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Math Help - Finding improper integral

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    Finding improper integral

    Integral

    I=\int t^{2/3} \exp^{a t} dt, where a < 0
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Konstantin View Post
    I=\int t^{2/3} \exp^{a t} dt, where a < 0

    Perhaps you meant I=\int_{0}^{+\infty}t^{2/3}e^{at}\;dt . In such case, use the substitution u=-at and the Gamma function.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Konstantin View Post
    Integral

    I=\int t^{2/3} \exp^{a t} dt, where a < 0
    If the integral is from 0 to + \infty, then setting s=-a You have...

    \displaystyle \int_{0}^{+\infty} t^{\frac{2}{3}}\ e^{-s t}\ dt = \mathcal{L} \{t^\frac{2}{3}\}= \frac{\Gamma(\frac{5}{3})}{s^{\frac{5}{3}}} (1)

    Kind regards

    \chi \sigma
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    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
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Konstantin View Post
    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...

    \displaystyle I(z)= \int_{0}^{z} t^{\frac{2}{3}} e^{a t} dt = \sum_{n=0}^{\infty} \frac{a^{n}}{n!}\ \int_{0}^{z} t^{n+\frac{2}{3}}\ dt= z^{\frac{5}{3}}\ \sum_{n=0}^{\infty} \frac{a^{n}}{(n+\frac{5}{3})\ n!}\ z^{n}

    Kind regards

    \chi \sigma
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  6. #6
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    And it is NOT, for any finite z, an "improper" integral!
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