A confusion about the integral $\int_{0}^{1} x^l e^{-\lambda(1-x)} dx$

Question

A basic calculation shows: $$ \int_{0}^{1} x^l e^{-\lambda(1-x)} dx = \frac{l!}{(-\lambda)^{l+1}} \left( \sum\limits_{j=l+1}^{\infty} \frac{(-\lambda)^j}{j!}\right) $$ where $\lambda$ is a constant.

I have: $$ \int_{0}^{1} x^l e^{-\lambda(1-x)} dx = e^{-\lambda} \int_{0}^{1} x^l e^{\lambda x} dx = \frac{e^{-\lambda}}{\lambda^{l+1}} \int_{0}^{1} (\lambda x)^l e^{{\lambda x}} d(\lambda x) $$

$$ =\frac{e^{-\lambda}}{\lambda^{l+1}} \left[ \left( \sum\limits_{j=0}^{l}(-1)^j \frac{l!}{(l-j)!} (\lambda x)^{l-j} \right) e^{\lambda x}\right]_{0}^{1} $$

I can't obtain the equation. Can someone give me some suggestion to prove it? Thank you.

Answer 1

My first instinct looking at the expression was to turn it into Gamma density, so that's what I did

$$\int_0^1x^le^{-\lambda{(1-x)}}dx=e^{-\lambda}\frac{\Gamma(l+1)}{(-\lambda)^{l+1}}\int_0^1\frac{1}{\Gamma(l+1)}(-\lambda)^{l+1}x^{(l+1)-1}e^{-(-\lambda) x}dx$$

Then I realized there isn't a closed form for the cdf. Thankfully, Wikipedia came to the rescue with the following:

If $\alpha$ is a positive integer, then the cdf has the following series expansion $$F(x;\alpha,\beta) = 1-\sum_{i=0}^{\alpha-1} \frac{(\beta x)^i}{i!} e^{-\beta x} = e^{-\beta x} \sum_{i=\alpha}^{\infty} \frac{(\beta x)^i}{i!}$$

We plug in $x=1, \alpha=l+1, \beta=(-\lambda)$ to get

$$e^{-\lambda}\frac{\Gamma(l+1)}{(-\lambda)^{l+1}}e^{\lambda}\sum_{i=l+1}^{\infty}\frac{(-\lambda)^{i}}{i!}=\frac{l!}{(-\lambda)^{l+1}}\left( \sum_{i=l+1}^{\infty} \frac{(-\lambda)^i}{i!} \right)$$