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Using the substitution $t=s\sqrt y$, we have \begin{equation*} f(y,L)=\sqrt y\,g(L/\sqrt y,\sqrt y), \end{equation*} where \begin{equation*} g(r,x):=\int_r^\infty e^{-x(s+1/s)}\,ds, \end{equation*} $r>0$, $x>0$. So, it is enough to study $g(r,x)$. Using now the substitution $t=s+1/s$ and letting \begin{equation*} F(t):=\frac t{\sqrt{t^2-4}}\quad\text{and}\quad t_r:=r+1/r, \end{equation*} we have \begin{equation*} g(r,x)= \left\{ \begin{alignedat}{2} &\frac12\,\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\ge1, \\ &\frac12\,\int_2^{t_r} e^{-xt}(F(t)-1)\,dt && \\ &+\frac12\,\int_2^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\in(0,1]. \end{alignedat} \right. \tag{10}\label{10} \end{equation*}

Integrating by parts twice, we get \begin{equation*} \begin{aligned} I(r,x)&:=\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt \\ &=\frac1x\,e^{-xt_r}(F(t_r)+1)+\frac1x\,\int_{t_r}^\infty e^{-xt}F'(t)\,dt \\ &=\frac1x\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ &+\frac1{x^2}\,\int_{t_r}^\infty e^{-xt}F''(t)\,dt. \end{aligned} \tag{20}\label{20} \end{equation*} Note now that for $t>2$ \begin{equation*} F'(t)=-\frac{4}{\left(t^2-4\right)^{3/2}}<0,\quad F''(t)=\frac{12 t}{\left(t^2-4\right)^{5/2}}>0. \end{equation*} So, from our integration by parts we get \begin{equation*} 2l(r,x)<I(r,x)<2u(r,x) \end{equation*} $r\ge1$ and $x>0$, where \begin{equation*} u(r,x):=\frac1{2x}\,e^{-xt_r}(F(t_r)+1)=\frac1{2x}e^{-x(r+1/r)}\frac{2r^2}{r^2-1} \end{equation*} and \begin{equation*} l(r,x):=\frac1{2x}\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ = \frac1{2x}e^{-x(r+1/r)}\Big(\frac{2r^2}{r^2-1}-\frac1x\frac{4 r^3}{\left(r^2-1\right)^3}\Big). \end{equation*} So, by \eqref{10}, \begin{equation*} l(r,x)<g(r,x)<u(r,x) \quad\text{for }r>1. \end{equation*}

In the original version of the OP, particular interest was expressed in large values of $y$ and $L$, so that $x=\sqrt y$ is large. Accordingly, we have $l(r,x)\sim u(r,x)$ and hence \begin{equation*} l(r,x)\sim g(r,x)\sim u(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\ge r_0$, for each real $r_0>1$, which means that the upper and lower bounds $u(r,x)$ and $l(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Consider now the case $r\in(0,1)$. Then, by \eqref{10}, \begin{equation*} 2g(r,x)=I(r,x)+2J(r,x), \end{equation*} where $I(r,x)$ is as in \eqref{20} and \begin{equation*} J(r,x):=\int_2^{t_r} e^{-xt}F(t)\,dt=e^{-2x}\int_0^{z_r} e^{-xz}G(z)\,dz, \tag{30}\label{30} \end{equation*} where $z_r:=t_r-2=r+1/r-2=(1/\sqrt r-\sqrt r)^2$ and \begin{equation*} \frac1{\sqrt z}<G(z):=\frac1{\sqrt z}\frac{2+z}{\sqrt{4+z}}<\frac{1+3z/8}{\sqrt z} \tag{40}\label{40} \end{equation*} for real $z>0$.

