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I am trying to integrate exponent distribution which is defined as: $$ f(x)=\begin{cases}\lambda e^{-\lambda x} & x\ge 0 \\ 0 & x\le 0 \end{cases} $$ Now i remember that integrating improper integral would go something like this ? $$F(x)= \int_{-\infty}^{x}f(s)ds=\lim_{n\rightarrow (-\infty)} \int_{n}^{x}f(s)ds=\lim_{n\rightarrow (-\infty)}\left[\begin{matrix} \end{matrix} F(x)-F(n) \right] $$ I end up with result as following: $$ F(x)= \int_{-\infty}^{x}\lambda e^{-\lambda s}ds=\lim_{n\rightarrow(-\infty)}(-e^{-\lambda x})-(-e^{-\lambda n}) $$ Now the end result should be following:

$$F(x)=\int_{-\infty}^x f(s) \, ds=\begin{cases} 1-e^{-\lambda x}, & x> 0. \\ 0, & x\le 0\end{cases}$$

This hints that $$ \lim_{n \rightarrow (-\infty)}-(-e^{\lambda n})=1 $$ but i cant understand why.

Now if someon could provide some answer to this that would be greatly appreciated.

Thanks,

Tuki

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1 Answer 1

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Note that in this case $f(x)$ vanishes for $x<0$, so that

$$ F(x) = \int_{-\infty}^{x}{\rm d}x~f(x) $$

can be evaluated in two steps

$x<0$

$$ \int_{-\infty}^x{\rm d}x~ f(x) = \int_{-\infty}^x{\rm d}x~ 0 = 0 $$

$x\ge0$

$$ \int_{-\infty}^x{\rm d}x~ f(x) = \int_{0}^x{\rm d}x~ \lambda e^{-\lambda x} = 1 - e^{-\lambda x} $$

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  • $\begingroup$ $$ e^{-\lambda*0}=e^{0}=1 $$ i think this makes sense now. $\endgroup$
    – Tuki
    Oct 22, 2017 at 10:53
  • $\begingroup$ also is the $dx$ intentionally before $f(x)$ or is this accidental ? $\endgroup$
    – Tuki
    Oct 22, 2017 at 10:57
  • $\begingroup$ @Tuki It is a matter of notation $$ \int {\rm d}s~g(s) = \int g(s){\rm d}s $$ $\endgroup$
    – caverac
    Oct 22, 2017 at 10:58
  • $\begingroup$ Sorry i didn't know this was possible notation. $\endgroup$
    – Tuki
    Oct 22, 2017 at 10:59

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