Integral I=\int t^{2/3} \exp^{a t} dt, where a < 0
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Originally Posted by Konstantin I=\int t^{2/3} \exp^{a t} dt, where a < 0 Perhaps you meant . In such case, use the substitution and the Gamma function.
Originally Posted by Konstantin Integral I=\int t^{2/3} \exp^{a t} dt, where a < 0 If the integral is from to , then setting You have... (1) Kind regards
The problem that its limits not from 0 to +\infty. Actually, it is a solution of some eqation, depending on coordinate z, in the form I(z)=\int_0^z t^{2/3} \exp^{at} dt
Originally Posted by Konstantin The problem that its limits not from 0 to +\infty. Actually, it is a solution of some eqation, depending on coordinate z, in the form I(z)=\int_0^z t^{2/3} \exp^{at} dt In that case the integral is not an elementary function but it can be represented as Taylor series... Kind regards
And it is NOT, for any finite z, an "improper" integral!