# Evaluation of $I_{a,b} = \int_{1}^{+\infty} \frac{\ \exp{(-at)}}{ 1-bt} \ \mathrm{d}t$

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, en.wikipedia.org/wiki/Exponential_integral – GEdgar May 26 '12 at 0:11
Consider the substitution $u=1-bt$ first and $v=au/b$ second; compare with $\mathrm{Ei}(\cdot)$. – anon May 26 '12 at 0:32
Thanks GEdgar, anon this was very helpful. – jack May 26 '12 at 0:55
@jack Could you write your own answer and accept it so that this question gets answered? – user17762 May 26 '12 at 2:39

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