Notice that $H_c(m,n)=H_c(n,m)$.

\[constantterm\] The constant term of $F_m(\tau,s)$ is equal to $$b_m(0,s)=\frac{4 \pi}{(2s-1)} \frac{\sigma_m(2s-1)}{\pi^{-s}\Gamma(s)\zeta(2s)}.$$ Here $\zeta(s)$ denotes the Riemann zeta function and $\sigma_m(s)$ the divisor sum $$\sigma_m(s)=m^{(1-s)/2}\sum_{d\mid m} d^s.$$

By Theorem \[fourierF\] we have $$\begin{aligned}
b_m(0,s)&= \frac{4 \pi^{1+s} m^{s}}{(2s-1)\Gamma(s)} \sum_{c=1}^\infty c^{-2s} \sum_{d\,(c)^*}
e\left(\frac{md}{c}\right).\end{aligned}$$ If we insert the formula for the Ramanujan sum (see [@Ap] Chapter 8.3), $$\sum_{d\,(c)^*} e\left(\frac{md}{c}\right) = \sum_{a\mid (c,m)}\mu(c/a)a,$$ where $\mu$ is the Moebius function, we obtain $$\begin{aligned}
b_m(0,s)&=\frac{4 \pi^{1+s} m^{s}}{(2s-1)\Gamma(s)} \sum_{a\mid m} \sum_{\substack{c=1\\a\mid c}}^\infty c^{-2s}\mu(c/a)a\\
&=\frac{4 \pi^{1+s}}{(2s-1)\Gamma(s)\zeta(2s)}  m^{s} \sum_{a\mid m} a^{1-2s}\\
&=\frac{4 \pi^{1+s}\sigma_m(2s-1)}{(2s-1)\Gamma(s)\zeta(2s)}.\end{aligned}$$ This proves the Proposition.

Recall that a weakly holomorphic modular form for $\Gamma'$ is a meromorphic modular form for $\Gamma'$ which is holomorphic outside the cusp $\infty$. In particular, the space of weakly holomorphic modular forms for $\Gamma'$ of weight $0$ is ${\mathbb{C}}[j]$, the polynomial ring in the $j$-function.

\[fourierF1\] The special value $F_{m}(\tau,1)$ has the Fourier expansion $$\begin{aligned}
F_{m}(\tau,1)&=q^{-m}
 +\sum_{n\geq 0} b_m(n,1) q^n,\end{aligned}$$ where $q=e^{2\pi i\tau}$ as usual. In particular, $F_{m}(\tau,1)$ is the unique weakly holomorphic modular form of weight $0$ for $\Gamma'$ whose Fourier expansion starts $q^{-m}+b_m(0,1)+O(q)$.

If we insert the special values and in the Fourier expansion given in Theorem \[fourierF1\], we find $$F_{m}(\tau,1)= q^{-m}+\sum_{n\geq 0} b_m(n,1) q^n + \sum_{\substack{ 