Using these lower and upper bounds on $G(z)$ together with \eqref{30}, we get \begin{equation*} l_1(r,x)<J(r,x)<u_1(r,x), \end{equation*} where \begin{equation*} l_1(r,x):=e^{-2x}\frac{\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{\sqrt{x}}, \end{equation*} \begin{equation*} u_1(r,x):=l_1(r,x) +e^{-2x}\frac{3\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{16x^{3/2}} -e^{-2x}\frac{3 \sqrt{z_r} e^{-x z_r}}{8x}. \end{equation*} So, \begin{equation*} L(r,x):=l(r,x)+l_1(r,x)<g(r,x)<U(r,x):=u(r,x)+u_1(r,x) \quad\text{for }r\in(0,1) \end{equation*} and \begin{equation*} L(r,x)\sim g(r,x)\sim U(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Using the inequality $G(z)<1+1/\sqrt z$ instead of the second inequality in \eqref{40}, we get \begin{equation*} J(r,x)<\tilde u_1(r,x):=l_1(r,x)+ \frac{1-e^{-xz_r}}x. \end{equation*} So, \begin{equation*} L(r,x)<g(r,x)<\tilde U(r,x):=u(r,x)+\tilde u_1(r,x) \quad\text{for }r\in(0,1) \end{equation*} and \begin{equation*} L(r,x)\sim g(r,x)\sim \tilde U(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $\tilde U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$. The upper bound $\tilde U(r,x)$ on $g(r,x)$ is simpler than $U(r,x)$, but not as accurate for large $x$.

Using the substitution $t=s\sqrt y$, we have \begin{equation*} f(y,L)=\sqrt y\,g(L/\sqrt y,\sqrt y), \end{equation*} where \begin{equation*} g(r,x):=\int_r^\infty e^{-x(s+1/s)}\,ds, \end{equation*} $r>0$, $x>0$. So, it is enough to study $g(r,x)$. Using now the substitution $t=s+1/s$ and letting \begin{equation*} F(t):=\frac t{\sqrt{t^2-4}}\quad\text{and}\quad t_r:=r+1/r, \end{equation*} we have \begin{equation*} g(r,x)= \left\{ \begin{alignedat}{2} &\frac12\,\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\ge1, \\ &\frac12\,\int_2^{t_r} e^{-xt}(F(t)-1)\,dt && \\ &+\frac12\,\int_2^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\in(0,1]. \end{alignedat} \right. \tag{10}\label{10} \end{equation*}

Integrating by parts twice, we get \begin{equation*} \begin{aligned} I(r,x)&:=\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt \\ &=\frac1x\,e^{-xt_r}(F(t_r)+1)+\frac1x\,\int_{t_r}^\infty e^{-xt}F'(t)\,dt \\ &=\frac1x\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ &+\frac1{x^2}\,\int_{t_r}^\infty e^{-xt}F''(t)\,dt. \end{aligned} \tag{20}\label{20} \end{equation*} Note now that for $t>2$ \begin{equation*} F'(t)=-\frac{4}{\left(t^2-4\right)^{3/2}}<0,\quad F''(t)=\frac{12 t}{\left(t^2-4\right)^{5/2}}>0. \end{equation*} So, from our integration by parts we get \begin{equation*} 2l(r,x)<I(r,x)<2u(r,x) \end{equation*} $r\ge1$ and $x>0$, where \begin{equation*} u(r,x):=\frac1{2x}\,e^{-xt_r}(F(t_r)+1)=\frac1{2x}e^{-x(r+1/r)}\frac{2r^2}{r^2-1} \end{equation*} and \begin{equation*} l(r,x):=\frac1{2x}\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ = \frac1{2x}e^{-x(r+1/r)}\Big(\frac{2r^2}{r^2-1}-\frac1x\frac{4 r^3}{\left(r^2-1\right)^3}\Big). \end{equation*} So, by \eqref{10}, \begin{equation*} l(r,x)<g(r,x)<u(r,x) \quad\text{for }r>1. \end{equation*}

In the original version of the OP, particular interest was expressed in large values of $y$ and $L$, so that $x=\sqrt y$ is large. Accordingly, we have $l(r,x)\sim u(r,x)$ and hence \begin{equation*} l(r,x)\sim g(r,x)\sim u(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\ge r_0$, for each real $r_0>1$, which means that the upper and lower bounds $u(r,x)$ and $l(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Consider now the case $r\in(0,1)$. Then, by \eqref{10}, \begin{equation*} 2g(r,x)=I(r,x)+2J(r,x), \end{equation*} where $I(r,x)$ is as in \eqref{20} and \begin{equation*} J(r,x):=\int_2^{t_r} e^{-xt}F(t)\,dt=e^{-2x}\int_0^{z_r} e^{-xz}G(z)\,dz, \tag{30}\label{30} \end{equation*} where $z_r:=t_r-2=r+1/r-2=(1/\sqrt r-\sqrt r)^2$ and \begin{equation*} \frac1{\sqrt z}<G(z):=\frac1{\sqrt z}\frac{2+z}{\sqrt{4+z}}<\frac{1+3z/8}{\sqrt z} \tag{40}\label{40} \end{equation*} for real $z>0$.

