How to integrate the function $\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}$

Question

How to integrate the following family of functions? I really do not have too many ideas. The context of this is to find moment generating functions but the context is not too important here.

$$I_n:=\int_{0}^{\infty}n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}\,dx$$

I wanted to try this inductively/iteratively but I got nowhere (by the way I don't have to do this iteratively). I should also stress that $\lambda>t$ so that I think $I_n$ converges for each $n$. Could someone please given me a hint or two? I know how to do $n=1$ btw.

Many thanks!

Answer 1

lets collect the terms first: $$f(x)=\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}=\lambda e^{-(t-\lambda)x}(1-e^{-\lambda x})^{n-1}$$ $$I_n=n\lambda\int_0^\infty (e^{-x})^{t-\lambda}(1-(e^{-x})^\lambda)^{n-1}dx\tag{1}$$ now if we make the substitution $u=e^{-x}\Rightarrow dx=-\frac{du}{u}$ and our integral becomes: $$I_n=n\lambda\int_0^1 u^{t-\lambda-1}(1-u^\lambda)^{n-1}$$ now make the substitution $v=u^\lambda\Rightarrow dv=\lambda u^{\lambda-1}du=\lambda v^{(\lambda-1)/\lambda}du\Rightarrow du=\frac{dv}{\lambda}v^{(1-\lambda)/\lambda}$ now sub this in: $$I_n=n\int_0^1 v^{(t-\lambda-1)/\lambda}\,v^{(1-\lambda)/\lambda}(1-v)^{n-1}dv$$ $$I_n=n\int_0^1 v^{(t-2\lambda)/\lambda}(1-v)^{n-1}dv\tag{2}$$

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Now, the incomplete beta function is defined as: $$B(x;a,b)=\int_0^x y^{a-1}(1-y)^{b-1}dy\tag{3}$$ and the beta function is defined as: $$B(a,b)=\int_0^1x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\tag{4,5}$$ Your integral fits this form nicely, you could also use binomial expansion.

Answer 2

Building off of the comment, it is actually not too terrible to use binomial theorem. However, first observe that

$$ne^{-\lambda x}(1-e^{-\lambda x})^{n-1} = \frac{d}{dx}(1-e^{-\lambda x})^n\text{.} $$

Then you can apply binomial theorem to

$$(1-e^{-\lambda x})^n $$

and take the derivative of the resulting sum. Once substituted into the integrand, evaluating the integral (term by term) is straightforward. The result will be a sum of $n$ terms and will not be pretty.

Answer 3

Using hypergeometric functions$$J=\int\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}\,dx$$ $$J=\frac{\lambda e^{t x} \left(1-e^{-\lambda x}\right)^n \left(1-e^{\lambda x}\right)^{-n} \, _2F_1\left(1-n,\frac{t}{\lambda }-n;-n+\frac{t}{\lambda }+1;e^{x \lambda }\right)}{\lambda n-t}$$ if $\Re(t)<\Re(\lambda )\land \Re(\lambda )>0\land \Re(n)>0$ $$K=\int_0^\infty\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}\,dx=\frac{\Gamma (n) \Gamma \left(1-\frac{t}{\lambda }\right)}{\Gamma \left(n-\frac{t}{\lambda }+1\right)}$$ $$I_n=n\frac{\Gamma (n) \Gamma \left(1-\frac{t}{\lambda }\right)}{\Gamma \left(n-\frac{t}{\lambda }+1\right)}$$

Answer 4

If you are familiar with the exponential distribution, its trivial to notice that $\lambda e^{-\lambda x}$ is the PDF and $1 - e^{-\lambda x}$ is the CDF. Therefore, a substitution $u = 1 - e^{-\lambda x}$ would simplify it.

$$ u = 1 - e^{-\lambda x}$$ $$du = \lambda e^{-\lambda x} dx \tag{1}$$ $$ 1 - u = e^{-\lambda x}$$ $$ \frac{-\ln(1 - u)}{\lambda} = x$$ $$ \frac{-t \hspace{0.05cm} \ln(1 - u)}{\lambda} = tx$$ $$ e^{\frac{-t \hspace{0.05cm} \ln(1 - u)}{\lambda}} = e^{tx}$$ $$ (1 - u)^{\frac{-t}{\lambda}} = e^{tx} \tag{2}$$ Plugging $(1)$ and $(2)$ into this expression, $$ I_n = \int_0^{\infty} n (1 - e^{-\lambda x})^{n-1} e^{tx} \lambda e^{-\lambda x} dx$$ and changing the limits, we get, $$ I_n = \int_0^{1} n u^{n-1} (1 - u)^{\frac{-t}{\lambda}} dt$$ This can be re-written in terms of the Beta function as follows- $$ I_n = n \operatorname{B}(n, 1 - \frac{t}{\lambda})$$