Determinantal Structures in the O’Connell-Yor Directed Random Polymer Model

Abstract

We study the semi-discrete directed random polymer model introduced by O’Connell and Yor. We obtain a representation for the moment generating function of the polymer partition function in terms of a determinantal measure. This measure is an extension of the probability measure of the eigenvalues for the Gaussian unitary ensemble in random matrix theory. To establish the relation, we introduce another determinantal measure on larger degrees of freedom and consider its few properties, from which the representation above follows immediately.

Appendices

Appendix 1: Proof of Lemma 5

First we give a proof of (2.29). For this purpose, it is sufficient to show the case of $m=1,~x=0$,

$$\begin{aligned} \int _{-\infty }^{\infty }dx\, e^{-ax}f_B(x)=\frac{\pi }{\beta }\cot \left( \frac{\pi a}{\beta }\right) . \end{aligned}$$

(7.1)

Furthermore setting $e^{\beta x}=y$, $a/\beta =b$, one sees that (7.1) is rewritten as

$$\begin{aligned} \int _0^\infty dy\, h_b(y)=\pi \cot \pi b, \end{aligned}$$

(7.2)

where $h_b(y)=y^{-b-1}/(y-1)$ and we take the branch cut of $h_b(y)$ to be the positive real axis. Hence here we prove (7.2). We set the contour C as depicted in Fig. 2 with $\alpha =1$. From the Cauchy integral theorem, we find

$$\begin{aligned} \int _{C} dy\, h_b(y)=0. \end{aligned}$$

(7.3)

Dividing the contour C into $C_i,~i=1,\ldots ,6$ as in Fig. 2, we find that by simple calculations,

$$\begin{aligned}&\lim _{\scriptstyle \begin{array}{c} \delta \rightarrow 0, \epsilon \rightarrow 0 \\ R\rightarrow \infty \end{array}} \int _{C_1} dy\, h_b(y) =-\lim _{\scriptstyle \begin{array}{c} \delta \rightarrow 0, \epsilon \rightarrow 0 \\ R\rightarrow \infty \end{array}} e^{2\pi i b}\int _{C_4}dy\, h_b(y) =\int _0^\infty dy\, h_b(y),\nonumber \\&\lim _{\epsilon \rightarrow 0}\int _{C_2dy\,}dy\, h_b(y)=\lim _{\epsilon \rightarrow 0}e^{2\pi i b} \int _{C_5}dy\, h_b(y)=-\pi i,\nonumber \\&\lim _{R\rightarrow \infty }\int _{C_3}dy\, h_b(y)=\lim _{\delta \rightarrow 0} \int _{C_6}dy\, h_b(y)=0, \end{aligned}$$

(7.4)

where note that the factors $e^{2\pi i b}$s come from the cut locus of $y^{-b}$.

From (7.3),(7.4), we get

$$\begin{aligned} 0=\lim _{\scriptstyle \begin{array}{c} \delta \rightarrow 0, \epsilon \rightarrow 0 \\ R\rightarrow \infty \end{array}} \sum _{j=1}^6\int _{C_j}dy\, h_b(y)=(1-e^{-2\pi i b})\int _0^\infty dy\, h_b(y) \, -(1+e^{-2\pi i b})\pi i, \end{aligned}$$

(7.5)

which leads to (7.2).

Fig. 2

The contour C on $\mathbb {C}$, where $\alpha \in (0,\infty )$. It consists of the paths $C_1,\ldots ,C_6$

Next we give a proof of (2.30). For this purpose we first show the following relation. Let $I_j(x),~j=1,2,\ldots$, $x\in (0,\infty )$, be

$$\begin{aligned} I_j(x)=\int _0^\infty dw\, \frac{1}{x-w}\frac{(\log w)^{j-1}}{w+1}. \end{aligned}$$

(7.6)

Then we have

$$\begin{aligned} I_j(x)=\frac{1}{x+1}\,r_{j}(\log x), \end{aligned}$$

(7.7)

where $r_k(x)~(k=0,1,2,\ldots )$ is a kth order polynomial of x where the coefficient of the highest degree is 1 / k. This relation (7.7) can be derived by considering the integration of $m_j(w;x):=(\log w)^j/((x-w)(w+1)),~x>0$ with respect to w along the contour C in Fig. 2 with $\alpha =x$ and $R>1$.

