I am trying to work out this integral. If there is no closed form, can you think of any approximations to it?
$$\int_0^T e^{a (T-x)} (T-x)^{1+m+n} x^k \, _1F_1\Big(1+n;2+m+n;a (x-T) \Big) \, dx$$
Thanks for your help!
update:
I found a couple of relations that helps. That is the closest I got.
At first step, I can use following relation from [1]:
$$_1F_1\Big(1+n;2+m+n;a (x-T)\Big)=e^{a(x-T)} \, _1F_1\Big(1+m;2+m+n;a(T-x)\Big)$$
so I can write the integral as: $$\int_0^T (T-x)^{1+m+n} x^k \, _1F_1\Big(1+m;2+m+n;a(T-x) \Big)dx$$ with change of variable $y=T-x$, I can rewrite as: $$\int_0^T y^{1+m+n} (T-y)^k \, _1F_1\Big(1+m;2+m+n;ay \Big) \, dy$$ A relation I found on [1] has closed form solution to following integral: $$\int_0^T y^{1+m+n} (T-y)^k \, _1F_1\Big(1+m;2+m+n;\color{red}y \Big)dy=T^{2+m+n+k}\frac{\Gamma(2+m+n)\Gamma(k+1)}{\Gamma(3+m+n+k)} \, _1F_1\Big(1+m;3+m+n+k;T\Big)$$ since $a>1$ would the answer be an upper bound or a good approximation to my integral? How far this would be?
Thanks!
[1] Gradshteyn and Ryzhik, "Table of Integrals, Series, and Products", 2007 (page 821 and 1023)