Prove some formulas for integrals of e-t tn

Prove the following integral formulas.

$\displaystyle{ \int_0^x e^{-t} t \, dt = e^{-x} \big( e^x - 1 - x \big)}$ .
$\displaystyle{ \int_0^x e^{-t} t^2 \, dt = 2! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} \right)}$ .
$\displaystyle{ \int_0^x e^{-t} t^3 \, dt = 3! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!} \right)}$ .
Guess and prove a general formula based on parts (a) – (c).

Proof. We use integration by parts with
$\begin{align*} u &= t & du &= dt \\ dv &= e^{-t} \, dt & v &= -e^{-t}. \end{align*}$

This gives us

$\begin{align*} \int_0^x e^{-t} t \, dt &= -te^{-t} \Bigr \rvert_0^x + \int_0^x e^{-t} \, dt \\[9pt] &= -xe^{-x} + (-e^{-t})\Bigr \rvert_0^x \\[9pt] &= -xe^{-x} - e^{-x} + 1 \\[9pt] &= e^{-x} \left( \frac{1}{e^{-x}} - 1 - x \right) \\[9pt] &= e^{-x} \left( e^x - 1 - x \right). \qquad \blacksquare \end{align*}$
Proof. We use integration by parts and the result of part (a). Let
$\begin{align*} u &= t^2 & du &= 2t dt \\ dv &= e^{-t} dt & v &= -e^{-t}. \end{align*}$

This gives us

$\begin{align*} \int_0^x e^{-t} t^2 \, dt &= -t^2 e^{-t} \Bigr \rvert_0^x + 2 \int_0^x e^{-t} t \, dt \\[9pt] &= -x^2 e^{-x} + 2 \left( e^{-x} ( e^x - 1 - x ) \right) \\[9pt] &= 2! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} \right). \qquad \blacksquare \end{align*}$
Proof. Again, we use integration by parts, and this time part (b). Let
$\begin{align*} u &= t^3 & du &= 3t^2 \, dt \\ dv &= e^{-t} \, dt & v &= -e^{-t}. \end{align*}$

This gives us

$\begin{align*} \int_0^x e^{-t} t^3 \, dt &= -t^3 e^{-t} \Bigr \rvert_0^x + 3 \int_0^x e^{-t} t^2 \, dt \\[9pt] &= -x^3 e^{-x} + 3 \left( 2! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} \right) \right) \\[9pt] &= 3! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!} \right). \qquad \blacksquare \end{align*}$
Claim:
$\int_0^x e^{-t} t^n \, dt = n! e^{-x} \left( e^x - \sum_{k=0}^n \frac{x^k}{k!} \right).$

Proof. The proof is by induction. We have already established the case $n = 1$ (and $n = 2$ and $n = 3$ ). Assume then that the formula holds for some positive integer $m$ . We then consider the integral

$\int_0^x e^{-t} t^{m+1} \, dt.$

Integrating by parts, we let

$\begin{align*} u &= t^{m+1} & du &= (m+1)t^m \, dt \\ dv &= e^{-t} \, dt & v &= -e^{-t}. \end{align*}$

Therefore, integrating by parts and using the induction hypothesis we have,

$\begin{align*} \int_0^x e^{-t} t^{m+1} \, dt &= -t^{m+1} e^{-t} \Bigr \rvert_0^x + (m+1) \int_0^x e^{-t} t^m \, dt \\[9pt] &= -x^{m+1}e^{-x} + (m+1) \left( m! e^{-x} \left( e^x - \sum_{k=0}^m \frac{x^k}{k!} \right) \right) \\[9pt] &= -x^{m+1} e^{-x} (m+1)! e^{-x} \left( e^x - \sum_{k=0}^m \frac{x^k}{k!} \right) \\[9pt] &= (m+1)! e^{-x} \left( e^x - \sum_{k=0}^m \frac{x^k}{k!} - \frac{x^{m+1}}{(m+1)!} \right) \\[9pt] &= (m+1)! e^{-x} \left(e^x - \sum_{k=0}^{m+1} \frac{x^k}{k!} \right). \end{align*}$

Therefore, the formula holds for $m+1$ , and hence, for all positive integers $n. \qquad \blacksquare$