# Integrals with fractions in exponentials

I came across this integral in a research paper while trying to understand Bernstein's inequalities. I want to upper bound or evaluate this integral but it looks too complicated. The integral is: \begin{align} I=\int_0^\infty \exp \left ( \frac{-at^2}{bt+c} \right ) dt, \quad a,b,c >0. \end{align} Is there any procedure to upper bound the above integral? My attempt is to simplify the fraction so that I would get $$I= \int_0^\infty \exp \left ( \frac{-at}{b} + \frac{ac}{b} - \frac{ac^2}{b^3}\left(\frac{1}{t+c/b} \right) \right) dt.$$ But this doesn't seem too useful either. Any suggestions?

• Do you have any constraints on the values of $a, b$ and $c$? $a>0?$ – caverac May 7 '17 at 12:42
• @caverac: You can assume that they are all greater than zero. – pikachuchameleon May 7 '17 at 18:53

I don't know how strict a bound you need, but let $x>b$, then $$xt \ge bt+c\qquad\text{for}\qquad t\ge\frac{c}{x-b}\ .$$ Write $k:=\tfrac{c}{x-b}$. It follows that $$\frac{-at^2}{bt+c}\le-\frac{a}{x}t\qquad\text{for}\qquad t\ge k$$ so that we can bound \begin{align} \int_0^\infty\exp\left(\frac{-at^2}{bt+c}\right)dt\le&\ k+\int_k^\infty\exp\left(-\frac{a}{x}t\right)dt\\ =&\ \frac{c}{x-b}+\frac{a}{x}\exp\left(-\frac{a}{x}\frac{c}{x-b}\right)\ , \end{align} where in the first line we bounded the function by $1$ for $t\in[0,k]$. Let's call $$F(x) := \frac{c}{x-b}+\frac{a}{x}\exp\left(-\frac{a}{x}\frac{c}{x-b}\right)\ .$$ Now if you want to optimize the bound obtained this way you have to find the minimum of $F(x)$ for $x>b$, which looks like a difficult problem to me (and will probably involve the Lambert $W$ function).

$\int_0^\infty e^\frac{-at^2}{bt+c}~dt$

$=\int_c^\infty e^\frac{-a\left(\frac{t-c}{b}\right)^2}{t}~d\left(\dfrac{t-c}{b}\right)$

$=\dfrac{e^\frac{2ac}{b^2}}{b}\int_c^\infty e^{-\frac{at}{b^2}-\frac{ac^2}{b^2t}}~dt$

$=\dfrac{e^\frac{2ac}{b^2}}{b}\int_1^\infty e^{-\frac{act}{b^2}-\frac{ac^2}{b^2ct}}~d(ct)$

$=\dfrac{ce^\frac{2ac}{b^2}}{b}\int_1^\infty e^{-\frac{act}{b^2}-\frac{ac}{b^2t}}~dt$

$=\dfrac{ce^\frac{2ac}{b^2}}{b}\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$

Consider $\int_0^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$ ,

$\int_0^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$

$=\int_0^1e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt+\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$

$=\int_\infty^1e^{-\frac{ac}{b^2}\left(\frac{1}{t}+t\right)}~d\left(\dfrac{1}{t}\right)+\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$

$=\int_1^\infty\dfrac{1}{t^2}e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt+\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$

$=\int_1^\infty\left(1+\dfrac{1}{t^2}\right)e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$

$=2\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt-\int_1^\infty\left(1-\dfrac{1}{t^2}\right)e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$

$\therefore\dfrac{ce^\frac{2ac}{b^2}}{b}\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$

$=\dfrac{ce^\frac{2ac}{b^2}}{2b}\int_0^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt+\dfrac{ce^\frac{2ac}{b^2}}{2b}\int_1^\infty\left(1-\dfrac{1}{t^2}\right)e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$

$=\dfrac{ce^\frac{2ac}{b^2}}{2b}K_1\left(\dfrac{2ac}{b^2}\right)+\dfrac{ce^\frac{2ac}{b^2}}{2b}\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~d\left(t+\dfrac{1}{t}\right)$

$=\dfrac{ce^\frac{2ac}{b^2}}{2b}K_1\left(\dfrac{2ac}{b^2}\right)-\dfrac{ce^\frac{2ac}{b^2}}{2b}\left[\dfrac{b^2}{ac}e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}\right]_1^\infty$

$=\dfrac{ce^\frac{2ac}{b^2}}{2b}K_1\left(\dfrac{2ac}{b^2}\right)+\dfrac{b}{2a}$