Consider the integral
$$\mathcal{I}=\int_0^t\left(\frac{Ae^{-\lambda t}-Ae^{-\lambda s}}{2}\right)^{2m+1}e^{-\epsilon(t-s)}ds,\tag{1}$$
for constants $A,\lambda,\epsilon,t\in\mathbb{R}$ and $m\in\mathbb{Z}^+$.
The intention of evaluating $\mathcal{I}$, is to find
\begin{equation}
\begin{split}
\mathcal{S}&=\int_0^tJ_1\left(Ae^{-\lambda t}-Ae^{-\lambda s}\right)e^{-\epsilon(t-s)}ds,
\\&=\int_0^t\sum_{m=0}^{\infty}\frac{
(-1)^m}{m!(m+1)!}\left(\frac{Ae^{-\lambda t}-Ae^{-\lambda s}}{2}\right)^{2m+1}e^{-\epsilon(t-s)}ds,\\
&=\sum_{m=0}^\infty \frac{
(-1)^m}{m!(m+1)!}\mathcal{I}_m
\end{split}\end{equation}
where $J_1$ denotes the first order Bessel function of first kind. Does a closed form exist for $\mathcal{S}$ (or Taylor series)? Could an asymptotic bound simplify things?
According to Mathematica, we have
\begin{equation}
\mathcal{I}=\frac{(A-Ae^{-\lambda t})^2 (Ae^{-\lambda t}-A)^{2m} 2^{-2m-1}e^{-\epsilon t}\cdot{}_2F_1\left(1,\frac{\epsilon+\lambda}{\lambda};\frac{\epsilon}{\lambda}-2m,e^{-\lambda t}\right)}{A\left(\epsilon-\lambda(2m+1)\right)}.
\end{equation}
From this, is it possible to prove that
$2\mathcal{S}\sim A\lambda^2\kappa e^{-\lambda t}-A\lambda e^{-\lambda t}$ as $t\rightarrow\infty$, for some $\kappa$?
Best Answer
If $\lambda>0$ and $\lambda>\epsilon$ one has $$\lim_{t\rightarrow\infty}e^{\epsilon t}\mathcal{S}=\int_0^\infty J_1\left(-Ae^{-\lambda s}\right)e^{\epsilon s}ds$$ $$\qquad=\frac{A/2}{\epsilon-\lambda} \, _1F_2\left(\frac{1}{2}-\frac{\epsilon}{2 \lambda};2,\frac{3}{2}-\frac{\epsilon}{2 \lambda};-A^2/4\right).$$ The asymptotics is different for other ranges of $\lambda,\epsilon$, for example, for $\lambda,\epsilon<0$ one has ${\cal S}\rightarrow -e^{-\epsilon t}J_1(Ae^{-\lambda t})/\epsilon$.