relative of exponential integral
Let a and b be positive numbers.β We want to calculate the value of the improper integral
β«β0e-ax-e-bxxπx | (1) |
related to the exponential integral.
The value may be found e.g. by utilising the derivative of the integral
I(y):=β«β0e-xyβ e-ax-e-bxxπx |
which can be formed by differentiating under the integral sign (http://planetmath.org/DifferentiationUnderIntegralSign):
Iβ²(y) | β=β«β0e-xy(-x)e-ax-e-bxxπx | ||
β=β«β0(e-(y+b)x-e-(y+a)x)πx | |||
β=β/x=0(e-(y+b)x-(y+b)-e-(y+a)x-(y+a)) | |||
β=1y+b-1y+a |
Thus,
I(y)=ln(y+b)-ln(y+a)=lny+by+a, |
and the integral (1) has the valueβ I(0)=lnba.
There is another method via Laplace transforms.β By the table of Laplace transforms, we have
β{e-at-e-bt}=1s+a-1s+b |
and therefore
β{e-at-e-btt}=β«βs(1u+a-1u+b)πu=β/u=slnu+au+b=lns+bs+a, |
i.e.
β«β0e-stβ e-at-e-bttπt=lns+bs+a. |
Lettingβ sβ0+,β this yields the equation
β«β0e-ax-e-bxxπx=lnba. |
Title | relative of exponential integral |
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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 |