Ratio of two integrals of exponential functions I would like to somehow simplify the following expression:
$$\frac{\int_0^\infty e^{f\left(x\right)+x}\;dx}{\int_0^\infty e^{f\left(x\right)}\;dx}$$
If someone could give me a hint if this is at all possible, or if there is a (symbolic) approximation for this expression, it would be very much appreciated.
Based on the book by Severini, T. - Elements of Distribution Theory (page 279) Laplace's method might be a good way of approximating this integral, but I am not familiar with it, so before delving into it, it would be nice to get some confirmation.
Thank you very much in advance!
 A: First of all we note that if $f$ is even function then $
{\int_0^\infty  {e^{f\left( x \right)} dx} }
\to \infty $ (i.e, diverges). So we will assume that $
g\left( x \right) = e^{f\left( x \right)}$, assuming that $\int_0^\infty  {\left| {g\left( x \right)} \right| dx}  < \infty.$ If we want to use "Your book hint", we should assume that $g$ is of exponential order i.e., $
\left| {g\left( x \right)} \right| \le Me^{\alpha x} ,x \in \left( {0,\infty } \right)$, for some constants $M>0$ and $\alpha \in \mathbb{R}$.
Now, 
let $x=-st$, $s<0$ and $t>0$, which gives $dx=-s\, dt$. Then
\begin{align}
\frac{{\int_0^\infty  {e^{x + f\left( x \right)} } dx}}{{\int_0^\infty  {e^{f\left( x \right)} dx} }} = \frac{{\int_0^\infty  {e^x g\left( x \right)dx} }}{{\int_0^\infty  {g\left( x \right)dx} }} = \frac{{ - s\int_0^\infty  {e^{ - st} g\left( { - st} \right)dt} }}{{ - s\int_0^\infty  {g\left( { - st} \right)dt} }} = \frac{{\int_0^\infty  {e^{ - st} g\left( { - st} \right)dt} }}{{\int_0^\infty  {g\left( { - st} \right)dt} }}
\end{align}
which imples
\begin{align}
\left| {\frac{{\int_0^\infty  {e^{ - st} g\left( { - st} \right)dt} }}{{\int_0^\infty  {g\left( { - st} \right)dt} }}} \right| = \frac{{\left| {\int_0^\infty  {e^{ - st} g\left( { - st} \right)dt} } \right|}}{{\left| {\int_0^\infty  {g\left( { - st} \right)dt} } \right|}} &\le \frac{{\int_0^\infty  {\left| {e^{ - st} } \right|\left| {g\left( { - st} \right)} \right|dt} }}{{\int_0^\infty  {\left| {g\left( { - st} \right)} \right|dt} }} 
\\
&\le \mathop {\sup }\limits_{t \in \left( {0,\infty } \right)} \left| {e^{ - st} } \right| \cdot \frac{{\int_0^\infty  {\left| {g\left( { - st} \right)} \right|dt} }}{{\int_0^\infty  {\left| {g\left( { - st} \right)} \right|dt} }} \le 1
\end{align}
