$

$$\int_{z_2}\pi e^{\pi \overline{z_2}}\ dz_2 = \int_0^1\pi e^{\pi (1-ti)}\ i \ dt = \int_0^1 \pi e^{\pi }e^{-\pi t i}i \ dt = \pi e^{\pi}\int_0^1\cos (-\pi t) + i\sin (-\pi t)\ dt = -i\pi\frac{2i}{\pi} = 2\\$$

$$\int_{z_3}\pi e^{\pi \overline{z_3}}\ dz_3 = \int_0^1 \pi e^{\pi(t-i)\ dt} = \pi\int_0^1e^{t\pi}e^{-i\pi}\ dt = \pi e^{-i\pi}\int_0^1 e^{t\pi} \ dt = \pi e^{-i\pi}\frac{e^{\pi}-1}{\pi}$$

$$\int_{z_4}\pi e^{\pi \overline{z_4}}\ dz_4 = \pi\int_0^1e^{-\pi ti}\ dt = \pi \int_0^1 \cos(-\pi t)+i\sin (-\pi t) \ dt = -2i$$

Am I right? Because the 2nd result seems strange, and I don't know how to verify integrals along paths in Wolfram Alpha. Thank you so much.

There's an error in your $z_2$ integral. You're missing the ${\rm e}^{\pi}$ factor that came out of the integral after the third equals sign,
$$\int _{z_2} dz\;{\rm e}^{\pi \bar{z}} = -2 i {\rm e}^\pi$$