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# Integral of Exponential Functions

Posted 11 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
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Posted 11 years ago
 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 11 years ago
 Hi Actually, I was looking for Laplace transform, but I think it's difficult.  Do you have any suggestions? Thanks
Posted 11 years ago
 HelloThanks 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 11 years ago
 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
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