SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \frac {\sqrt {\pi}}{2} \operatorname {E r f c} (x) = \int_ {x} ^ {\infty} \exp (- t ^ {2}) d t = \left. \left(- \frac {\exp (- t ^ {2})}{2 t}\right) \right| _ {x} ^ {\infty} - \int_ {x} ^ {\infty} \frac {\exp (- t ^ {2})}{2 t ^ {2}} d t \\ = \frac {\exp (- x ^ {2})}{2 x} - \left(\left(- \frac {\exp (- t ^ {2})}{4 t ^ {3}}\right) \Bigg | _ {x} ^ {\infty} - \int_ {x} ^ {\infty} \frac {3 \exp (- t ^ {2})}{4 t ^ {4}} d t\right) \\ = \frac {\exp (- x ^ {2})}{2 x} - \frac {\exp (- x ^ {2})}{4 x ^ {3}} + \underbrace {\int_ {x} ^ {\infty} \frac {3 \exp (- t ^ {2})}{4 t ^ {4}} d t} _ {\geq 0} (48) \\ = \frac {\exp (- x ^ {2})}{2 x} - \frac {\exp (- x ^ {2})}{4 x ^ {3}} + \left(- \frac {3 \exp (- t ^ {2})}{8 t ^ {5}}\right) \bigg | _ {x} ^ {\infty} - \int_ {x} ^ {\infty} \frac {1 5 \exp (- t ^ {2})}{8 t ^ {6}} d t \\ = \frac {\exp (- x ^ {2})}{2 x} - \frac {\exp (- x ^ {2})}{4 x ^ {3}} + \frac {3 \exp (- x ^ {2})}{8 x ^ {5}} \underbrace {- \int_ {x} ^ {\infty} \frac {1 5 \exp (- t ^ {2})}{8 t ^ {6}} d t} _ {\leq 0} (49) \\ \end{array}
$$
The results follow by (48) and (49).
Our next result, states an expression for the function  $p(n_{i})$  and its derivative as defined in (23).
Lemma 13 (Value and derivative of penalty function at  $s^{\star}$ ). Let  $p(n_i) = p_i(M_{\mathrm{c3D}},((n_i,f_i^\star ,h_i^\star),s_{-i}^\star))$  (see (23)) and  $s_i^\star, f_i^\star, h_i^\star$  be as specified in (8). The penalty of agent  $i$  in Algorithm 1 satisfies:
$$
p \left(n _ {i} ^ {\star}\right) = \frac {\sqrt {\frac {\alpha^ {2}}{m n _ {i} ^ {\star}}} \sigma^ {2} \left(2 m \sqrt {2 \pi} \sqrt {\frac {\alpha^ {2}}{m n _ {i} ^ {\star}}} - \exp \left(\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}\right) (m - 2) \pi \operatorname {E r f c} \left(\frac {1}{2 \sqrt {2} \sqrt {\frac {\alpha^ {2}}{m n _ {i} ^ {\star}}}}\right)\right)}{4 \sqrt {2 \pi} \alpha^ {2}} + c n _ {i} ^ {\star} \tag {50}
$$
$$
\begin{array}{l} p ^ {\prime} (n _ {i} ^ {\star}) = - \frac {\sigma^ {2}}{6 4 \frac {\alpha^ {2}}{m - 2} \frac {\alpha}{\sqrt {m n _ {i} ^ {\star}}} m n _ {i} ^ {\star}} \left(\frac {4 \alpha}{\sqrt {m n _ {i} ^ {\star}}} \left(\frac {4 \alpha^ {2} m}{(m - 2) n _ {i} ^ {\star}} - 1\right) \right. \\ \left. - \exp \left(\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}\right) \left(\frac {4 \alpha^ {2}}{m n _ {i} ^ {\star}} (m + 1) - 1\right) \sqrt {2 \pi} \operatorname {E r f c} \left(\frac {1}{2 \sqrt {2} \sqrt {\frac {\alpha^ {2}}{m n _ {i} ^ {\star}}}}\right)\right) + c. \tag {51} \\ \end{array}
$$
