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Let $B_t:=B(t)$. By the Itô formula $$f(B_1)-f(B_0)=\int_0^1 f'(B_t)\,dB_t+\frac12\,\int_0^1 f''(B_t)\,dt$$ with $f(b):=\int_0^b e^{a^2}da$ (with $\int_0^b:=-\int_b^0$ for $b<0$), we have \begin{equation*} X_1=\int_0^1 f'(B_t)\,dB_t =f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt =f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt. \tag{1} \end{equation*} From here, it is not hard to see that $EX_1$ does not exist.

Indeed, consider the event \begin{equation*} A:=\{B_u\in[b,b+1/b],B_1-B_u<-2/b,M_u>-b\}, \end{equation*} where $b\to\infty$, \begin{equation*} u:=1-1/b^2, \end{equation*} \begin{equation*} M_u:=\min_{t\in[u,1]}(B_t-B_u). \end{equation*} Note that on the event $A$ we have $B_u\ge b$, $B_1<b-1/b$ and $B_t>0$ for all $t\in[u,1]$. Therefore and because (by the l'Hospital rule) $f(b)\sim e^{b^2}/(2b)$, it follows from (1) that on $A$ \begin{align*} X_1&\le\frac{e^{(b-1/b)^2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &=\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &\le\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u (\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}\,dt, \end{align*} where $B_t^u:=B_t-\frac tu\,B_u$; for the latter, displayed inequality, we recall that $B_u\ge b$ on $A$ and use the fact that $se^{s^2}$ is increasing in real $s$.

The Brownian bridge $(B_t^u)_{t\in[0,u]}$ is a zero-mean (Gaussian) process independent of $(B_u,B_1-B_u,M_u)$; so, $(B_t^u)_{t\in[0,u]}$ is independent of the event $A$. So, \begin{equation*} EX_1\,1_A\le\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u E(\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}\,dt\Big)\,P(A). \tag{2} \end{equation*} Because the distribution of $B_t^u$ is symmetric and $(a+b) e^{(a+b)^2}+(a-b)e^{(a-b)^2}$ is convex in real $b$ for each real $a\ge0$, we can use Jensen's inequality to get \begin{align*} &2E(a+B_t^u) \exp\{(a+B_t^u)^2\} \\ & =E[(a+B_t^u) \exp\{(a+B_t^u)^2\}+(a-B_t^u) \exp\{(a-B_t^u)^2\}] \\ & \ge2ae^{a^2} \end{align*} for $a\ge0$. So, by (2), \begin{align*} EX_1\,1_A&\le\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u \tfrac tu\,b \exp\{(\tfrac tu\,b)^2\}\,dt\Big)\,P(A) \\ &=\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\frac u{2b} \, (e^{b^2}-1)\Big)\,P(A) \\ &=-e^{b^2(1+o(1))}\,P(A). \tag{3} \end{align*} On the other hand, \begin{align*} P(A)&=P(B_u\in[b,b+1/b])P(B_1-B_u<-2/b,M_u>-b) \\ &\ge P(B_u\in[b,b+1/b])[P(B_1-B_u<-2/b)-P(M_u\le-b)] \\ &=e^{-b^2/(2+o(1))}[P(B_1<-2)-o(1)]=e^{-b^2/(2+o(1))}. \end{align*} Thus, \begin{align*} EX_1\,1_A&\le-e^{b^2(1+o(1))}\,e^{-b^2/(2+o(1))}=-e^{b^2/(2+o(1))}\to-\infty, \end{align*} which shows that indeed $EX_1$ does not exist.

Let $B_t:=B(t)$. By the Itô formula $$f(B_1)-f(B_0)=\int_0^1 f'(B_t)\,dB_t+\frac12\,\int_0^1 f''(B_t)\,dt$$ with $f(b):=\int_0^b e^{a^2}da$ (with $\int_0^b:=-\int_b^0$ for $b<0$), we have \begin{equation*} X_1=\int_0^1 f'(B_t)\,dB_t =f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt =f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt. \tag{1} \end{equation*} From here, it is not hard to see that $EX_1$ does not exist.

Indeed, consider the event \begin{equation*} A:=\{B_u\in[b,b+1/b],B_1-B_u<-2/b,M_u>-b\}, \end{equation*} where $b\to\infty$, \begin{equation*} u:=1-1/b^2, \end{equation*} \begin{equation*} M_u:=\min_{t\in[u,1]}(B_t-B_u). \end{equation*} Note that on the event $A$ we have $B_u\ge b$, $B_1<b-1/b$ and $B_t>0$ for all $t\in[u,1]$. Therefore and because (by the l'Hospital rule) $f(b)\sim e^{b^2}/(2b)$, it follows from (1) that on $A$ \begin{align*} X_1&\le\frac{e^{(b-1/b)^2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &=\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &\le\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u (\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}\,dt, \end{align*} where $B_t^u:=B_t-\frac tu\,B_u$; for the latter, displayed inequality, we recall that $B_u\ge b$ on $A$ and use the fact that $se^{s^2}$ is increasing in real $s$.