Using these lower and upper bounds on $G(z)$ together with \eqref{30}, we get \begin{equation*} l_1(r,x)<J(r,x)<u_1(r,x), \end{equation*} where \begin{equation*} l_1(r,x):=e^{-2x}\frac{\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{\sqrt{x}}, \end{equation*} \begin{equation*} u_1(r,x):=l_1(r,x) +e^{-2x}\frac{3\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{16x^{3/2}} -e^{-2x}\frac{3 \sqrt{z_r} e^{-x z_r}}{8x}. \end{equation*} So, \begin{equation*} L(r,x):=l(r,x)+l_1(r,x)<g(r,x)<U(r,x):=u(r,x)+u_1(r,x) \quad\text{for }r\in(0,1) \end{equation*} and \begin{equation*} L(r,x)\sim g(r,x)\sim U(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Using the inequality $G(z)<1+1/\sqrt z$ instead of the second inequality in \eqref{40}, we get \begin{equation*} J(r,x)<\tilde u_1(r,x):=l_1(r,x)+ e^{-2x}\frac{1-e^{-xz_r}}x. \end{equation*} So, \begin{equation*} L(r,x)<g(r,x)<\tilde U(r,x):=u(r,x)+\tilde u_1(r,x) \quad\text{for }r\in(0,1) \end{equation*} and \begin{equation*} L(r,x)\sim g(r,x)\sim \tilde U(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $\tilde U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$. The upper bound $\tilde U(r,x)$ on $g(r,x)$ is simpler than $U(r,x)$, but not as accurate for large $x$.

Using the substitution $t=s\sqrt y$, we have \begin{equation*} f(y,L)=\sqrt y\,g(L/\sqrt y,\sqrt y), \end{equation*} where \begin{equation*} g(r,x):=\int_r^\infty e^{-x(s+1/s)}\,ds, \end{equation*} $r>0$, $x>0$. So, it is enough to study $g(r,x)$. Using now the substitution $t=s+1/s$ and letting \begin{equation*} F(t):=\frac t{\sqrt{t^2-4}}\quad\text{and}\quad t_r:=r+1/r, \end{equation*} we have \begin{equation*} g(r,x)= \left\{ \begin{alignedat}{2} &\frac12\,\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\ge1, \\ &\frac12\,\int_2^{t_r} e^{-xt}(F(t)-1)\,dt && \\ &+\frac12\,\int_2^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\in(0,1]. \end{alignedat} \right. \tag{10}\label{10} \end{equation*}

Integrating by parts twice, we get \begin{equation*} \begin{aligned} I(r,x)&:=\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt \\ &=\frac1x\,e^{-xt_r}(F(t_r)+1)+\frac1x\,\int_{t_r}^\infty e^{-xt}F'(t)\,dt \\ &=\frac1x\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ &+\frac1{x^2}\,\int_{t_r}^\infty e^{-xt}F''(t)\,dt. \end{aligned} \tag{20}\label{20} \end{equation*} Note now that for $t>2$ \begin{equation*} F'(t)=-\frac{4}{\left(t^2-4\right)^{3/2}}<0,\quad F''(t)=\frac{12 t}{\left(t^2-4\right)^{5/2}}>0. \end{equation*} So, from our integration by parts we get \begin{equation*} 2l(r,x)<I(r,x)<2u(r,x) \end{equation*} $r\ge1$ and $x>0$, where \begin{equation*} u(r,x):=\frac1{2x}\,e^{-xt_r}(F(t_r)+1)=\frac1{2x}e^{-x(r+1/r)}\frac{2r^2}{r^2-1} \end{equation*} and \begin{equation*} l(r,x):=\frac1{2x}\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ = \frac1{2x}e^{-x(r+1/r)}\Big(\frac{2r^2}{r^2-1}-\frac1x\frac{4 r^3}{\left(r^2-1\right)^3}\Big). \end{equation*} So, by \eqref{10}, \begin{equation*} l(r,x)<g(r,x)<u(r,x) \quad\text{for }r>1. \end{equation*}

