I'm trying to fill in the details in proposition 14 of this paper by Mazur, Tate, and Teitelbaum. In particular, I'd like to understand the following.
Let $f$ be a cuspidal eigenform of weight $k$ and level $\Gamma_1(N)$, with nebentypus $\epsilon$. Fix a prime $p\not\mid N$ and suppose that the polynomial $x^2-a_px+\epsilon(p)p^{k-1}$ has an "allowable root" $\alpha$ (i.e., $ord_p(\alpha)<k-1$), where $a_p$ is the $T_p$-eigenvalue of $f$. Let $\chi$ be a $p$-adic character of conductor $p^n$. I am trying to fill in the details on their claim that $$\tag{1} L_p(f,\alpha,\chi)=\frac{1}{\alpha^n}\frac{p^n}{\tau(\overline{\chi})}L(f_{\overline{\chi}},1) $$ where the $L$ function on the right is the classical $L$-function of the twisted form $f_{\overline{\chi}}(z)=\sum \overline{\chi}(n)a_nq^n$, and $\tau$ is a Gauss sum. In this case, they state in the article that (1) is a "straightforward computation" and cite a few earlier formulas, but I'm clearly having some trouble doing this computation.
In the article, they define the $p$-adic $L$-function by $$ L_p(f,\alpha,\chi):= \int_{\mathbb{Z}_p^\times}\chi d\mu_{f,\alpha}, $$ where $\mu_{f,\alpha}$ is the measure on $\mathbb{Z}_p^\times$ defined in terms of modular symbols $\lambda_{f,P}$ ($P$ is a one variable complex polynomial): $$ \mu_{f,\alpha}(P;a,p^n)=\frac{1}{\alpha^n}\lambda_{f,P}(a,p^n)-\frac{\epsilon(p)p^{k-2}}{\alpha^{n+1}}\lambda_{f,P}(a,p^{n-1}). $$ Essentially, this is the measure of the function $P$ on the open set $a+p^n\mathbb{Z}_p$. In other words, the integral satisfies: $$ \int_{a+p^n\mathbb{Z}_p}x^j d\mu_{f,\alpha}=\mu_{f,\alpha}(z^j;a,p^n). $$ So we can compute the $p$-adic $L$-function by computing modular symbols. My attempt at (1) is as follows. By 8.2., 8.5, and 8.6 in the article (I've put these below), it's easy to see that $$ L(f_{\overline{\chi}},1)=\lambda_{f_{\overline{\chi}},1}(0,1), $$ so we want to relate $L_p(f,\alpha,\chi)$ to $\lambda_{f_{\overline{\chi}},1}(0,1)$. Since $\chi$ is a $p$-adic character of conductor $p^n$, it is constant on the open sets $a+p^n\mathbb{Z}_p$. So by definition we have \begin{align*} L_p(f,\alpha,\chi) &=\sum_{a\in (\mathbb{Z}/p^n\mathbb{Z})^\times}\chi(a)\int_{a+p^n\mathbb{Z}_p}d\mu_{f,\alpha}\\ &=\sum_{a\in (\mathbb{Z}/p^n\mathbb{Z})^\times}\chi(a)\mu_{f,\alpha}(1;a,p^n)\\ &=\frac{1}{\alpha^n}\sum_{a\in (\mathbb{Z}/p^n\mathbb{Z})^\times}\chi(a)\big(\lambda_{f,1}(a,p^n)-\frac{\epsilon(p)p^{k-2}}{\alpha}\lambda_{f,1}(a,p^{n-1})\big), \end{align*} and this is where I'm stuck. Clearly, it suffices to show that $$ \sum_{a\in (\mathbb{Z}/p^n\mathbb{Z})^\times}\chi(a)\big(\lambda_{f,1}(a,p^n)-\frac{\epsilon(p)p^{k-2}}{\alpha}\lambda_{f,1}(a,p^{n-1})\big)=\frac{p^n}{\tau(\overline{\chi})}\lambda_{f_{\overline{\chi}},1}(0,1), $$ but this just doesn't seem to be true, since, for example, it would appear that formula 8.5 in the article shows that $$ \sum_{a\in (\mathbb{Z}/p^n\mathbb{Z})^\times}\chi(a)\lambda_{f,1}(a,p^n)=\frac{p^n}{\tau(\overline{\chi})}\lambda_{f_{\overline{\chi}},1}(0,1). $$
Where am I going wrong here?
Formulae: Let $\chi$ be a character of conductor $m$.
(8.2) $\chi(-1)\tau(\chi)\tau(\overline{\chi})=m$
(8.5) $\lambda_{f_{\overline \chi}, P(mz)}(b,n)=\frac{1}{\tau(\chi)}\sum_{a\in (\mathbb{Z}/m\mathbb{Z})^\times}\chi(a)\lambda_{f,P}(mb-na,mn)$
(8.6) $L(f_{\overline \chi},n+1)=\frac{1}{n+1}\frac{(-2\pi i)^n}{m^{n+1}}\tau(\bar\chi)\sum_{a\in (\mathbb{Z}/m\mathbb{Z})^\times} \chi(a)\lambda_{f,z^n}(a,m)$