relative of exponential integral

Let $a$ and $b$ be positive numbers.  We want to calculate the value of the improper integral

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{e^{-ax}-e^{-bx}}{x}}𝑑x$(1)

related to the exponential integral.

The value may be found e.g. by utilising the derivative of the integral

$$I(y):={\int }_0^{\mathrm{\infty }}{e}^{-xy}\cdot \frac{e^{-ax}-e^{-bx}}{x}𝑑x$$

which can be formed by differentiating under the integral sign (http://planetmath.org/DifferentiationUnderIntegralSign):

$I^{\prime }(y)$$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-xy}(-x){\displaystyle \frac{e^{-ax}-e^{-bx}}{x}}𝑑x$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^{\mathrm{\infty }}}\left(e^{-(y+b)x}-e^{-(y+a)x}\right)𝑑x$

$\mathrm{\hspace{0.33em}}=\underset{x=0}{\overset{\mathrm{\infty }}{/}}\left({\displaystyle \frac{e^{-(y+b)x}}{-(y+b)}}-{\displaystyle \frac{e^{-(y+a)x}}{-(y+a)}}\right)$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{y+b}}-{\displaystyle \frac{1}{y+a}}$

Thus,

$$I(y)=\ln (y+b)-\ln (y+a)=\ln \frac{y+b}{y+a},$$

and the integral (1) has the value  $I(0)=\ln {\displaystyle \frac{b}{a}}$.

There is another method via Laplace transforms.  By the table of Laplace transforms, we have

$$ℒ\{e^{-at}-e^{-bt}\}=\frac{1}{s+a}-\frac{1}{s+b}$$

and therefore

$$ℒ\{\frac{e^{-at}-e^{-bt}}{t}\}={\int }_s^{\mathrm{\infty }}\left(\frac{1}{u+a}-\frac{1}{u+b}\right)𝑑u=\underset{u=s}{\overset{\mathrm{\infty }}{/}}\ln \frac{u+a}{u+b}=\ln \frac{s+b}{s+a},$$

i.e.

$${\int }_0^{\mathrm{\infty }}{e}^{-st}\cdot \frac{e^{-at}-e^{-bt}}{t}𝑑t=\ln \frac{s+b}{s+a}.$$

Letting  $s\to 0+$,  this yields the equation

$${\int }_0^{\mathrm{\infty }}\frac{e^{-ax}-e^{-bx}}{x}𝑑x=\ln \frac{b}{a}.$$