In the original version of the OP, particular interest was expressed in large values of $y$ and $L$, so that $x=\sqrt y$ is large. Accordingly, we have $l(r,x)\sim u(r,x)$ and hence \begin{equation*} l(r,x)\sim g(r,x)\sim u(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\ge r_0$, for each real $r_0>1$, which means that the upper and lower bounds $u(r,x)$ and $l(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Consider now the case $r\in(0,1)$. Then, by \eqref{10}, \begin{equation*} 2g(r,x)=I(r,x)+J(r,x), \end{equation*} where $I(r,x)$ is as in \eqref{20} and \begin{equation*} J(r,x):=\int_2^{t_r} e^{-xt}F(t)\,dt=e^{-2x}\int_0^{z_r} e^{-xz}G(z)\,dz, \tag{30}\label{30} \end{equation*} where $z_r:=t_r-2=r+1/r-2=(1/\sqrt r-\sqrt r)^2$ and \begin{equation*} \frac1{\sqrt z}<G(z):=\frac1{\sqrt z}\frac{2+z}{\sqrt{4+z}}<\frac{1+3z/8}{\sqrt z} \end{equation*} for real $z>0$.

Using these lower and upper bounds on $G(z)$ together with \eqref{30}, we get \begin{equation*} 2l_1(r,x)<J(r,x)<2u_1(r,x), \end{equation*} where \begin{equation*} l_1(r,x):=e^{-2x}\frac{\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{2\sqrt{x}}, \end{equation*} \begin{equation*} u_1(r,x):=l_1(r,x) +e^{-2x}\frac{3\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{32x^{3/2}} -e^{-2x}\frac{3 \sqrt{z_r} e^{-x z_r}}{16x}. \end{equation*} So, \begin{equation*} L(r,x):=l(r,x)+l_1(r,x)<g(r,x)<U(r,x):=u(r,x)+u_1(r,x), \quad\text{for }r\in(0,1) \end{equation*} and \begin{equation*} L(r,x)\sim g(r,x)\sim U(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Using the substitution $t=s\sqrt y$, we have \begin{equation*} f(y,L)=\sqrt y\,g(L/\sqrt y,\sqrt y), \end{equation*} where \begin{equation*} g(r,x):=\int_r^\infty e^{-x(s+1/s)}\,ds, \end{equation*} $r>0$, $x>0$. So, it is enough to study $g(r,x)$. Using now the substitution $t=s+1/s$ and letting \begin{equation*} F(t):=\frac t{\sqrt{t^2-4}}\quad\text{and}\quad t_r:=r+1/r, \end{equation*} we have \begin{equation*} g(r,x)= \left\{ \begin{alignedat}{2} &\frac12\,\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\ge1, \\ &\frac12\,\int_2^{t_r} e^{-xt}(F(t)-1)\,dt && \\ &+\frac12\,\int_2^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\in(0,1]. \end{alignedat} \right. \tag{10}\label{10} \end{equation*}

Integrating by parts twice, we get \begin{equation*} \begin{aligned} I(r,x)&:=\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt \\ &=\frac1x\,e^{-xt_r}(F(t_r)+1)+\frac1x\,\int_{t_r}^\infty e^{-xt}F'(t)\,dt \\ &=\frac1x\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ &+\frac1{x^2}\,\int_{t_r}^\infty e^{-xt}F''(t)\,dt. \end{aligned} \tag{20}\label{20} \end{equation*} Note now that for $t>2$ \begin{equation*} F'(t)=-\frac{4}{\left(t^2-4\right)^{3/2}}<0,\quad F''(t)=\frac{12 t}{\left(t^2-4\right)^{5/2}}>0. \end{equation*} So, from our integration by parts we get \begin{equation*} 2l(r,x)<I(r,x)<2u(r,x) \end{equation*} $r\ge1$ and $x>0$, where \begin{equation*} u(r,x):=\frac1{2x}\,e^{-xt_r}(F(t_r)+1)=\frac1{2x}e^{-x(r+1/r)}\frac{2r^2}{r^2-1} \end{equation*} and \begin{equation*} l(r,x):=\frac1{2x}\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ = \frac1{2x}e^{-x(r+1/r)}\Big(\frac{2r^2}{r^2-1}-\frac1x\frac{4 r^3}{\left(r^2-1\right)^3}\Big). \end{equation*} So, by \eqref{10}, \begin{equation*} l(r,x)<g(r,x)<u(r,x) \quad\text{for }r>1. \end{equation*}

