relative of exponential integral


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

0e-ax-e-bxx𝑑x (1)

related to the exponential integralDlmfDlmfDlmfMathworldPlanetmath.

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

I(y):=0e-xye-ax-e-bxx𝑑x

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

I(y) =0e-xy(-x)e-ax-e-bxx𝑑x
=0(e-(y+b)x-e-(y+a)x)𝑑x
=/x=0(e-(y+b)x-(y+b)-e-(y+a)x-(y+a))
=1y+b-1y+a

Thus,

I(y)=ln(y+b)-ln(y+a)=lny+by+a,

and the integral (1) has the value  I(0)=lnba.

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

{e-at-e-bt}=1s+a-1s+b

and therefore

{e-at-e-btt}=s(1u+a-1u+b)𝑑u=/u=slnu+au+b=lns+bs+a,

i.e.

0e-ste-at-e-btt𝑑t=lns+bs+a.

Letting  s0+,  this yields the equation

0e-ax-e-bxx𝑑x=lnba.
Title relative of exponential integral
Canonical name RelativeOfExponentialIntegral
Date of creation 2013-03-22 18:44:20
Last modified on 2013-03-22 18:44:20
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 12
Author pahio (2872)
Entry type Example
Classification msc 44A10
Classification msc 26A36
Related topic SubstitutionNotation
Related topic RelativeOfCosineIntegral
Related topic IntegrationOfLaplaceTransformWithRespectToParameter
Related topic IntegrationUnderIntegralSign