# relative of exponential integral

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

 $\displaystyle\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\,dx$ (1)

The value may be found e.g. by utilising the derivative of the integral  $I(y)\;:=\,\int_{0}^{\infty}e^{-xy}\!\cdot\!\frac{e^{-ax}-e^{-bx}}{x}\,dx$

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

 $\displaystyle I^{\prime}(y)$ $\displaystyle\;=\;\int_{0}^{\infty}e^{-xy}(-x)\frac{e^{-ax}-e^{-bx}}{x}\,dx$ $\displaystyle\;=\;\int_{0}^{\infty}\left(e^{-(y+b)x}-e^{-(y+a)x}\right)\,dx$ $\displaystyle\;=\;\operatornamewithlimits{\Big{/}}_{\!\!\!x=0}^{\,\quad\infty}% \!\left(\frac{e^{-(y+b)x}}{-(y\!+\!b)}-\frac{e^{-(y+a)x}}{-(y\!+\!a)}\right)$ $\displaystyle\;=\;\frac{1}{y\!+\!b}-\frac{1}{y\!+\!a}$

Thus,

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

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

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

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

and therefore

 $\mathcal{L}\{\frac{e^{-at}-e^{-bt}}{t}\}\;=\;\int_{s}^{\infty}\left(\frac{1}{u% \!+\!a}-\frac{1}{u\!+\!b}\right)\,du\;=\;\operatornamewithlimits{\Big{/}}_{\!% \!\!u=s}^{\,\quad\infty}\ln\frac{u\!+\!a}{u\!+\!b}\;=\;\ln\frac{s\!+\!b}{s\!+% \!a},$

i.e.

 $\int_{0}^{\infty}e^{-st}\!\cdot\!\frac{e^{-at}-e^{-bt}}{t}\,dt\;=\;\ln\frac{s% \!+\!b}{s\!+\!a}.$

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

 $\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\,dx\;=\;\ln\frac{b}{a}.$
Title relative of exponential integral RelativeOfExponentialIntegral 2013-03-22 18:44:20 2013-03-22 18:44:20 pahio (2872) pahio (2872) 12 pahio (2872) Example msc 44A10 msc 26A36 SubstitutionNotation RelativeOfCosineIntegral IntegrationOfLaplaceTransformWithRespectToParameter IntegrationUnderIntegralSign