In the original version of the OP, particular interest was expressed in large values of $y$ and $L$, so that $x=\sqrt y$ is large. Accordingly, we have $l(r,x)\sim u(r,x)$ and hence \begin{equation*} l(r,x)\sim g(r,x)\sim u(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\ge r_0$, for each real $r_0>1$, which means that the upper and lower bounds $u(r,x)$ and $l(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Consider now the case $r\in(0,1)$. Then, by \eqref{10}, \begin{equation*} 2g(r,x)=I(r,x)+2J(r,x), \end{equation*} where $I(r,x)$ is as in \eqref{20} and \begin{equation*} J(r,x):=\int_2^{t_r} e^{-xt}F(t)\,dt=e^{-2x}\int_0^{z_r} e^{-xz}G(z)\,dz, \tag{30}\label{30} \end{equation*} where $z_r:=t_r-2=r+1/r-2=(1/\sqrt r-\sqrt r)^2$ and \begin{equation*} \frac1{\sqrt z}<G(z):=\frac1{\sqrt z}\frac{2+z}{\sqrt{4+z}}<\frac{1+3z/8}{\sqrt z} \tag{40}\label{40} \end{equation*} for real $z>0$.

Using these lower and upper bounds on $G(z)$ together with \eqref{30}, we get \begin{equation*} l_1(r,x)<J(r,x)<u_1(r,x), \end{equation*} where \begin{equation*} l_1(r,x):=e^{-2x}\frac{\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{\sqrt{x}}, \end{equation*} \begin{equation*} u_1(r,x):=l_1(r,x) +e^{-2x}\frac{3\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{16x^{3/2}} -e^{-2x}\frac{3 \sqrt{z_r} e^{-x z_r}}{8x}. \end{equation*} So, \begin{equation*} L(r,x):=l(r,x)+l_1(r,x)<g(r,x)<U(r,x):=u(r,x)+u_1(r,x) \quad\text{for }r\in(0,1) \end{equation*} and \begin{equation*} L(r,x)\sim g(r,x)\sim U(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Using the inequality $G(z)<1+1/\sqrt z$ instead of the second inequality in \eqref{40}, we get \begin{equation*} J(r,x)<\tilde u_1(r,x):=l_1(r,x)+ \frac{1-e^{-xz_r}}x. \end{equation*} So, \begin{equation*} L(r,x)<g(r,x)<\tilde U(r,x):=u(r,x)+\tilde u_1(r,x) \quad\text{for }r\in(0,1) \end{equation*} and \begin{equation*} L(r,x)\sim g(r,x)\sim \tilde U(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $\tilde U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$. The upper bound $\tilde U(r,x)$ on $g(r,x)$ is simpler than $U(r,x)$, but not as accurate for large $x$.

Using the substitution $t=s\sqrt y$, we have \begin{equation*} f(y,L)=\sqrt y\,g(L/\sqrt y,\sqrt y), \end{equation*} where \begin{equation*} g(r,x):=\int_r^\infty e^{-x(s+1/s)}\,ds, \end{equation*} $r>0$, $x>0$. So, it is enough to study $g(r,x)$. Using now the substitution $t=s+1/s$ and letting \begin{equation*} F(t):=\frac t{\sqrt{t^2-4}}\quad\text{and}\quad t_r:=r+1/r, \end{equation*} we have \begin{equation*} g(r,x)= \left\{ \begin{alignedat}{2} &\frac12\,\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\ge1, \\ &\frac12\,\int_2^{t_r} e^{-xt}(F(t)-1)\,dt && \\ &+\frac12\,\int_2^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\in(0,1]. \end{alignedat} \right. \tag{10}\label{10} \end{equation*}