This proof involves several algebraic manipulations, so we will provide an outline of our proof strategy. First, we will rearrange the denominator inside the expectation in (23), to write the LHS of (50) as  $J + K\mathbb{E}\left[\frac{1}{L + x^2}\right]$ , and the LHS of (51) as  $J' + K'\mathbb{E}\left[\frac{1}{L + x^2}\right] + K''\mathbb{E}\left[\frac{1}{(L + x^2)^2}\right]$ , where the expectation is with respect to a standard normal  $\mathcal{N}(0,1)$  variable,  $J, K, K', K'', L$  are quantities that depend on  $n_i, m, c, \sigma^2, \alpha^2$ , and importantly,  $L$  is strictly larger than 0. Using properties of the normal distribution, in Lemma 14, we prove the following result:
$$
\mathbb {E} \left[ \frac {1}{L + x ^ {2}} \right] = \sqrt {\frac {\pi}{2 L}} \exp \left(\frac {L}{2}\right) \operatorname {E r f c} \left(\sqrt {\frac {L}{2}}\right) \tag {52}
$$
$$
\mathbb {E} \left[ \frac {1}{(L + x ^ {2}) ^ {2}} \right] = \frac {\sqrt {\pi}}{2 \sqrt {2} L ^ {3 / 2}} (1 - L) \exp \left(\frac {L}{2}\right) \operatorname {E r f c} \left(\sqrt {\frac {L}{2}}\right) + \frac {1}{2 L} \tag {53}
$$
By plugging in these expressions and then substituting  $n_i = n_i^\star$ , we obtain (50) and (51).
Proof of Lemma 13. We will rewrite  $p(n_i^\star)$  and  $p'(n_i^\star)$  as the Gaussian integral of rational functions and use (52) to calculate their values. By (23),
$$
\begin{array}{l} p (n _ {i} ^ {\star}) = \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {1}{\frac {(m - 2) n _ {i} ^ {\star}}{\sigma^ {2} + \alpha^ {2} \left(\frac {\sigma^ {2}}{n _ {i} ^ {\star}} + \frac {\sigma^ {2}}{n _ {i} ^ {\star}}\right) x ^ {2}} + \frac {n _ {i} ^ {\star} + n _ {i} ^ {\star}}{\sigma^ {2}}} \right] + c n _ {i} ^ {\star} \\ = \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {1}{\frac {(m - 2) n _ {i} ^ {\star}}{\sigma^ {2} + \alpha^ {2} \frac {2 \sigma^ {2}}{n _ {i} ^ {\star}} x ^ {2}} + \frac {2 n _ {i} ^ {\star}}{\sigma^ {2}}} \right] + c n _ {i} ^ {\star} = \frac {\sigma^ {2}}{n _ {i} ^ {\star}} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {1}{\frac {m - 2}{1 + \frac {2 \alpha^ {2}}{n _ {i} ^ {\star}} x ^ {2}} + 2} \right] + c n _ {i} ^ {\star} \\ = \frac {\sigma^ {2}}{n _ {i} ^ {\star}} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {1}{2} - \frac {m - 2}{2} \frac {1}{\frac {4 \alpha^ {2}}{n _ {i} ^ {\star}} x ^ {2} + m} \right] + c n _ {i} ^ {\star} \\ = \frac {\sigma^ {2}}{2 n _ {i} ^ {\star}} - \frac {\sigma^ {2}}{n _ {i} ^ {\star}} \frac {m - 2}{2} \frac {n _ {i} ^ {\star}}{4 \alpha^ {2}} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {1}{x ^ {2} + \frac {m n _ {i} ^ {\star}}{4 \alpha^ {2}}} \right] + c n _ {i} ^ {\star} \\ = \frac {\sigma^ {2}}{2 n _ {i} ^ {\star}} - \frac {\sigma^ {2}}{4 \alpha^ {2}} \frac {m - 2}{2} \exp \left(\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}\right) \mathrm {E r f c} \left(\sqrt {\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}}\right) \sqrt {\frac {\pi}{\frac {m n _ {i} ^ {\star}}{2 \alpha^ {2}}}} + c n _ {i} ^ {\star} \\ \left(\text {I n} (5 2), \text {l e t} L = \frac {m n _ {i} ^ {\star}}{4 \alpha^ {2}}\right) \\ = \text {R H S} \text {o f} (5 0). \\ \end{array}
$$
