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How to calculate $\int_{0}^{\infty}e^{ -\lambda x}(1-e^{ -\alpha x})(1-e^{ -\beta x})dx$?

Is there a way to calculate this integral?
I know that
$\int_{0}^{\infty}e^{ -\lambda x}(1-e^{ -\alpha x})dx=\frac{1}{\alpha}$
Thanks!

A:

Using your integral (the initial $\alpha x$ makes no difference, as we can factor out $\alpha$) one can rearrange the function as follows :
$$\int_{0}^{\infty}(1-e^{ -(x+\beta)})e^{ -\lambda x}\,dx = \int_{0}^{\infty}e^{ -\lambda x-\beta x}-e^{ -\beta x}e^{ -\lambda x}\,dx$$
Now the integral on the right has a closed form in terms of a Barnes G-function, and hence the integral on the left is a linear combination of hypergeometric functions of the same type. It’s my impression that you’re dealing with some special function, so I’ll leave it to you to work out the coefficients.

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