Closed form of $\int_{\delta_1}^{\delta_2}(1+Ax)^{-L}x^{L}\exp\left(-Bx\right)dx$

Question

Is there a closed-form expression for the following definite integral? \begin{equation} \mathcal{I} = \int_{\delta_1}^{\delta_2}(1+Ax)^{-L}x^{L}\exp\left(-Bx\right)dx, \end{equation} where $A$, $B$, $\delta_1$, and $\delta_2$ are positive constant. $L$ is a positive integer.

I am facing problem due to finite limits $\delta_1$ and $\delta_2$. I know the answer when "$\delta_1 = 0$ and $\delta_2 = \infty$."

Answer 1

Since your endpoints are arbitrary, what you need is an antiderivative.
For convenience, apply scaling so that $A=1$. Let

$$ f_L(x) = \int \left( \dfrac{x}{1+x}\right)^L e^{-Bx}\; dx $$

The generating function is

$$ \eqalign{g(t,x) &= \sum_{L=0}^\infty f_L(x) t^L\cr &= \int \sum_{L=0}^\infty \left( \dfrac{xt}{1+x}\right)^L e^{-Bx}\; dx\cr &= \int \dfrac{1+x}{1+ (1-t)x} e^{-Bx}\; dx \cr &= \int \left(\dfrac{1}{1-t} - \dfrac{t}{(1-t)(1 + (1-t) x)}\right) e^{-Bx}\; dx\cr &= - \dfrac{e^{-Bx}}{B(1-t)} - \dfrac{t}{(1-t)^2} e^{B/(1-t)} \text{Ei}\left(-Bx - B/(1-t) \right) }$$

I don't know if there is a single closed form for all $L$, but you can get arbitrarily many terms by taking derivatives of this. Thus

$$ \eqalign{f_0(x) &= - \dfrac{- e^{-Bx}}{B} \cr f_1(x) &= - - \dfrac{- e^{-Bx}}{B} - e^B \text{Ei}(-B(x+1))\cr f_2(x) &= - \dfrac{e^{-Bx}}{B} - \dfrac{e^{-Bx}}{x+1} - (B+2) e^{B} \text{Ei}(-B(x+1))\cr f_3(x) &= - \dfrac{e^{-Bx}}{B} - \dfrac{B+6}{2(x+1)} e^{-Bx} + \dfrac{e^{-Bx}}{2(x+1)^2} - \dfrac{B^2 + B + 6}{2} e^B \text{Ei}(-B(x+1)) }$$

In general, it seems you have $$ f_L(x) = -\dfrac{e^{-Bx}}{B} + p_L(B) e^B \text{Ei}(-B(x+1)) + e^{-Bx} \sum_{j=1}^{L-1} \dfrac{q_{L,j}(B)}{(x+1)^j}$$

where $p_L$ and $q_{L,j}$ are polynomials of degree $L-1$ and $L-1-j$ respectively.

EDIT: Indeed, if you take this form and differentiate it, you find

$$ f'_L(x) = e^{-Bx} \left( 1 + \dfrac{p_L(B)}{x+1} - \sum_{j=1}^{L-1} \dfrac{B q_{L,j}(B) }{(x+1)^j}- \sum_{j=1}^{L-1} \dfrac{j q_{L,j}(B)}{(x+1)^{j+1}}\right) $$

which you want to be $$ e^{-Bx} \left( 1 - \dfrac{1}{1+x}\right)^L = e^{-Bx} \sum_{j=0}^L {L \choose j} \dfrac{(-1)^j}{(1+x)^j}$$

Equating coefficients of powers of $1+x$ and solving gives you the $p_L$ and $q_{L,j}$.

Answer 2

Use the identity $$ \frac{1}{(1+A x)^{L}}=\frac{1}{\Gamma(L)}\int_0^\infty dy\ y^{L-1}e^{-y}e^{-y A x} $$ and swapping the two integrals (using $\partial_B^{k} e^{-B x}=(-1)^k x^k e^{-B x}$) you get $$ \mathcal{I}=\frac{(-1)^L}{\Gamma(L)}\partial_B^{(L)}\int_0^\infty dy\ y^{L-1}e^{-y}\underbrace{\int_{\delta_1}^{\delta_2}dx\ e^{-x (B+A y)}}_{\frac{e^{\text{$\delta_1 $} (-(A y+B))}-e^{\text{$\delta_2 $} (-(A y+B))}}{A y+B}}\ , $$ therefore $$ \mathcal{I}=\frac{(-1)^L}{\Gamma(L)}\partial_B^{(L)}\left[\frac{e^{B/A} \Gamma (L) \left(\frac{A}{B}\right)^{-L} \left(\Gamma \left(1-L,B \left(\text{$\delta_1 $}+\frac{1}{A}\right)\right)-\Gamma \left(1-L,B \left(\text{$\delta_2 $}+\frac{1}{A}\right)\right)\right)}{B}\right]\ . $$