How to evaluate the definite integral $\int_0^x t^{n-1}e^{-(a+bt)}dt$

Question

How to evaluate the following definite integral $\int_0^x t^{n-1}e^{-(a+bt)}dt$, where $n\in\mathbb{N}$ and $a,b>0$.

Answer 1

Let $y=bt$ then $t=\cfrac{y}{b}$ and $dt=\cfrac{dy}{b}$. \begin{align} \int_0^x t^{n-1}e^{-a}e^{-bt}\,dt &=e^{-a}\int_0^x \left(\cfrac{y}{b}\right)^{n-1}e^{-y}\,\cfrac{dy}{b}\\ &=\frac{e^{-a}}{b^n}\int_0^xy^{n-1}e^{-y}\,dy \end{align} Since we cannot use gamma function, the only way is using integration by parts to yield integral by reduction formula. Let $u=y^{n-1}$, $du=(n-1)y^{n-2}$, $dv=e^{-y}\,dy$, and $v=-e^{-y}$. Then \begin{align} \int_0^xy^{n-1}e^{-y}\,dy&=\left.-y^{n-1}e^{-y}\right|_0^x+(n-1)\int_0^xy^{n-2}e^{-y}\,dy\\ I_n&=-x^{n-1}e^{-x}+(n-1)I_{n-1} \end{align} Hence \begin{align} \int_0^x t^{n-1}e^{-(a+bt)}\,dt &=\frac{e^{-a}}{b^n}\left(-x^{n-1}e^{-x}+(n-1)I_{n-1}\right) \end{align}