convergence of integral function I am considering the function
$$F(x) = \int_0^\infty \frac{e^{-t}}{1+xt} \mathrm dt$$
and would like to show that $F(x) \to 0$ as $x\to\infty$ which, at least though my numerical investigations, appears to be true.
This function (specifically it's asymtotic expansion) seems well understood when $x<<1$ but I cannot find any material for $x>>1$.
Am I missing a simple proof of this, or alternatively does anyone have any resource I could read through proving this?
 A: With elementary steps, we can take this approach. Define
$$
F_M(x)=\int_0^M\frac{e^{-t}}{1+xt}dt,
$$
hence
$$
F_M(x){\le \int_0^M\frac{(1+t)^{-1}}{1+xt}dt
\\=
\int_0^M\frac{1}{(1+t)(1+xt)}dt
\\=
\frac{1}{x-1}\int_0^M\frac{x}{1+xt}-\frac{1}{1+t}dt
\\=
\frac{1}{x-1}\ln \frac{1+Mx}{1+M}
}
$$
therefore by tending $M$ to $\infty$
$$
F(x)\le \frac{\ln x}{x-1}
$$
where the upper bound tends to $0$ as $x\to\infty$ $\blacksquare$
A: It is possible to obtain a clean expansion that describes the behaviour of $F(x)$ for large $x$ completely. We can write $F(x)$ in terms of the exponential integral:
$$
F(x) = \int_0^{ + \infty } {\frac{{e^{ - t} }}{{1 + xt}}dt}  = \frac{1}{x}\int_0^{ + \infty } {\frac{{e^{ - s/x} }}{{1 + s}}ds}  = \frac{1}{x}e^{1/x} E_1 \!\left( {\frac{1}{x}} \right).
$$
Using http://dlmf.nist.gov/6.6.E3, we find
\begin{align*}
F(x) & = \frac{{e^{1/x} }}{x}\log x + \sum\limits_{n = 1}^\infty  {\frac{{\psi (n)}}{{(n - 1)!}}\frac{1}{{x^n }}} 
\\ & = \log x\sum\limits_{n = 1}^\infty  {\frac{1}{{(n - 1)!}}\frac{1}{{x^n }}}  + \sum\limits_{n = 1}^\infty  {\frac{{\psi (n)}}{{(n - 1)!}}\frac{1}{{x^n }}}
\end{align*}
for $|\arg x|<\pi$. Here $\psi$ denotes the logarithmic derivative of the gamma function. Note that
$$
\psi(1)=-\gamma, \quad\psi (n) =  - \gamma  + \sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} \qquad (n\geq 2),
$$
where $\gamma=0.577215\ldots$ is the Euler–Mascheroni constant.
A: We can get the closed form of $F(x)$ and its asymptotics for $x>>1$.
Thanks to @Maxim who pointed out a mistake in previous calculations.
$$F(x) = \int_0^\infty \frac{e^{-t}}{1+xt} \mathrm dt=\frac{1}{x}\int_0^\infty \frac{e^{-t}}{\frac{1}{x}+t} \mathrm dt=\alpha\int_0^\infty \frac{e^{-t}}{\alpha+t} \mathrm dt,\,\, \text {where  } \alpha=\frac{1}{x}$$
$$I(\alpha)=\alpha\int_0^\infty \frac{e^{-t}}{\alpha+t} \mathrm dt=\alpha e^\alpha\int_\alpha^\infty \frac{e^{-t}}{t} \mathrm dt=\alpha e^\alpha e^{-t}\log t|_\alpha^\infty+\alpha e^\alpha\int_\alpha^\infty e^{-t}\log t \,\mathrm dt$$
$$=-\alpha\log\alpha+\alpha e^\alpha\int_0^\infty e^{-t}\log t\,\mathrm dt-\alpha e^\alpha\int_0^\alpha e^{-t}\log t\,\mathrm dt$$
$$=-\alpha\log\alpha-\gamma\alpha e^\alpha-\alpha e^\alpha\int_0^\alpha \bigl(1-t+\frac{t^2}{2!}-+...\bigr)\log t\,\mathrm dt\,\,(\gamma \text { is Euler constant})$$
$$=-\alpha\log\alpha -\gamma\alpha e^\alpha+\alpha\log\alpha e^\alpha\sum_{n=1}^\infty(-1)^n\frac{\alpha^n}{n!}-\alpha e^\alpha\sum_{n=1}^\infty(-1)^n\frac{\alpha^n}{(n!)^2}$$
$$=-\alpha e^\alpha\log\alpha +\alpha e^\alpha(1-\gamma)-\alpha e^\alpha\sum_{n=0}^\infty(-1)^n\frac{\alpha^n}{(n!)^2}$$
$$I(\alpha)=-\alpha e^\alpha 
\log\alpha +\alpha e^\alpha(1-\gamma)-\alpha e^\alpha J_0(2\sqrt\alpha)$$
where $J_0$ is Bessel function of the first kind
$\,\,\bigl(J_0(t)=\sum_{n=0}^\infty(-1)^n\frac{t^{2n}}{(n!)^2}\bigr)$
For $\alpha<<1 \,\,\, I(\alpha)=-\alpha\log\alpha-\alpha\gamma+\alpha^2(1-\gamma)-\alpha^2\log\alpha+O(\alpha^3\log\alpha)$, and
$$F(x)=\frac{\log x}{x}-\frac{\gamma}{x}+\frac{\log x}{x^2}+\frac{1-\gamma}{x^2}+O\Bigl(\frac{\log x}{x^3}\Bigr),\,\,x>>1$$
