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How to evaluate this integral:

$$I_{a,b} = \int_{1}^{\infty} \frac{\ \exp\left(-a t\right)}{ 1-b t} \mathrm{d}t $$
where $a, b \in R^*_+$ ?

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    In terms of the "exponential integral" function, http://en.wikipedia.org/wiki/Exponential_integral2012-05-26
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    Consider the substitution $u=1-bt$ first and $v=au/b$ second; compare with [$\mathrm{Ei}(\cdot)$](http://en.wikipedia.org/wiki/Exponential_integral).2012-05-26
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    Thanks GEdgar, anon this was very helpful.2012-05-26
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    @jack Could you write your own answer and accept it so that this question gets answered?2012-05-26

1 Answers 1

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Assuming that $b<1$ so that $1-bt >0 $.

Thanks to @anon this can be done by making the substitution :
1). $u=1-bt$, with $\mathrm{d}t = (-1/b) \mathrm{d}u$, so : $$I_{a,b} =\frac{\exp(-\frac{a}{b})}{b} \ \int\limits_{-\infty}^{1-b} \frac{\ \exp\left(\frac{a}{b} u\right)}{u} \mathrm{d}u$$

2.) $v=\frac{a}{b}u$ with $\frac{\mathrm{d}v}{v}=\frac{\mathrm{d}u}{u}$ , then: $$I_{a,b} =\frac{\exp(-\frac{a}{b})}{b} \ \int\limits_{-\infty}^{(1-b)\frac{a}{b}} \frac{\ \exp(v)}{v} \mathrm{d}v$$

Hence. $$I_{a,b} =\frac{\exp(-\frac{a}{b})}{b} \ Ei({(1-b)\frac{a}{b}})$$ where $Ei(x) =\int\limits_{-\infty}^{x} \frac{\ \exp(t)}{t} \mathrm{d}t $ : The Exponential Integral Function