So we have a function $$f(\tau)=\sum_{d\in\mathbb{Z}}\frac{1}{(\tau+d)^k},\ k\geq 2$$ defined on complex upper half plane $\mathbb{H}$, which has a Fourier expansion $$f(\tau)=\sum_{m=1}^{\infty}a_mq^m=g(q),$$
where $q=e^{2\pi i\tau}$, $$a_m=\frac{1}{2\pi i}\int_{\gamma}\frac{g(q)}{q^{m+1}}dq.$$
Here the path $\gamma$ is taken once counterclockwise over a circle about $0$ in the punctured disk $D'=\{z\in\mathbb{C}:0<|z|<1\}.$
Using change of variable, I got $$q=e^{2\pi i\tau}\implies dq=2\pi ie^{2\pi i\tau}d\tau=2\pi iqd\tau$$
so that $$a_m=\int_{\gamma}f(\tau)\cdot\frac{d\tau}{q^m}=\int_{\gamma}f(\tau)e^{-2\pi im\tau}d\tau.$$
What I'm confused about now is to fomulate/express $\gamma$ in terms of $\tau$ or $q$.
By the definition, I guess $\gamma$ is something like $\gamma(\tau)=r(\tau)q=r(\tau)e^{2\pi i\tau}$, like in the polar coordinate system with the radius $r$ depending on the angle $\tau$.
Since what I need to show is $$a_m=\int_{\tau=0+iy}^{1+iy}f(\tau)e^{-2\pi im\tau}d\tau$$ for any $y>0,$ I have to turn $\gamma$ into a segment of a line somehow.
Any hints would be appreciated.