To prove the second statement of Lemma 13, by (24) and the dominated convergence theorem, we have:
$$
p ^ {\prime} (n _ {i} ^ {\star}) = \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ - \sigma^ {2} \frac {1 + \frac {(m - 2) n _ {i} ^ {\star}}{\left(1 + \alpha^ {2} \left(\frac {1}{n _ {i} ^ {\star}} + \frac {1}{n _ {i} ^ {\star}}\right) x ^ {2}\right) ^ {2}} \frac {\alpha^ {2} x ^ {2}}{n _ {i} ^ {\star 2}}}{\left(\frac {(m - 2) n _ {i} ^ {\star}}{1 + \alpha^ {2} \left(\frac {1}{n _ {i} ^ {\star}} + \frac {1}{n _ {i} ^ {\star}}\right) x ^ {2}} + n _ {i} ^ {\star} + n _ {i} ^ {\star}\right) ^ {2}} \right] + c
$$
(By (24) and dominated convergence theorem)
$$
\begin{array}{l} = - \sigma^ {2} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {1 + \frac {(m - 2) n _ {i} ^ {\star}}{\left(1 + \frac {2 \alpha^ {2}}{n _ {i} ^ {\star}} x ^ {2}\right) ^ {2}} \frac {\alpha^ {2} x ^ {2}}{n _ {i} ^ {\star 2}}}{\left(\frac {(m - 2) n _ {i} ^ {\star}}{1 + \frac {2 \alpha^ {2}}{n _ {i} ^ {\star}} x ^ {2}} + 2 n _ {i} ^ {\star}\right) ^ {2}} \right] + c \\ = - \frac {\sigma^ {2}}{n _ {i} ^ {\star 2}} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {1 + \frac {(m - 2) n _ {i} ^ {\star} \alpha^ {2} x ^ {2}}{\left(n _ {i} ^ {\star} + 2 \alpha^ {2} x ^ {2}\right) ^ {2}}}{\left(\frac {(m - 2) n _ {i} ^ {\star}}{n _ {i} ^ {\star} + 2 \alpha^ {2} x ^ {2}} + 2\right) ^ {2}} \right] + c \\ = - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {4 (n _ {i} ^ {\star} + 2 \alpha^ {2} x ^ {2}) ^ {2} + 4 (m - 2) n _ {i} ^ {\star} \alpha^ {2} x ^ {2}}{((m - 2) n _ {i} ^ {\star} + 2 (n _ {i} ^ {\star} + 2 \alpha^ {2} x ^ {2})) ^ {2}} \right] + c \\ = - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ 1 + \frac {- (m - 2) ^ {2} n _ {i} ^ {\star 2} - 4 (m - 2) n _ {i} ^ {\star} \left(n _ {i} ^ {\star} + 2 \alpha^ {2} x ^ {2}\right) + 4 (m - 2) n _ {i} ^ {\star} \alpha^ {2} x ^ {2}}{\left((m - 2) n _ {i} ^ {\star} + 2 (n _ {i} ^ {\star} + 2 \alpha^ {2} x ^ {2})\right) ^ {2}} \right] + c \\ \end{array}
$$
$$
\begin{array}{l} = - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ 1 + (m - 2) n _ {i} ^ {\star} \frac {- (m - 2) n _ {i} ^ {\star} - 4 \left(n _ {i} ^ {\star} + 2 \alpha^ {2} x ^ {2}\right) + 4 \alpha^ {2} x ^ {2}}{\left(4 \alpha^ {2} x ^ {2} + m n _ {i} ^ {\star}\right) ^ {2}} \right] + c \\ = - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ 1 + (m - 2) n _ {i} ^ {\star} \frac {- (m + 2) n _ {i} ^ {\star} - 4 \alpha^ {2} x ^ {2}}{\left(4 \alpha^ {2} x ^ {2} + m n _ {i} ^ {\star}\right) ^ {2}} \right] + c \\ = - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} + \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} (m - 2) n _ {i} ^ {\star} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {(4 \alpha^ {2} x ^ {2} + m n _ {i} ^ {\star}) + 2 n _ {i} ^ {\star}}{\left(4 \alpha^ {2} x ^ {2} + m n _ {i} ^ {\star}\right) ^ {2}} \right] + c \\ = - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} + \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} (m - 2) n _ {i} ^ {\star} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {1}{4 \alpha^ {2} x ^ {2} + m n _ {i} ^ {\star}} + \frac {2 n _ {i} ^ {\star}}{\left(4 \alpha^ {2} x ^ {2} + m n _ {i} ^ {\star}\right) ^ {2}} \right] + c \\ = - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} + \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} (m - 2) n _ {i} ^ {\star} \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \left[ \frac {1}{4 \alpha^ {2}} \frac {1}{x ^ {2} + \frac {m n _ {i} ^ {\star}}{4 \alpha^ {2}}} + \frac {2 n _ {i} ^ {\star}}{1 6 \alpha^ {4}} \frac {1}{\left(x ^ {2} + \frac {m n _ {i} ^ {\star}}{4 \alpha^ {2}}\right) ^ {2}} \right] + c \\ = c - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} + \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} (m - 2) n _ {i} ^ {\star} \left(\frac {1}{4 \alpha^ {2}} + \frac {2 n _ {i} ^ {\star}}{1 6 \alpha^ {4}} \frac {1 - \frac {m n _ {i} ^ {\star}}{4 \alpha^ {2}}}{\frac {m n _ {i} ^ {\star}}{2 \alpha^ {2}}}\right) \exp \left(\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}\right) \mathrm {E r f c} \left(\sqrt {\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}}\right) \sqrt {\frac {\pi}{\frac {m n _ {i} ^ {\star}}{2 \alpha^ {2}}}} \\ + \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} (m - 2) n _ {i} ^ {\star} \frac {2 n _ {i} ^ {\star}}{1 6 \alpha^ {4}} \frac {1}{\frac {m n _ {i} ^ {\star}}{2 \alpha^ {2}}} \\ \end{array}
$$
$$
\begin{array}{l} \left(\text {I n} (5 2) \text {a n d} (5 3) \text {a n d} L = \frac {m n _ {i} ^ {\star}}{4 \alpha^ {2}}\right) \\ = c - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} \left(1 - \frac {(m - 2) n _ {i} ^ {\star}}{4 \alpha^ {2} m}\right) \\ + \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} (m - 2) n _ {i} ^ {\star} \left(\frac {1}{4 \alpha^ {2}} + \frac {1}{4 m \alpha^ {2}} - \frac {n _ {i} ^ {\star}}{1 6 \alpha^ {4}}\right) \exp \left(\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}\right) \operatorname {E r f c} \left(\sqrt {\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}}\right) \sqrt {\frac {\pi}{\frac {m n _ {i} ^ {\star}}{2 \alpha^ {2}}}} \\ = c - \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} \left(1 - \frac {(m - 2) n _ {i} ^ {\star}}{4 \alpha^ {2} m}\right) \\ + \frac {\sigma^ {2}}{4 n _ {i} ^ {\star 2}} (m - 2) n _ {i} ^ {\star} \frac {\alpha \sqrt {2 \pi}}{\sqrt {m n _ {i} ^ {\star}}} \frac {n _ {i} ^ {\star}}{1 6 \alpha^ {4}} \left(\frac {4 \alpha^ {2}}{m n _ {i} ^ {\star}} (m + 1) - 1\right) \exp \left(\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}\right) \operatorname {E r f c} \left(\sqrt {\frac {m n _ {i} ^ {\star}}{8 \alpha^ {2}}}\right) \\ = \text {R H S} \tag {51} \\ \end{array}
$$
We will now prove the statements in (52) and (53). Both statements follow from the Lemma below by substituting  $t = 1 / 2$ .
