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Integral of Exponential Functions

Posted 12 years ago
Is there any body can help me to solve the following integral? 
 
Integrate[ x^h*E^-((\[Alpha]*x) + (\[Eta]*x^n))*((DiracDelta[x]*x^t* E^(-\[Lambda]*x^m)) + 1), {x, 0, \[Infinity]}]

Thanks a lot
POSTED BY: John G
4 Replies
Hi,

Tne integral can be expanded as:
Integrate[ x^h*E^-((?*x) + (?*x^n))*(DiracDelta[x]*x^t* E^(-?*x^m)), {x, 0, ?}] +
Integrate[ x^h*E^-((?*x) + (?*x^n)), {x, 0, ?}]
The first integral can be easily evaluated using the properties of the DiracDelta function. However, Mathematica does not evaluate the second integral in closed form for general parameters n and h. Closed form results may be obtained if the parameters are given specific values.

Youngjoo Chung
POSTED BY: Youngjoo Chung
Posted 12 years ago
Hello
Thanks for your response. However, n is real numbers 0 <= n <= 1 and h is real numbers > 0 

In fact, I have tried the first integral and got results as 
-0^(h + t) E^(-0^n \[Eta] - 0^m \[Lambda]) (-1 + HeavisideTheta[0])

so, what does the above expression mean?. Thanks for your support
POSTED BY: John G
Hi,

Unless n and h are given specific values like n = 1/2 and h = 2, Integrate will not evaluate the integral.

About the result
-0^(h + t) E^(-0^n ? - 0^m ?) (-1 + HeavisideTheta[0])
^ means power. When x is real, 0^x is 0 for x>0, 1 for x=0, and infinity for x<0. For HeavisideTheta, please refer to Mathematica documentation.

Youngjoo Chung
POSTED BY: Youngjoo Chung
Posted 12 years ago
Hi 

Actually, I was looking for Laplace transform, but I think it's difficult.  Do you have any suggestions? 

Thanks
POSTED BY: John G
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