Note that

$$\begin{aligned} \int _{C}dw\, m_j(w;x)=2\pi i\,\frac{(\pi i)^{j}}{x+1}, \end{aligned}$$

(7.8)

where RHS corresponds to the residue of $m_j(w;x)$ at $w=-1$. As in the previous case (7.4), one easily gets

$$\begin{aligned} \lim _{\scriptstyle \begin{array}{c} \delta \rightarrow 0, \epsilon \rightarrow 0 \\ R\rightarrow \infty \end{array}} \int _{C_1}dw\,m_j(w;x)&=I_{j+1}(x),\nonumber \\ \lim _{\scriptstyle \begin{array}{c} \delta \rightarrow 0, \epsilon \rightarrow 0 \\ R\rightarrow \infty \end{array}} \int _{C_4}dw\,m_j(w;x)&=-\int _0^\infty dw\, m_j(we^{2\pi i};x) =-\int _0^\infty dw\, \frac{(\log w+2\pi i)^j}{(x-w)(w+1)}\nonumber \\&\quad =-\sum _{k=0}^{j}\left( {\begin{array}{c}j\\ k\end{array}}\right) (2\pi i)^{j-k}I_{k+1}(x),\nonumber \\ \lim _{\epsilon \rightarrow 0}\int _{C_2} dw\, m_j(w;x)&=\pi i\,\frac{(\log x)^j}{x+1},~~ \lim _{\epsilon \rightarrow 0}\int _{C_5}m_j(w;x)dw=\pi i\,\frac{(\log x+2\pi i)^j}{x+1},\nonumber \\ \lim _{R\rightarrow \infty }\int _{C_3} dw\, m_j(w;x)&=\lim _{\delta \rightarrow 0} \int _{C_6}dw\,m_j(w;x)=0. \end{aligned}$$

(7.9)

Substituting (7.9) into (7.8), we find

$$\begin{aligned} 2\pi i\frac{(\pi i)^j}{x+1}&=\lim _{\scriptstyle \begin{array}{c} \delta \rightarrow 0, \epsilon \rightarrow 0 \\ R\rightarrow \infty \end{array}}\sum _{k=1}^6\int _{C_k}dw\, m_j(w;x)\nonumber \\&=I_{j+1}(x)-\sum _{k=0}^j\left( {\begin{array}{c}j\\ k\end{array}}\right) (2\pi i)^{j-k}I_{k+1}(x)+\frac{\pi i}{x+1} \left( (\log x)^j+\left( \log x+2\pi i\right) ^j\right) . \end{aligned}$$

(7.10)

Thus we obtain

$$\begin{aligned} I_{j}(x)=\frac{(\log x)^{j}+(\log x+2\pi i)^j-2(\pi i)^j}{2j(x+1)}-\frac{1}{j}\sum _{k=0}^{j-2}\left( {\begin{array}{c}j\\ k\end{array}}\right) (2\pi i)^{j-1-k}I_{k+1}(x) \end{aligned}$$

(7.11)

which leads to (7.7).

Here we show (2.30). We find that (2.30) is rewritten as

$$\begin{aligned} J_B^{*(m)}J_F(x)=q_m\left( \frac{\log x}{\beta }\right) J_F(x), \end{aligned}$$

(7.12)

where $q_m(x)$ is defined below (2.30) and the functions $J_F(x)$ and $J_B(x)$ on $\mathbb {R}_+$ are defined by $J_F(x)=1/(x+1)$ and $J_B(x)=1/\beta x$.

We prove (7.12) by using (7.7) and by mathematical induction: suppose that (7.12) holds for $m=N-1$. Then we get

$$\begin{aligned} J^{*(N)}_BJ_F(x)&=\frac{1}{\beta }\int _{0}^\infty dy\,\frac{1}{x-y}J^{*(N-1)}_BJ_F(y) =\frac{1}{\beta ^N(N-1)!}I_{N}(y)+\sum _{k=0}^{N-2}\frac{c_k}{\beta ^{k+1}}I_{k+1}(y) \end{aligned}$$

(7.13)

where $c_k (k=0,1,\ldots ,N-2)$ is the coefficient of $x^k$ in $q_{N-1}(x)$ and in the last equality we used the assumption for the mathematical induction and (7.6). Considering (7.7), we arrive at (7.12). $\square$