The Brownian bridge $(B_t^u)_{t\in[0,u]}$ is a zero-mean (Gaussian) process independent of $(B_u,B_1-B_u,M_u)$; so, $(B_t^u)_{t\in[0,u]}$ is independent of the event $A$. Introduce also the event $$C_x:=\{\max_{t\in[0,u]}|B_t^u|\le x\}$$ for real $x>0$, which allows us to use the Fubini theorem to get \begin{align*} EX_1\,1_{A\cap C_x}\le\Big(&\frac{e^{b^2-2}}{(2+o(1))b}P(C_x) \\ &-\int_0^u E(\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}1_{C_x}\,dt\Big)\,P(A). \tag{2} \end{align*} Because $g_a(b):=(a+b) e^{(a+b)^2}+(a-b)e^{(a-b)^2}$ is convex and even in real $b$ for each real $a\ge0$, we have $g_a(b)\ge g_a(0)=2ae^{a^2}$. Therefore and because the distribution of the Brownian bridge $(B_t^u)_{t\in[0,u]}$ is symmetric,
\begin{align*} E(a+B_t^u) \exp\{(a+B_t^u)^2\}1_{C_x} =\tfrac12\,Eg_a(B_t^u)\,1_{C_x} \ge ae^{a^2}\,P(C_x) \end{align*} for $a\ge0$. So, by (2), \begin{align*} EX_1\,1_{A\cap C_x}&\le\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u \tfrac tu\,b \exp\{(\tfrac tu\,b)^2\}\,dt\Big)\,P(C_x)P(A) \\ &=\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\frac u{2b} \, (e^{b^2}-1)\Big)\,P(C_x)P(A) \\ &=-e^{b^2(1+o(1))}\,P(C_x)P(A). \end{align*}

On the other hand, letting now $x\to\infty$, we have $P(C_x)\to1$ and, still with $b\to\infty$, \begin{align*} P(A)&=P(B_u\in[b,b+1/b])P(B_1-B_u<-2/b,M_u>-b) \\ &\ge P(B_u\in[b,b+1/b])[P(B_1-B_u<-2/b)-P(M_u\le-b)] \\ &=e^{-b^2/(2+o(1))}[P(B_1<-2)-o(1)]=e^{-b^2/(2+o(1))}. \end{align*} Thus, \begin{align*} EX_1\,1_{A\cap C_x}&\le-e^{b^2(1+o(1))}\,(1-o(1))\,e^{-b^2/(2+o(1))}=-e^{b^2/(2+o(1))}\to-\infty, \end{align*} which shows that indeed $EX_1$ does not exist.

Let $B_t:=B(t)$. By the Itô formula $$f(B_1)-f(B_0)=\int_0^1 f'(B_t)\,dB_t+\frac12\,\int_0^1 f''(B_t)\,dt$$ with $f(b):=\int_0^b e^{a^2}da$ (with $\int_0^b:=-\int_b^0$ for $b<0$), we have $$X_1=\int_0^1 f'(B_t)\,dB_t =f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt =f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt.$$ From here, it should be not too hard to see that $EX_1$ does not exist.

One way to do this may be to recall that $B_1$ and the Brownian bridge $(Y_t):=(B_t-tB_1)$ are independent and write this for a real $c>0$: $$EX_1\,I\{Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1)\}\le P(Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1))\ Q,$$ $$Q:=EF(B_1),$$ $$F(b):=f(b)-\int_0^1 (tb+c\sqrt{(1-t)t})\, \exp((tb+c\sqrt{(1-t)t})^2)\,dt. $$ Then $P(Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1))$ should be $>0$ and hopefully $Q=-\infty$.

Here is the graph $\{(b,\ln(-F(b))/b^2)\colon1\le b\le10\}$ for $c=1$, which strongly suggests that $F(b)=-e^{b^2(1+o(1))}$ as $b\to\infty$, which would indeed imply $Q=-\infty$.

Let $B_t:=B(t)$. By the Itô formula $$f(B_1)-f(B_0)=\int_0^1 f'(B_t)\,dB_t+\frac12\,\int_0^1 f''(B_t)\,dt$$ with $f(b):=\int_0^b e^{a^2}da$ (with $\int_0^b:=-\int_b^0$ for $b<0$), we have \begin{equation*} X_1=\int_0^1 f'(B_t)\,dB_t =f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt =f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt. \tag{1} \end{equation*} From here, it is not hard to see that $EX_1$ does not exist.

