# Improper integral of $\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x}\ dx$ [duplicate]

For $a>b>0$, calculate $$\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x}\ dx$$

My try : By Taylor series, $$\int\ \frac{e^{-ax} - e^{-bx}}{x} \ dx= \sum_{n=1}^\infty \frac{[\ (-a)^n -(-b)^n\ ]}{n}\frac{x^n}{n!} +C$$

Note that from ratio test, this series converges absolutely for $x\in [0,\infty)$. So give me a hint. Thanks

Seocond Try : Recall $$\Gamma(t) = \int_0^\infty s^{t-1}e^{-s}\ ds\ (t>0)$$

So $$\Gamma(t) = a^t \int_0^\infty \frac{e^{-as}}{s^{1-t}}\ ds$$

So integral we want to calculate is $$\lim_{t\rightarrow 0}\ [a^{-t} - b^{-t}]\ \Gamma(t)$$ Right ?

See I found the following article proving of Integral $\int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}dx = \ln\left(\frac{a}{b}\right)$

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## marked as duplicate by mrf, user1337, Daniel Fischer♦, mau, egregDec 3 '13 at 13:02

$$\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}dx = \int_{0}^{\infty}\int_{a}^{b}e^{-rx}dr\,dx = \int_{a}^{b}\int_{0}^{\infty}e^{-rx}dxdr = \int_{a}^{b}\frac{dr}{r}$$
This can be solved by differentiating under the integral sign. Writing $$I(t) = \int_0^\infty \frac{e^{-tx}}{x} dx$$ We immediately see that $$\frac{dI}{dt} = -\int_0^\infty e^{-tx} dx$$ which equals $\frac{-1}{a}$. So $$I = -log{t} + C$$ And the given integral is therefore, $$-loga - (-logb) = log(\frac{b}{a})$$