solution of another definite integral Does the following integral converge or not?
\begin{align}
&&
\sum_{k=0}^{\infty} (-\varphi)^k \binom{\frac1\varphi+k}{k}\int_{-\infty}^\infty\beta x^n e^{-\beta x(k+1)}dx&&
\end{align}
where $e,\beta$ and $\varphi>0$ and $n$ is a positive integer.
Well I just simplified the following integral

and I got the above integral.
 A: Consider $\int_{-\infty}^\infty e^{-\beta x}\left(1+\varphi e^{-bx}\right)^{-\frac{1}{\varphi}-1}~dx$ ,
$\int_{-\infty}^\infty e^{-\beta x}\left(1+\varphi e^{-bx}\right)^{-\frac{1}{\varphi}-1}~dx$
$=\int_\infty^0x^\frac{\beta}{b}~(1+\varphi x)^{-\frac{1}{\varphi}-1}~d\left(-\dfrac{\ln x}{b}\right)$
$=\dfrac{1}{b}\int_0^\infty x^{\frac{\beta}{b}-1}(1+\varphi x)^{-\frac{1}{\varphi}-1}~dx$
$=\dfrac{1}{b}\int_0^\infty\left(\dfrac{x}{\varphi}\right)^{\frac{\beta}{b}-1}(1+x)^{-\frac{1}{\varphi}-1}~d\left(\dfrac{x}{\varphi}\right)$
$=\dfrac{1}{\varphi^\frac{\beta}{b}~b}\int_0^\infty x^{\frac{\beta}{b}-1}(1+x)^{-\frac{1}{\varphi}-1}~dx$
$=\dfrac{1}{\varphi^\frac{\beta}{b}~b}B\left(\dfrac{\beta}{b},\dfrac{1}{\varphi}-\dfrac{\beta}{b}+1\right)$
$\therefore\int_{-\infty}^\infty\beta x^ne^{-\beta x}\left(1+\varphi e^{-bx}\right)^{-\frac{1}{\varphi}-1}~dx=(-1)^n\beta\dfrac{d^n}{d\beta^n}\left(\dfrac{1}{\varphi^\frac{\beta}{b}~b}B\left(\dfrac{\beta}{b},\dfrac{1}{\varphi}-\dfrac{\beta}{b}+1\right)\right)(b=\beta)$
A: $$
 \beta \frac{x^n e^{-\beta x}}
{\left(1+\varphi e^{-\beta x}\right)^{\frac{1}{\varphi}+1}}\approx \frac{\beta}{\varphi^{1+\varphi^{-1}}} x^n \exp(\frac{\beta}{\varphi}x) \quad\mbox{ for } x\to-\infty, 
$$
so it is clearly convergent at this end, and
$$
 \beta \frac{x^n e^{-\beta x}}
{\left(1+\varphi e^{-\beta x}\right)^{\frac{1}{\varphi}+1}}\approx \beta x^n \exp(- \beta x) \quad\mbox{ for } x\to \infty, 
$$
so it is also clearly convergent at the other end, so it is indeed convergent.