Appendix 2: Proof of Lemma 10

To show Lemma 10, we will use the following identity. For $(x_1,\ldots , x_{N-1})\in \mathbb {R}^{N-1}$ satisfying $x_1>\cdots >x_{N-1}$ and $(y_1,\ldots ,y_N)\in \mathbb {R}^N$, we have

$$\begin{aligned}&\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{j=2}^N1_{>0}\left( x_{j-1}-y_{\sigma (j)}\right) \nonumber \\&\qquad \quad =\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{j=2}^N1_{>0}\left( x_{j-1}-y_{\sigma (j)}\right) 1_{>0}\left( y_{\sigma (j-1)}-x_{j-1}\right) , \end{aligned}$$

(8.1)

where $S_N$ is the permutation of $(1,2,\ldots ,N)$. For the proof of (8.1), it is sufficient to show for $m=1,2, \cdots , N$,

$$\begin{aligned}&\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{j=2}^N1_{>0}(x_{j-1}-y_{\sigma (j)})\cdot \prod _{k=2}^{m}1_{>0}(y_{\sigma (k-1)}-x_{k-1})\cdot 1_{>0}(x_{m}-y_{\sigma (m)}) =0, \end{aligned}$$

(8.2)

where, as in (8.1), we assume the condition $x_1>x_2>\cdots >x_{N-1}$. This can easily be obtained by noting that

$$\begin{aligned}&\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{j=2}^N1_{>0}(x_{j-1}-y_{\sigma (j)})\cdot \prod _{k=2}^{m}1_{>0}(y_{\sigma (k-1)}-x_{k-1})\cdot 1_{>0}(x_{m}-y_{\sigma (m)})\nonumber \\&\quad =\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{\begin{array}{c} j=2\\ j\ne m \end{array}}^N1_{>0}(x_{j-1}-y_{\sigma (j)})\cdot \prod _{k=2}^{m}1_{>0}(y_{\sigma (k-1)}-x_{k-1})\cdot 1_{>0}(x_{m}-y_{\sigma (m)})\nonumber \\&\quad =\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{\begin{array}{c} j=2\\ j\ne m,~m+1 \end{array}}^N 1_{>0}(x_{j-1}-y_{\sigma (j)})\nonumber \\&\qquad \times \prod _{k=2}^{m}1_{>0}(y_{\sigma (k-1)}-x_{k-1})\cdot 1_{>0}(x_{m}-y_{\sigma (m)})1_{>0}(x_{m}-y_{\sigma (m+1)})\nonumber \\&\quad =0 \end{aligned}$$

(8.3)

where in the first equality we used the fact that the factor $1_{>0}(x_{m-1}-y_{\sigma (m)})$ can be omitted in this equation thanks to the factor $1_{>0}(x_m-y_{\sigma (m)})$ with the condition $x_{m-1}>x_{m}$ and the last equality follows from the fact that the term with $\sigma$ cancels the term with $\sigma '$ where $\sigma '$ is defined in terms of $\sigma$ as $\sigma '(m)=\sigma (m+1)$ and $\sigma '(m+1)=\sigma (m)$ with $\sigma '(k)=\sigma (k)$ for $k\ne m,~m+1$.

Using (8.2), we have for $x_1>x_2>\cdots >x_{N-1}$

$$\begin{aligned}&\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{j=2}^N1_{>0}\left( x_{j-1}-y_{\sigma (j)}\right) \nonumber \\&\quad = \sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{j=2}^N1_{>0}\left( x_{j-1}-y_{\sigma (j)}\right) \cdot (1-1_{>0}(x_{1}-y_{\sigma (1)}))\nonumber \\&\quad =\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{j=2}^N1_{>0}\left( x_{j-1}-y_{\sigma (j)}\right) \cdot 1_{>0}(y_{\sigma (1)}-x_1)\nonumber \\&\quad =\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{j=2}^N1_{>0}\left( x_{j-1}-y_{\sigma (j)}\right) \cdot 1_{>0}(y_{\sigma (1)}-x_1)(1-1_{>0}(x_{2}-y_{\sigma (2)}))\nonumber \\&\quad =\sum _{\sigma \in S_N}\mathrm{{sgn}}\sigma \prod _{j=2}^N1_{>0}\left( x_{j-1}-y_{\sigma (j)}\right) \cdot \prod _{k=1}^2 1_{>0}(y_{\sigma (k)}-x_k), \end{aligned}$$

(8.4)

where for the first and the third equality, we used (8.2) with $m=1$ and $m=2$ respectively. Performing the procedure in (8.4) repeatedly, we arrive at (8.1).

