When proving functional equation for Riemann zeta function one starts at the definition of gamma function $$\Gamma(s) = \int_0^{\infty} x^{s-1} e^x\mathrm dx\tag1$$ After a few steps we arrive at $$ \pi^{-\frac{s}{2}} \Gamma \left(\frac{s}{2} \right) \zeta(s) = \frac{1}{s(1-s)} - \int_1^{\infty} \left( x^{\frac{s}{2}} + x^{\frac{1-s}{2}} \right) \frac{\psi(x)}{x}\mathrm dx\tag2$$ And observing the RHS of the above equation doesn’t change when we replace $s$ with $(1-s)$ in equation $(2)$ to get functional equation below: $$ \pi^{-\frac{s}{2}} \Gamma \left(\frac{s}{2} \right) \zeta(s) = \pi^{-\frac{1-s}{2}} \Gamma \left(\frac{1-s}{2} \right) \zeta(1-s) \tag3 $$
You can watch detail proof here
I’m trying to find similar generalized proof for any Dirichlet $L$-function. In particular I want to see what equation $(2)$ looks like.
I tried searching using google search but could not find it. Does anybody know a reference for this proof? Thank you!
Once you know the proof for $\zeta(s)$ you want to look at $\Theta(x,\chi) = \sum_n \chi(n) e^{-\pi n^2 x}$ for $\chi$ an even primitive Dirichlet character modulo $q> 1$ having the discrete Fourier transform $\chi(n) = \frac1q \sum_{k=1}^q \hat{\chi}(k) e^{2i \pi nk/q}$ where $\hat{\chi}(k) = \sum_{n=1}^q \chi(n) e^{-2i\pi nk /q} = \overline{\chi(n)} \hat{\chi}(1)$.
Together with the Fourier transform pair $$h_x(t) =e^{-\pi t^2 x}, \hat{h_x}(u) = x^{-1/2} h_{1/x}(u)$$ and the Fourier series $$ f_x(t) = \sum_n h_x(t+n) = \sum_m e^{2i \pi m t} \hat{h_x}(m) =\sum_m e^{2i \pi m t} x^{-1/2} h_{1/x}(m)$$ you'll obtain $$\Theta(x,\chi) = q^{-1} \hat{\chi}(1) x^{-1/2} \Theta(1/(qx),\overline{\chi})$$ and $$\Lambda(s,\chi)=q^{s/2}\pi^{-s/2} \Gamma(s/2) L(s,\chi)=\int_0^\infty x^{s/2-1}\Theta(x/q,\chi) dx\\= \hat{\chi}(1)q^{-1/2} \int_0^\infty x^{(1-s)/2-1}\Theta(x/q,\overline{\chi}) dx=\hat{\chi}(1)q^{-1/2} \Lambda(1-s,\overline{\chi})$$
For $\chi$ an odd primitive character you'll look at $\Theta(x,\chi) = \sum_n \chi(n)n e^{-\pi n^2 x}$ using the Fourier transform pair $g_x(t) =2i\pi t e^{-\pi t^2 x}, \hat{g_x}(u) = \hat{h_x}'(u) = -ix^{1/2}g_{1/x}(u)$.