Let $y>0$, $L>0$. Has the (special?) function given by $$f(y,L) = \int_{L}^\infty e^{- t - y/t} dt$$ been studied? Are there precise, simple bounds?

Let me try to attempt to reinvent the wheel, briefly:

The integrand clearly takes its maximum when $t = \sqrt{y}$, and so we have two kinds of behavior depending on whether $L\leq \sqrt{y}$ or $L>\sqrt{y}$. If $L\leq \sqrt{y}$, then the main contribution is that of $t\sim \sqrt{y}$: around there, $-t-y/t = -2\sqrt{y} - \frac{1}{\sqrt{y}} (t - \sqrt{y})^2 + \dotsc$, and so the main term of $f(y,L)$ should be $$e^{-2 \sqrt{y}} \int_{-\infty}^\infty e^{-\frac{x^2}{\sqrt{y}}} dx = \sqrt{\pi} y^{1/4} e^{-2 \sqrt{y}}.$$ If $L>\sqrt{y}$, then the main contribution comes from $t$ close to $L$. Around there, $$-t-y/t = -L - \frac{y}{L} - \left(1-\frac{y}{L^2}\right) (t-L) + \dotsb,$$ and so the main term of $f(y,L)$ should be $$e^{-L-\frac{y}{L}} \int_0^\infty e^{-\left(1-\frac{y}{L^2}\right) x} dx = \frac{e^{-L - \frac{y}{L}}}{1 - \frac{y}{L^2}}.$$

Motivation: this integral comes up in smoothed variants of the prime number theorem.

Let $y>0$, $L>0$. Has the (special?) function given by $$f(y,L) = \int_{L}^\infty e^{- t - y/t} \, dt$$ been studied? Are there precise, simple bounds?

Let me try to attempt to reinvent the wheel, briefly:

The integrand clearly takes its maximum when $t = \sqrt{y}$, and so we have two kinds of behavior depending on whether $L\leq \sqrt{y}$ or $L>\sqrt{y}$. If $L\leq \sqrt{y}$, then the main contribution is that of $t\sim \sqrt{y}$: around there, $-t-y/t = -2\sqrt{y} - \frac{1}{\sqrt{y}} (t - \sqrt{y})^2 + \dotsc$, and so the main term of $f(y,L)$ should be $$e^{-2 \sqrt{y}} \int_{-\infty}^\infty e^{-\frac{x^2}{\sqrt{y}}} \, dx = \sqrt{\pi} y^{1/4} e^{-2 \sqrt{y}}.$$ If $L>\sqrt{y}$, then the main contribution comes from $t$ close to $L$. Around there, $$-t-y/t = -L - \frac{y}{L} - \left(1-\frac{y}{L^2}\right) (t-L) + \dotsb,$$ and so the main term of $f(y,L)$ should be $$e^{-L-\frac{y}{L}} \int_0^\infty e^{-\left(1-\frac{y}{L^2}\right) x} dx = \frac{e^{-L - \frac{y}{L}}}{1 - \frac{y}{L^2}}.$$

Motivation: this integral comes up in smoothed variants of the prime number theorem.

Let $y>0$, $L>0$. Has the (special?) function given by $$f(y,L) = \int_{L}^\infty e^{- t - y/t} dt$$ been studied? Does it have, say, a series development for large $y$ and $L$? Are there precise, simple bounds?

Let me try to attempt to reinvent the wheel, briefly:

The integrand clearly takes its maximum when $t = \sqrt{y}$, and so we have two kinds of behavior depending on whether $L\leq \sqrt{y}$ or $L>\sqrt{y}$. If $L\leq \sqrt{y}$, then the main contribution is that of $t\sim \sqrt{y}$: around there, $-t-y/t = -2\sqrt{y} - \frac{1}{\sqrt{y}} (t - \sqrt{y})^2 + \dotsc$, and so the main term of $f(y,L)$ should be $$e^{-2 \sqrt{y}} \int_{-\infty}^\infty e^{-\frac{x^2}{\sqrt{y}}} dx = \sqrt{\pi} y^{1/4} e^{-2 \sqrt{y}}.$$ If $L>\sqrt{y}$, then the main contribution comes from $t$ close to $L$. There, $-t-y/t = -L - \frac{y}{L} - \left(1-\frac{y}{L^2}\right) (t-L)$, and so the main term of $f(y,L)$ should be $$e^{-L-\frac{y}{L}} \int_0^\infty e^{-\left(1-\frac{y}{L^2}\right) x} dx = \frac{e^{-L - \frac{y}{L}}}{1 - \frac{y}{L^2}}.$$

Motivation: this integral comes up in smoothed variants of the prime number theorem.

Let $y>0$, $L>0$. Has the (special?) function given by $$f(y,L) = \int_{L}^\infty e^{- t - y/t} dt$$ been studied? Are there precise, simple bounds?

Let me try to attempt to reinvent the wheel, briefly:

The integrand clearly takes its maximum when $t = \sqrt{y}$, and so we have two kinds of behavior depending on whether $L\leq \sqrt{y}$ or $L>\sqrt{y}$. If $L\leq \sqrt{y}$, then the main contribution is that of $t\sim \sqrt{y}$: around there, $-t-y/t = -2\sqrt{y} - \frac{1}{\sqrt{y}} (t - \sqrt{y})^2 + \dotsc$, and so the main term of $f(y,L)$ should be $$e^{-2 \sqrt{y}} \int_{-\infty}^\infty e^{-\frac{x^2}{\sqrt{y}}} dx = \sqrt{\pi} y^{1/4} e^{-2 \sqrt{y}}.$$ If $L>\sqrt{y}$, then the main contribution comes from $t$ close to $L$. Around there, $$-t-y/t = -L - \frac{y}{L} - \left(1-\frac{y}{L^2}\right) (t-L) + \dotsb,$$ and so the main term of $f(y,L)$ should be $$e^{-L-\frac{y}{L}} \int_0^\infty e^{-\left(1-\frac{y}{L^2}\right) x} dx = \frac{e^{-L - \frac{y}{L}}}{1 - \frac{y}{L^2}}.$$

Motivation: this integral comes up in smoothed variants of the prime number theorem.