Now we give a proof of the lemma by the mathematical induction. The case $N=1$ in is trivial. Suppose that it holds for $N-1$. Then noticing

$$\begin{aligned} 1_{V_N}(\underline{x}_N)=\prod _{k=2}^N\prod _{j=2}^k1_{>0}\left( x^{(k-1)}_{j-1}-x^{(k)}_j\right) =1_{V_{N-1}}(\underline{x}_{N-1})\prod _{j=2}^N1_{>0}\left( x^{(N-1)}_{j-1}-x^{(N)}_j\right) , \end{aligned}$$

(8.5)

we see that LHS of (3.41) is written as

$$\begin{aligned}&\sum _{\sigma ^{(j)}\in S_j,j=1,\ldots ,N}\mathrm{{sgn}}\sigma ^{(N)}\, 1_{V_{N-1}}(\underline{x}_{N-1}^{\sigma }) \prod _{j=2}^N 1_{>0}\left( x^{(N-1)}_{\sigma ^{(N-1)}(j-1)}-x^{(N)}_{\sigma ^{(N)}(j)}\right) \nonumber \\&\qquad =\sum _{\sigma ^{(j)}\in S_j,j=1,\ldots ,N}\mathrm{{sgn}}\sigma ^{(N-1)}\, 1_{V_{N-1}}(\underline{x}_{N-1}^{\sigma })\cdot \mathrm{{sgn}}\sigma ^{(N)} \prod _{j=2}^N 1_{>0}\left( x^{(N-1)}_{j-1}-x^{(N)}_{\sigma ^{(N)}(j)}\right) \nonumber \\&\qquad =\sum _{\sigma ^{(j)}\in S_j,j=1,\ldots ,N}\mathrm{{sgn}}\sigma ^{(N-1)}\, 1_{\text {GT}}(\underline{x}_{N-1}^{\sigma })\cdot \mathrm{{sgn}}\sigma ^{(N)} \prod _{j=2}^N 1_{>0}\left( x^{(N-1)}_{j-1}-x^{(N)}_{\sigma ^{(N)}(j)}\right) \nonumber \\&\qquad =\sum _{\sigma ^{(j)}\in S_j,j=1,\ldots ,N}\mathrm{{sgn}}\sigma ^{(N)}\, 1_{\text {GT}}(\underline{x}_{N-1}^{\sigma }) \prod _{j=2}^N 1_{>0}\left( x^{(N-1)}_{\sigma ^{(N-1)}(j-1)}-x^{(N)}_{\sigma ^{(N)}(j)}\right) , \end{aligned}$$

(8.6)

where in the second equality we used the assumption for $N-1$. Note that in the rightmost side of (8.6), the condition $x^{(N-1)}_{\sigma ^{(N-1)}(1)}>x^{(N-1)}_{\sigma ^{(N-1)}(2)}>\cdots > x^{(N-1)}_{\sigma ^{(N-1)}(N-1)}$ holds for the support of $1_{\text {GT}}(\underline{x}_{N-1}^{\sigma })$. Thus we can apply (8.1) to the rightmost side. We see that it becomes

$$\begin{aligned}&\sum _{\sigma ^{(j)}\in S_j,j=1,\ldots ,N}\mathrm{{sgn}}\sigma ^{(N)}\, 1_{\text {GT}}(\underline{x}_{N-1}^{\sigma })\nonumber \\&\qquad \times \prod _{j=2}^N 1_{>0}\left( x^{(N-1)}_{\sigma ^{(N-1)}(j-1)}-x^{(N)}_{\sigma ^{(N)}(j)}\right) 1_{>0}\left( x^{(N)}_{\sigma ^{(N)}(j-1)}-x^{(N-1)}_{\sigma ^{(N-1)}(j-1)}\right) \nonumber \\&\quad = \sum _{\sigma ^{(j)}\in S_j,j=1,\ldots ,N}\mathrm{{sgn}}\sigma ^{(N)} \, 1_\mathrm{{GT}}(\underline{x}_N^\sigma ), \end{aligned}$$