Lemma 14. For all  $t \geq 0$  and some  $L > 0$ ,
$$
I (t) := \int_ {- \infty} ^ {\infty} \frac {1}{L + x ^ {2}} \frac {1}{\sqrt {2 \pi}} \exp (- t x ^ {2}) d x = \exp (L t) \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}}
$$
$$
J (t) := \int_ {- \infty} ^ {\infty} \frac {1}{(L + x ^ {2}) ^ {2}} \frac {1}{\sqrt {2 \pi}} \exp (- t x ^ {2}) d x = \sqrt {\frac {\pi}{2 L}} \left(\frac {1}{2 L} - t\right) \exp (L t) \operatorname {E r f c} (\sqrt {L t}) + \frac {\sqrt {t}}{\sqrt {2} L}
$$
Proof. We derive  $I(t)$  and  $J(t)$  as the solutions to two ODEs and solve the ODEs to obtain the results. Firstly, by calculation:
$$
- I ^ {\prime} (t) + L I (t) = \int_ {- \infty} ^ {\infty} \frac {x ^ {2} + L}{L + x ^ {2}} \frac {1}{\sqrt {2 \pi}} \exp \left(- t x ^ {2}\right) d x = \frac {1}{\sqrt {2 t}}.
$$
$$
I (0) = \int_ {- \infty} ^ {\infty} \frac {1}{L + x ^ {2}} \frac {1}{\sqrt {2 \pi}} d x = \sqrt {\frac {\pi}{2 L}}.
$$
This means  $I(t)$  satisfies the following ODE:
$$
\left\{ \begin{array}{l} - I ^ {\prime} (t) + L I (t) = \frac {1}{\sqrt {2 t}} \\ I (0) = \sqrt {\frac {\pi}{2 L}} \end{array} \right.. \tag {54}
$$
We solve (54) by multiplying integrating factor  $-\exp(-Lt)$ :
$$
\exp (- L t) I ^ {\prime} (t) - L \exp (- L t) I (t) = - \frac {1}{\sqrt {2 t}} \exp (- L t)
$$
Note that the LHS is the derivative of  $\exp(-Lt)I(t)$ , the ODE becomes:
$$
\frac {d}{d t} (\exp (- L t) I (t)) = - \frac {1}{\sqrt {2 t}} \exp (- L t)
$$
Integrating both sides over  $t$ , we get:
$$
\exp (- L t) I (t) = - \int \frac {1}{\sqrt {2 t}} \exp (- L t) d t = - \int \frac {2}{\sqrt {2 L}} \exp (- L t) d \sqrt {L t} = \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}} + C,
$$
where we use integration by substitution for the last two equalities and  $C$  is some constant that does not depend on  $t$ . This means  $I(t)$  satisfies the following form:
$$
I (t) = \exp (L t) \left(\operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}} + C\right)
$$
Using the initial condition  $I(0) = \sqrt{\frac{\pi}{2L}}$  and the fact that  $\operatorname{Erfc}(0) = 0$ , we conclude that  $C = 0$ . Thus
$$
I (t) = \exp (L t) \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}}.
