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I have to calculate Integral with Romberg's method of function representing density of Erlang's Distribution: $f(x)=\frac{\lambda^{k} x^{k-1} \mathrm{e}^{-\lambda x}}{(k-1) !}, \quad x \in \mathbb{R}$

For $t>0$

$G(t)=\int_{0}^{t} f(x) d x$

My friend gave me solution(that has to be proven by induction) :

$G_{k}(t)=1-e^{-\lambda t}\left(1+\frac{\lambda t}{1 !}+\cdots+\frac{(\lambda t)^{k-1}}{(k-1) !}\right)$,

But I have no idea how to get that.

user3231387
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    $f(x)$ is a density for $x>0$. See https://math.stackexchange.com/q/467341/321264, https://math.stackexchange.com/q/1960133/321264 and their linked posts. – StubbornAtom Apr 14 '21 at 21:11
  • Your friend's solution is $Pr(X\ge k)$ with $X\sim \text{Poisson}(\lambda t)$. This has not much to do with the problem. Btw, it should be $G(t)=\int_0^t f(x) dx$. Check on Wikipedia to see the cdf of a gamma distribution, hopefully it has some closed form xD – Vons Apr 15 '21 at 00:09

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