(8.7)

which completes the proof of Lemma 10. $\square$

Appendix 3: The Saddle Point Analysis of $\psi _k(x;t)$

In this Appendix, we give a proof of (5.42) based on the saddle point method in a similar way to Sect. 5.4.3 in [10]. Here we deal with the case of general Y while the case of $Y=0$ was considered in [10]. We focus mainly on the limit about $\psi _k(x;t)$ (2.1) in (5.42) since the case $\phi _k(x;t)$ (4.1) can also be estimated in a parallel way. Changing the variable as $w=-i\sqrt{N}z$, (2.1) becomes

$$\begin{aligned} \psi _k(x,t)=\frac{\sqrt{N}}{2\pi i}\int _{-i\infty }^{i\infty } dz\, e^{f_N(z;t,x)}, \end{aligned}$$

(9.1)

where

$$\begin{aligned} f_N(z;t,x)=-\sqrt{N}zx+N\frac{z^2t}{2}+(k-N)\log (\sqrt{N}z) -N\log \Gamma (\sqrt{N} z). \end{aligned}$$

(9.2)

Substituting (5.41) and (5.43) into (9.2), we arrange the first three terms in ascending order of powers of N as

$$\begin{aligned}&-\sqrt{N}zx_i+N\frac{z^2t}{2}+(k-N)\log (\sqrt{N}z)\nonumber \\&\quad =N^{3/2}\log N \cdot \frac{z}{2} +N^{3/2}\left( -z+\frac{T^{1/2}z^2}{2}-\frac{z}{2}\log T\right) +N\left( \gamma _T^2Yz^2-2\gamma _T^2YT^{-1/2}z -\frac{T^{1/2}z}{2}\right) \nonumber \\&\qquad +N^{1/2}\log N\cdot \frac{\lambda }{2(2\gamma _T)^{1/2}} +N^{1/2}\left( \frac{\gamma _T^3}{12}z-\gamma _T\xi _i z +\gamma _TY^2z-\gamma _T^2Yz +\frac{\lambda }{(2\gamma _T)^{1/2}}\log z\right) . \end{aligned}$$

(9.3)

Using the Stirling formula

$$\begin{aligned} \log \Gamma (n)=n\log n-n-\frac{\log 2\pi n}{2}+\frac{1}{12n}+O(n^{-3}) \end{aligned}$$

(9.4)

for the last term in (9.2), we have

$$\begin{aligned} -N\log \Gamma (\sqrt{N}z)&=-N^{3/2}\log N\cdot \frac{z}{2}+ N^{3/2}\left( z-z\log z\right) +\frac{N}{4}\log N+ N \frac{\log 2\pi z}{2}\nonumber \\&~~~~-N^{1/2}\frac{1}{12z}+O(N^{-1/2}). \end{aligned}$$

(9.5)

Thus from (9.3) and (9.5), $f_N(z)$ (9.2) can be expressed as

$$\begin{aligned}&f_N(z;t,x)=N^{3/2}f(z)+Ng(z)+N^{1/2}h(z)+C_1+O(N^{-1/2}), \end{aligned}$$

(9.6)

$$\begin{aligned}&f(z)=\frac{T^{1/2}z^2}{2}-z\log z-\frac{z}{2}\log T, \end{aligned}$$

(9.7)

$$\begin{aligned}&g(z)=-\frac{T^{1/2}z}{2}+\frac{\log z}{2}+\gamma _T^2Yz^2-\frac{2\gamma _T^2Y}{T^{1/2}}z, \end{aligned}$$

(9.8)

$$\begin{aligned}&h(z)=-\frac{1}{12z}+\left( \frac{\gamma _T^3}{12}-\gamma _T^2Y-\gamma _T(\xi _i-Y^2) \right) z +\frac{\lambda }{(2\gamma _T)^{1/2}}\log z. \end{aligned}$$

(9.9)

Here $C_1$, which does not depend on z is written as

$$\begin{aligned} C_1=N^{1/2}\log N\cdot \frac{\lambda }{2(2\gamma _T)^{1/2}}+\frac{N}{2}\log 2\pi \sqrt{N}. \end{aligned}$$

(9.10)

