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Determining the general form

For brevity, we define

$\displaystyle I_n(x) := \int_0^x t^n {\rm e}^t {\rm d}t
$

Integrating by parts,

$\displaystyle I_n(x) = \int_0^x t^n {\rm d}({\rm e}^t) = x^n {\rm e}^x - n \int_0^x t^{n-1} {\rm e}^t {\rm d}t =: x^n {\rm e}^x - n I_{n-1}(x)$ (1)

By direct integration,

$\displaystyle I_0(x) = {\rm e}^x - 1
$

Hence, by (1),
$\displaystyle I_1(x)$ $\displaystyle =$ $\displaystyle x {\rm e}^x - {\rm e}^x + 1$  
$\displaystyle I_2(x)$ $\displaystyle =$ $\displaystyle x^2 {\rm e}^x - 2 x {\rm e}^x + 2 {\rm e}^x - 2$  

Looking at the above results, we can make the conjecture

$\displaystyle I_n(x) = F_n(x) {\rm e}^x - (-1)^n n!$ (2)

where $ F_n(x)$ is a polynomial of degree $ n$ in $ x$. We can prove our conjecture by induction,
$\displaystyle I_{n+1}(x)$ $\displaystyle =$ $\displaystyle x^{n+1} {\rm e}^x - (n+1) I_n$  
  $\displaystyle =$ $\displaystyle x^{n+1} {\rm e}^x - (n+1) (F_n(x) {\rm e}^x - (-1)^n n!)$  
  $\displaystyle =$ $\displaystyle (x^{n+1} - (n+1) F_n(x)) {\rm e}^x + (-1)^n (n+1) n!$  
  $\displaystyle =$ $\displaystyle F_{n+1}(x) {\rm e}^x - (-1)^{n+1} (n+1)!$  
  $\displaystyle =$ $\displaystyle I_{n+1}(x)$  

Hence, our conjecture is proved and so $ I_n(x)$ has the general form as stated in (2).
next up previous
Next: Deducing Up: Finding Previous: Introduction
Alexander Frolkin 2001-06-02