Integrating by parts twice, we get \begin{equation*} \begin{aligned} I(r,x)&:=\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt \\ &=\frac1x\,e^{-xt_r}(F(t_r)+1)+\frac1x\,\int_{t_r}^\infty e^{-xt}F'(t)\,dt \\ &=\frac1x\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ &+\frac1{x^2}\,\int_{t_r}^\infty e^{-xt}F''(t)\,dt. \end{aligned} \tag{20}\label{20} \end{equation*} Note now that for $t>2$ \begin{equation*} F'(t)=-\frac{4}{\left(t^2-4\right)^{3/2}}<0,\quad F''(t)=\frac{12 t}{\left(t^2-4\right)^{5/2}}>0. \end{equation*} So, from our integration by parts we get \begin{equation*} 2l(r,x)<I(r,x)<2u(r,x) \end{equation*} $r\ge1$ and $x>0$, where \begin{equation*} u(r,x):=\frac1{2x}\,e^{-xt_r}(F(t_r)+1)=\frac1{2x}e^{-x(r+1/r)}\frac{2r^2}{r^2-1} \end{equation*} and \begin{equation*} l(r,x):=\frac1{2x}\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ = \frac1{2x}e^{-x(r+1/r)}\Big(\frac{2r^2}{r^2-1}-\frac1x\frac{4 r^3}{\left(r^2-1\right)^3}\Big). \end{equation*} So, by \eqref{10}, \begin{equation*} l(r,x)<g(r,x)<u(r,x) \quad\text{for }r>1. \end{equation*}

In the original version of the OP, particular interest was expressed in large values of $y$ and $L$, so that $x=\sqrt y$ is large. Accordingly, we have $l(r,x)\sim u(r,x)$ and hence \begin{equation*} l(r,x)\sim g(r,x)\sim u(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\ge r_0$, for each real $r_0>1$, which means that the upper and lower bounds $u(r,x)$ and $l(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Consider now the case $r\in(0,1)$. Then, by \eqref{10}, \begin{equation*} 2g(r,x)=I(r,x)+J(r,x), \end{equation*} where $I(r,x)$ is as in \eqref{20} and \begin{equation*} J(r,x):=\int_2^{t_r} e^{-xt}F(t)\,dt=e^{-2x}\int_0^{z_r} e^{-xz}G(z)\,dz, \tag{30}\label{30} \end{equation*} where $z_r:=t_r-2=r+1/r-2=(1/\sqrt r-\sqrt r)^2$ and \begin{equation*} \frac1{\sqrt z}<G(z):=\frac1{\sqrt z}\frac{2+z}{\sqrt{4+z}}<\frac{1+3z/8}{\sqrt z} \end{equation*} for real $z>0$.

Using these lower and upper bounds on $G(z)$ together with \eqref{30}, we get \begin{equation*} 2l_1(r,x)<J(r,x)<2u_1(r,x), \end{equation*} where \begin{equation*} l_1(r,x):=\frac{\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{2\sqrt{x}}, \end{equation*} \begin{equation*} u_1(r,x):=l_1(r,x) +\frac{3\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{32x^{3/2}} -\frac{3 \sqrt{z_r} e^{-x z_r}}{16x}. \end{equation*} So, \begin{equation*} L(r,x):=l(r,x)+l_1(r,x)<g(r,x)<U(r,x):=u(r,x)+u_1(r,x), \quad\text{for }r\in(0,1) \end{equation*} and \begin{equation*} L(r,x)\sim g(r,x)\sim U(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Using the substitution $t=s\sqrt y$, we have \begin{equation*} f(y,L)=\sqrt y\,g(L/\sqrt y,\sqrt y), \end{equation*} where \begin{equation*} g(r,x):=\int_r^\infty e^{-x(s+1/s)}\,ds, \end{equation*} $r>0$, $x>0$. So, it is enough to study $g(r,x)$. Using now the substitution $t=s+1/s$ and letting \begin{equation*} F(t):=\frac t{\sqrt{t^2-4}}\quad\text{and}\quad t_r:=r+1/r, \end{equation*} we have \begin{equation*} g(r,x)= \left\{ \begin{alignedat}{2} &\frac12\,\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\ge1, \\ &\frac12\,\int_2^{t_r} e^{-xt}(F(t)-1)\,dt && \\ &+\frac12\,\int_2^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\in(0,1]. \end{alignedat} \right. \tag{10}\label{10} \end{equation*}

