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Extending the result

Above we have shown that

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

We now want to find

$\displaystyle J := \int_0^x t^n {\rm e}^{at} {\rm d}t

All we need to do to get this into the form of $ I_n$ is to substitute $ u = at$ and $ {\rm d}u = a{\rm d}t$,

$\displaystyle J = \int_0^{ax} \frac{u^n}{a^n} {\rm e}^u \frac{{\rm d}u}{a} =
\frac{1}{a^{n+1}} \int_0^{ax} u^n {\rm e}^u {\rm d}u =:
\frac{1}{a^{n+1}} I_n(ax)

$\displaystyle \frac{1}{a^{n+1}} I_n(ax)$ $\displaystyle =$ $\displaystyle {\rm e}^{ax} \sum_{r=0}^n (-1)^{n-r}
\frac{n!}{r!} \frac{a^r x^r}{a^{n+1}} - \frac{(-1)^n n!}{a^{n+1}}$  
  $\displaystyle =$ $\displaystyle {\rm e}^{ax} \sum_{r=0}^n (-1)^{n-r} \frac{n!}{r!}
\frac{x^r}{a^{n-r+1}} - \frac{(-1)^n n!}{a^{n+1}}$  

Alexander Frolkin 2001-06-02