2

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 $\int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}dx = \ln\left(\frac{a}{b}\right)$

Community
  • 1
HK Lee
  • 19,964

2 Answers2

18

$$ \int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}dx = \int_{0}^{\infty}\int_{a}^{b}e^{-rx}dr\,dx = \int_{a}^{b}\int_{0}^{\infty}e^{-rx}dxdr = \int_{a}^{b}\frac{dr}{r} $$

Disintegrating By Parts
  • 87,459
  • 5
  • 65
  • 149
7

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, $$-\log\ a - (-\log\ b) = \log\ (\frac{b}{a})$$

HK Lee
  • 19,964
StrangelyCorrelated
  • 111
  • This integral $I(t)$ is note exist, since it's diverges at $0$, since $\frac{e^{-tx}}{x} \sim \frac{1}{x}$ when $x \to 0$. – Mohamed Jan 13 '18 at 20:18