$$
We can similarly derive an ODE for  $J(t)$ . By calculation:
$$
- J ^ {\prime} (t) + L J (t) = \int_ {- \infty} ^ {\infty} \frac {x ^ {2} + L}{(L + x ^ {2}) ^ {2}} \frac {1}{\sqrt {2 \pi}} \exp (- t x ^ {2}) d x = I (t)
$$
$$
J (0) = \int_ {- \infty} ^ {\infty} \frac {1}{(L + x ^ {2}) ^ {2}} \frac {1}{\sqrt {2 \pi}} d x = \frac {1}{2 L ^ {3 / 2}} \sqrt {\frac {\pi}{2}}
$$
Thus  $J(t)$  satisfies the following ODE:
$$
\left\{ \begin{array}{l} - J ^ {\prime} (t) + L J (t) = I (t) \\ J (0) = \frac {1}{2 L ^ {3 / 2}} \sqrt {\frac {\pi}{2}} \end{array} \right.. \tag {55}
$$
We similarly multiply integrating factor  $-\exp (-Lt)$  and integrate both sides:
$$
\int_ {0} ^ {t} d \exp (- L x) J (x) = - \int_ {0} ^ {t} I (x) \exp (- L x) d x = - \int_ {0} ^ {t} \operatorname {E r f c} (\sqrt {L x}) \sqrt {\frac {\pi}{2 L}} d x
$$
$$
= - \left(x \operatorname {E r f c} (\sqrt {L x}) \sqrt {\frac {\pi}{2 L}} \bigg | _ {0} ^ {t} + \int_ {0} ^ {t} x \frac {\exp (- L x)}{\sqrt {2 x}} d x\right)
$$
(Integration by parts)
$$
= - t \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}} - \frac {\sqrt {2}}{L ^ {3 / 2}} \int_ {0} ^ {\sqrt {L t}} y ^ {2} \exp (- y ^ {2}) d y
$$
(Change of variable:  $y = \sqrt{Lx}$ )
$$
= - t \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}} + \frac {\sqrt {2}}{L ^ {3 / 2}} \left(\frac {1}{2} y \exp (- y ^ {2}) \Big | _ {0} ^ {\sqrt {L t}} - \int_ {0} ^ {\sqrt {L t}} \frac {1}{2} \exp (- y ^ {2}) d y\right)
$$
$$
= - t \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}} + \frac {\sqrt {2}}{L ^ {3 / 2}} \frac {1}{2} \sqrt {L t} \exp (- L t) - \frac {\sqrt {2}}{L ^ {3 / 2}} \int_ {0} ^ {\sqrt {L t}} \frac {1}{2} \exp (- y ^ {2}) d y
$$
$$
= - t \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}} + \frac {\sqrt {t}}{\sqrt {2} L} \exp (- L t) - \frac {\sqrt {\pi}}{2 \sqrt {2} L ^ {3 / 2}} \operatorname {E r f} \left(\sqrt {L t}\right)
$$
(By definition of Erf)
$$
= - t \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}} + \frac {\sqrt {t}}{\sqrt {2} L} \exp (- L t) - \frac {\sqrt {\pi}}{2 \sqrt {2} L ^ {3 / 2}} \left(1 - \operatorname {E r f c} (\sqrt {L t})\right)
$$
(By definition of Erfc)
$$
= \left(\frac {1}{2 L} - t\right) \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}} + \frac {\sqrt {t}}{\sqrt {2} L} \exp (- L t) - J (0)
$$
(By (55))
$$
\begin{array}{l} J (t) = \exp (L t) \left(\int_ {0} ^ {t} d \exp (- L x) J (x) + J (0)\right) \\ = \exp (L t) \left(\left(\frac {1}{2 L} - t\right) \operatorname {E r f c} (\sqrt {L t}) \sqrt {\frac {\pi}{2 L}} + \frac {\sqrt {t}}{\sqrt {2} L} \exp (- L t)\right) \\ = \sqrt {\frac {\pi}{2 L}} \left(\frac {1}{2 L} - t\right) \exp (L t) \operatorname {E r f c} (\sqrt {L t}) + \frac {\sqrt {t}}{\sqrt {2} L} \\ \end{array}
$$