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I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy
\langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67}
The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form
f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}
f_{s'}(u) is also given by this expression but with s \to s', where s, s' \in \mathbb{R}^+ and where a_s, a_{s'} \in \mathbb{C} are complex coefficients in terms of s, s', respectively, given by
a_s = \frac{\Gamma(-2i s)}{\Gamma\left(\frac{1}{2} - i s + l\right) \Gamma\left(\frac{1}{2} - i s + r\right)}.
Using these approximations, they immediately present in equation (70) the normalization condition
\langle f_s | f_{s'} \rangle = 4 \pi |a_s|^2 \delta(s - s'). \tag{70}
which is what I'm trying to obtain. My attempt at solving this involved splitting the integral into two, as follows
\int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du = \int_{0}^{1 - e^{-1/\varepsilon}} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) , du + \int_{1 - e^{-1/\varepsilon}}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du
and solving for the latter, which contains the divergent part. I used the u \to 1 approximation presented above and simplified the expression to a sum of 4 terms. Each term is an integral of the form (singular endpoint – problem arises here)
\mathcal{I} = \int_{1 - e^{-1/\varepsilon}}^{1} (1-u)^{i(s + s') - 1} , du = -\frac{i e^{-i(s + s')/\varepsilon}}{s + s'}.
After integrating all terms and using the following identities (in the distributional sense)
\lim_{a \to \infty} \frac{\sin(a x)}{\pi x} = \delta(x), \quad \lim_{a \to \infty} \cos(a x) = 0,
I got
\mathcal{I} = -4 \pi |a_s|^2 \delta(s - s').
I believe the problem comes from the way I solved the \mathcal{I} integral, which leads to a sign discrepancy between what I obtained and the paper.
I've also tried the integral
\int_{0}^{1 - e^{-1/\varepsilon}} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du
Eventhough we're using an asymptotic expansion for u \to 1, I considered these limits since the dominant contribution to the Dirac delta function lies near u=1. This produces the correct sign for the normalization as seen in Kitaev's paper, but it also yields non-Dirac delta terms that don't cancel out. Namely
\int_0^1 \frac{2 du}{(1-u)^2} f_s(u) f_{s'}^*(u) = 4 \pi |a_s|^2 \delta(s-s') + \frac{4 \Im(a_s a_{s'})}{s+s'} + \frac{4 \Im(a_s a_{s'}^*)}{s-s'}
\langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67}
The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form
f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}
f_{s'}(u) is also given by this expression but with s \to s', where s, s' \in \mathbb{R}^+ and where a_s, a_{s'} \in \mathbb{C} are complex coefficients in terms of s, s', respectively, given by
a_s = \frac{\Gamma(-2i s)}{\Gamma\left(\frac{1}{2} - i s + l\right) \Gamma\left(\frac{1}{2} - i s + r\right)}.
Using these approximations, they immediately present in equation (70) the normalization condition
\langle f_s | f_{s'} \rangle = 4 \pi |a_s|^2 \delta(s - s'). \tag{70}
which is what I'm trying to obtain. My attempt at solving this involved splitting the integral into two, as follows
\int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du = \int_{0}^{1 - e^{-1/\varepsilon}} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) , du + \int_{1 - e^{-1/\varepsilon}}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du
and solving for the latter, which contains the divergent part. I used the u \to 1 approximation presented above and simplified the expression to a sum of 4 terms. Each term is an integral of the form (singular endpoint – problem arises here)
\mathcal{I} = \int_{1 - e^{-1/\varepsilon}}^{1} (1-u)^{i(s + s') - 1} , du = -\frac{i e^{-i(s + s')/\varepsilon}}{s + s'}.
After integrating all terms and using the following identities (in the distributional sense)
\lim_{a \to \infty} \frac{\sin(a x)}{\pi x} = \delta(x), \quad \lim_{a \to \infty} \cos(a x) = 0,
I got
\mathcal{I} = -4 \pi |a_s|^2 \delta(s - s').
I believe the problem comes from the way I solved the \mathcal{I} integral, which leads to a sign discrepancy between what I obtained and the paper.
I've also tried the integral
\int_{0}^{1 - e^{-1/\varepsilon}} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du
Eventhough we're using an asymptotic expansion for u \to 1, I considered these limits since the dominant contribution to the Dirac delta function lies near u=1. This produces the correct sign for the normalization as seen in Kitaev's paper, but it also yields non-Dirac delta terms that don't cancel out. Namely
\int_0^1 \frac{2 du}{(1-u)^2} f_s(u) f_{s'}^*(u) = 4 \pi |a_s|^2 \delta(s-s') + \frac{4 \Im(a_s a_{s'})}{s+s'} + \frac{4 \Im(a_s a_{s'}^*)}{s-s'}
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