Integrating by parts twice, we get \begin{equation*} \begin{aligned} I(r,x)&:=\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt \\ &=\frac1x\,e^{-xt_r}(F(t_r)+1)+\frac1x\,\int_{t_r}^\infty e^{-xt}F'(t)\,dt \\ &=\frac1x\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ &+\frac1{x^2}\,\int_{t_r}^\infty e^{-xt}F''(t)\,dt. \end{aligned} \tag{20}\label{20} \end{equation*} Note now that for $t>2$ \begin{equation*} F'(t)=-\frac{4}{\left(t^2-4\right)^{3/2}}<0,\quad F''(t)=\frac{12 t}{\left(t^2-4\right)^{5/2}}>0. \end{equation*} So, from our integration by parts we get \begin{equation*} 2l(r,x)<I(r,x)<2u(r,x) \end{equation*} $r\ge1$ and $x>0$, where \begin{equation*} u(r,x):=\frac1{2x}\,e^{-xt_r}(F(t_r)+1)=\frac1{2x}e^{-x(r+1/r)}\frac{2r^2}{r^2-1} \end{equation*} and \begin{equation*} l(r,x):=\frac1{2x}\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\ = \frac1{2x}e^{-x(r+1/r)}\Big(\frac{2r^2}{r^2-1}-\frac1x\frac{4 r^3}{\left(r^2-1\right)^3}\Big). \end{equation*} So, by \eqref{10}, \begin{equation*} l(r,x)<g(r,x)<u(r,x) \quad\text{for }r>1. \end{equation*}

In the original version of the OP, particular interest was expressed in large values of $y$ and $L$, so that $x=\sqrt y$ is large. Accordingly, we have $l(r,x)\sim u(r,x)$ and hence \begin{equation*} l(r,x)\sim g(r,x)\sim u(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\ge r_0$, for each real $r_0>1$, which means that the upper and lower bounds $u(r,x)$ and $l(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.

Consider now the case $r\in(0,1)$. Then, by \eqref{10}, \begin{equation*} 2g(r,x)=I(r,x)+J(r,x), \end{equation*} where $I(r,x)$ is as in \eqref{20} and \begin{equation*} J(r,x):=\int_2^{t_r} e^{-xt}F(t)\,dt=e^{-2x}\int_0^{z_r} e^{-xz}G(z)\,dz, \tag{30}\label{30} \end{equation*} where $z_r:=t_r-2=r+1/r-2=(1/\sqrt r-\sqrt r)^2$ and \begin{equation*} \frac1{\sqrt z}<G(z):=\frac1{\sqrt z}\frac{2+z}{\sqrt{4+z}}<\frac{1+3z/8}{\sqrt z} \end{equation*} for real $z>0$.

Using these lower and upper bounds on $G(z)$ together with \eqref{30}, we get \begin{equation*} 2l_1(r,x)<J(r,x)<2u_1(r,x), \end{equation*} where \begin{equation*} l_1(r,x):=e^{-2x}\frac{\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{2\sqrt{x}}, \end{equation*} \begin{equation*} u_1(r,x):=l_1(r,x) +e^{-2x}\frac{3\sqrt{\pi }\, \text{erf}\left(\sqrt{x z_r}\right)}{32x^{3/2}} -e^{-2x}\frac{3 \sqrt{z_r} e^{-x z_r}}{16x}. \end{equation*} So, \begin{equation*} L(r,x):=l(r,x)+l_1(r,x)<g(r,x)<U(r,x):=u(r,x)+u_1(r,x), \quad\text{for }r\in(0,1) \end{equation*} and \begin{equation*} L(r,x)\sim g(r,x)\sim U(u,x) \end{equation*} as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.