Indeed, consider the event \begin{equation*} A:=\{B_u\in[b,b+1/b],B_1-B_u<-2/b,M_u>-b\}, \end{equation*} where $b\to\infty$, \begin{equation*} u:=1-1/b^2, \end{equation*} \begin{equation*} M_u:=\min_{t\in[u,1]}(B_t-B_u). \end{equation*} Note that on the event $A$ we have $B_u\ge b$, $B_1<b-1/b$ and $B_t>0$ for all $t\in[u,1]$. Therefore and because (by the l'Hospital rule) $f(b)\sim e^{b^2}/(2b)$, it follows from (1) that on $A$ \begin{align*} X_1&\le\frac{e^{(b-1/b)^2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &=\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &\le\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u (\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}\,dt, \end{align*} where $B_t^u:=B_t-\frac tu\,B_u$; for the latter, displayed inequality, we recall that $B_u\ge b$ on $A$ and use the fact that $se^{s^2}$ is increasing in real $s$.

The Brownian bridge $(B_t^u)_{t\in[0,u]}$ is a zero-mean (Gaussian) process independent of $(B_u,B_1-B_u,M_u)$; so, $(B_t^u)_{t\in[0,u]}$ is independent of the event $A$. So, \begin{equation*} EX_1\,1_A\le\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u E(\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}\,dt\Big)\,P(A). \tag{2} \end{equation*} Because the distribution of $B_t^u$ is symmetric and $(a+b) e^{(a+b)^2}+(a-b)e^{(a-b)^2}$ is convex in real $b$ for each real $a\ge0$, we can use Jensen's inequality to get \begin{align*} &2E(a+B_t^u) \exp\{(a+B_t^u)^2\} \\ & =E[(a+B_t^u) \exp\{(a+B_t^u)^2\}+(a-B_t^u) \exp\{(a-B_t^u)^2\}] \\ & \ge2ae^{a^2} \end{align*} for $a\ge0$. So, by (2), \begin{align*} EX_1\,1_A&\le\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u \tfrac tu\,b \exp\{(\tfrac tu\,b)^2\}\,dt\Big)\,P(A) \\ &=\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\frac u{2b} \, (e^{b^2}-1)\Big)\,P(A) \\ &=-e^{b^2(1+o(1))}\,P(A). \tag{3} \end{align*} On the other hand, \begin{align*} P(A)&=P(B_u\in[b,b+1/b])P(B_1-B_u<-2/b,M_u>-b) \\ &\ge P(B_u\in[b,b+1/b])[P(B_1-B_u<-2/b)-P(M_u\le-b)] \\ &=e^{-b^2/(2+o(1))}[P(B_1<-2)-o(1)]=e^{-b^2/(2+o(1))}. \end{align*} Thus, \begin{align*} EX_1\,1_A&\le-e^{b^2(1+o(1))}\,e^{-b^2/(2+o(1))}=-e^{b^2/(2+o(1))}\to-\infty, \end{align*} which shows that indeed $EX_1$ does not exist.

Let $B_t:=B(t)$. By the Itô formula $$f(B_1)-f(B_0)=\int_0^1 f'(B_t)\,dB_t+\frac12\,\int_0^1 f''(B_t)\,dt$$ with $f(b):=\int_0^b e^{a^2}da$ (with $\int_0^b:=-\int_b^0$ for $b<0$), we have $$X_1=\int_0^1 f'(B_t)\,dB_t =f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt =f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt.$$ From here, it should be not too hard to see that $EX_1$ does not exist.

One way to do this may be to recall that $B_1$ and the Brownian bridge $(Y_t):=(B_t-tB_1)$ are independent and write this for a real $c>0$: $$EX_1\,I\{Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1)\}\le P(Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1))\ Q,$$ $$Q:=E\Big(f(B_1)-\int_0^1 (tB_1+c\sqrt{(1-t)t})\, \exp((tB_1+c\sqrt{(1-t)t})^2)\,dt\Big).$$ Then $P(Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1))$ should be $>0$ and hopefully $Q=-\infty$.

Let $B_t:=B(t)$. By the Itô formula $$f(B_1)-f(B_0)=\int_0^1 f'(B_t)\,dB_t+\frac12\,\int_0^1 f''(B_t)\,dt$$ with $f(b):=\int_0^b e^{a^2}da$ (with $\int_0^b:=-\int_b^0$ for $b<0$), we have $$X_1=\int_0^1 f'(B_t)\,dB_t =f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt =f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt.$$ From here, it should be not too hard to see that $EX_1$ does not exist.

One way to do this may be to recall that $B_1$ and the Brownian bridge $(Y_t):=(B_t-tB_1)$ are independent and write this for a real $c>0$: $$EX_1\,I\{Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1)\}\le P(Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1))\ Q,$$ $$Q:=EF(B_1),$$ $$F(b):=f(b)-\int_0^1 (tb+c\sqrt{(1-t)t})\, \exp((tb+c\sqrt{(1-t)t})^2)\,dt. $$ Then $P(Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1))$ should be $>0$ and hopefully $Q=-\infty$.

Here is the graph $\{(b,\ln(-F(b))/b^2)\colon1\le b\le10\}$ for $c=1$, which strongly suggests that $F(b)=-e^{b^2(1+o(1))}$ as $b\to\infty$, which would indeed imply $Q=-\infty$.