Prove or disprove: $\int\limits_{0}^{+\infty}\frac{e^{-x}}{1+ax}\,dx=1-a,~~0<a<1.$

Question

Prove or disprove: $$\int\limits_{0}^{+\infty}\frac{e^{-x}}{1+ax}\,dx=1-a,~~0<a<1.$$

Attempt. I have seen function $$\frac{1}{1-a}\,\frac{e^{-x}}{1+ax}$$ as pdf of a continuous rv, so I guess the answer is positive, but I couldn't handle the manipulations regarding the calculation of the integral (these is no closed form of course for this integral).

Thanks in advance.

Answer 1

This is certainly false. Were it true, the dominated convergence theorem implies that $$\int_0^\infty\frac{e^{-x}}{1+x}\,dx =\lim_{a\to1^-}\int_0^\infty\frac{e^{-x}}{1+ax}\,dx =0$$ which is absurd.

Answer 2

Setting $1+ax = t$ and $t = ay$ gives, $$\int_0^{\infty}\frac{e^{-x}}{1+ax} dx = \frac{e^{\frac{1}{a}}}{a}\int_1^{\infty}\frac{e^{-\frac{t}{a}}}{t} dt = \frac{e^{\frac{1}{a}}}{a}\int_{\frac{1}{a}}^{\infty}\frac{e^{-y}}{y} dy = \frac{e^{\frac{1}{a}}}{a} \textrm{E1}\left(\frac{1}{a}\right)$$

where $\textrm{E1}$ is the exponential integral [1]

[1] http://mathworld.wolfram.com/ExponentialIntegral.html