We note that f(z) above has a double saddle point $z_c=T^{-1/2}$ such that $f'(z_c)=f''(z_c)=0$. We expand f(z),  g(z),  h(z) around $z_c$. Noting $f'''(z_c)=2\gamma _T^3$, $g'(z_c)=0,~g''(z_c)=2\gamma _T^2Y-\gamma _T^3$, $h'(z_c)=\gamma _T^3/4-\gamma _T^2Y+\gamma _T(\lambda -\xi _i+Y^2)$, we get

$$\begin{aligned} N^{3/2}f(z)&=N^{3/2}f(z_c)+\frac{N^{3/2}}{3!}f'''(z_c)(z-z_3)^3+ O(N^{3/2}(z-z_c)^4)\nonumber \\&=C_2+N^{3/2}\frac{\gamma _T^3}{3}(z-z_c)^3+O(N^{3/2}(z-z_c)^4), \end{aligned}$$

(9.11)

$$\begin{aligned} Ng(z)&=Ng(z_c)+Ng'(z_c)(z-z_c)+N\frac{g''(z_c)}{2}(z-z_c)^2 +O(N(z-z_c)^3)\nonumber \\&=C_3-N\left( \frac{\gamma _T^3}{2}-\gamma _T^2Y\right) (z-z_c)^2+O(N(z-z_c)^3), \end{aligned}$$

(9.12)

$$\begin{aligned} N^{1/2}h(z)&=N^{1/2}h(z_c)+N^{1/2}h'(z_c)(z-z_c)+C_4+O(N^{1/2}(z-z_c)^2) \nonumber \\&=C_4+N^{1/2}\left( \frac{\gamma _T^3}{4}-\gamma _T^2Y+\gamma _T(\lambda -\xi +Y^2) \right) (z-z_c)+O(N^{1/2}(z-z_c)^2), \end{aligned}$$

(9.13)

where $C_2,~C_3$ and $C_4$ are

$$\begin{aligned}&C_2:=N^{3/2}f(z_c)=N^{3/2}T^{-1/2}/2,~~C_3:=Ng(z_c)=-N \left( \frac{1+\log T}{2}+\frac{Y}{2\gamma _T}\right) , \nonumber \\&C_4:=N^{1/2}h(z_c)=-N^{1/2}\left( \frac{T^{1/2}}{24}+(\gamma _T^2Y+\gamma _T(\xi -Y^2))T^{-1/2} -\frac{\lambda \log T^{-1/2}}{(2\gamma _T)^{1/2}} \right) . \end{aligned}$$

(9.14)

Thus from (9.6)–(9.13) and under the scaling $\frac{z'}{\sqrt{N}}=(z-z_c)$, we have

$$\begin{aligned} f_N(z;t,x_i)=\frac{\gamma _T^3}{3} \left( z'+\frac{Y}{\gamma _T}-\frac{1}{2}\right) ^3-\gamma _T(\xi -\lambda ) \left( z'+\frac{Y}{\gamma _T}-\frac{1}{2}\right) +\sum _{j=1}^5C_j+O\left( N^{-1/2}\right) , \end{aligned}$$

(9.15)

where $C_1,\ldots ,C_4$ are defined in (9.10) and (9.14) and $C_5$ is

$$\begin{aligned} C_5=-\frac{1}{3}\left( Y-\frac{\gamma _T}{2}\right) ^3+\left( \xi -\lambda \right) \left( Y-\frac{\gamma _T}{2}\right) . \end{aligned}$$

(9.16)

Further changing the variable $z'+Y/\gamma _T-1/2=-iv/\gamma _T$, we obtain

$$\begin{aligned}&f_N(z;t,x_i)=\frac{i}{3}v^3+i(\xi _i-\lambda )v +\sum _{j=1}^5C_j+O\left( N^{-1/2}\right) . \end{aligned}$$

(9.17)

Hence from (9.1) and (9.17), we get the limiting form of $\psi _k(x_i;t)$

$$\begin{aligned} e^{-\sum _{j=1}^5C_j}\gamma _T \psi _k(x_i;t)\sim \frac{1}{2\pi }\int _{-\infty }^{\infty }dv\,e^{\frac{i}{3}v^3+i(\xi _i-\lambda )v} =\mathrm{Ai}(\xi _i-\lambda ), \end{aligned}$$

(9.18)

which is nothing but (5.42).

As with (9.1), we rewrite $\phi _k(x;t)$ (4.1) by the change of variable $v=\sqrt{N}z$,

$$\begin{aligned} \phi _k(x;t)=\frac{1}{2\pi i}\oint dz\, \frac{e^{-f_N(z;t,x)}}{z}, \end{aligned}$$

(9.19)

where $f_N(z;t,x)$ is given in (9.2). Applying the same techniques as above to this equation, we get the result for $\phi _k(x;t)$.