Evaluate $int-x^{1-n}e^{xt} dx$











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I have to evaluate $$largedisplaystyleint-x^{1-n}e^{xt} dx$$ with respect to x but I am not sure how. I have tried integration by parts but this gets very complicated, is there an easier way?
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calculus integration indefinite-integrals







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  • You probably won't like the answer, but it's $frac{x^{-n}(tx)Gamma (2-n,tx)}{t^2}+c$
    – UserX
    Dec 2 '14 at 12:55










  • For some basic information about writing math at this site see e.g. here, here, here and here.
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up vote
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down vote

favorite












I have to evaluate $$largedisplaystyleint-x^{1-n}e^{xt} dx$$ with respect to x but I am not sure how. I have tried integration by parts but this gets very complicated, is there an easier way?
Thank you!






calculus integration indefinite-integrals







share|cite|improve this question
























  • You probably won't like the answer, but it's $frac{x^{-n}(tx)Gamma (2-n,tx)}{t^2}+c$
    – UserX
    Dec 2 '14 at 12:55










  • For some basic information about writing math at this site see e.g. here, here, here and here.
    – user137731
    Dec 2 '14 at 12:59













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have to evaluate $$largedisplaystyleint-x^{1-n}e^{xt} dx$$ with respect to x but I am not sure how. I have tried integration by parts but this gets very complicated, is there an easier way?
Thank you!






calculus integration indefinite-integrals







share|cite|improve this question















I have to evaluate $$largedisplaystyleint-x^{1-n}e^{xt} dx$$ with respect to x but I am not sure how. I have tried integration by parts but this gets very complicated, is there an easier way?
Thank you!






calculus integration indefinite-integrals




calculus integration indefinite-integrals






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share|cite|improve this question













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edited Nov 24 at 6:18









Martin Sleziak

44.6k7115270




44.6k7115270










asked Dec 2 '14 at 12:50







user187039



















  • You probably won't like the answer, but it's $frac{x^{-n}(tx)Gamma (2-n,tx)}{t^2}+c$
    – UserX
    Dec 2 '14 at 12:55










  • For some basic information about writing math at this site see e.g. here, here, here and here.
    – user137731
    Dec 2 '14 at 12:59


















  • You probably won't like the answer, but it's $frac{x^{-n}(tx)Gamma (2-n,tx)}{t^2}+c$
    – UserX
    Dec 2 '14 at 12:55










  • For some basic information about writing math at this site see e.g. here, here, here and here.
    – user137731
    Dec 2 '14 at 12:59
















You probably won't like the answer, but it's $frac{x^{-n}(tx)Gamma (2-n,tx)}{t^2}+c$
– UserX
Dec 2 '14 at 12:55




You probably won't like the answer, but it's $frac{x^{-n}(tx)Gamma (2-n,tx)}{t^2}+c$
– UserX
Dec 2 '14 at 12:55












For some basic information about writing math at this site see e.g. here, here, here and here.
– user137731
Dec 2 '14 at 12:59




For some basic information about writing math at this site see e.g. here, here, here and here.
– user137731
Dec 2 '14 at 12:59










1 Answer
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$$I(x)=int-x^{1-n}cdotexp(xt),mathrm dx$$
Let $u=-xt$
$$I(x)=intleft(frac {-u}tright)^{1-n}exp(-u),mathrm du$$
$$I(x)=(-t)^{n-1}int u^{1-n}exp(-u),mathrm du$$
The lower incomplete gamma function $gamma(a,x)$ is defined by
$$gamma(a,x)=int_0^xt^{a-1}exp(-t),mathrm dt$$
If $a=2-n$ then the above integral is equal to the integral in our problem, up to a constant. Therefore
$$I(x)=(-t)^{n-1}gamma(2-n, -xt)+c_1$$
However the lower incomplete gamma function is very unknown to most people, however there is another function called the incomplete gamma function, which is more known, and defined as the following:
$$Gamma(a,x)=int_x^infty t^{a-1}exp(-t),mathrm dt$$
By definition, the following is true, where $Gamma(x)$ is the gamma function
$$gamma(a,x)+Gamma(a,x)=Gamma(a)$$
$$gamma(a,x)=Gamma(a)-Gamma(a,x)$$
Therefore we can say that
$$I(x)=(-t)^{n-1}left(Gamma(2-n)-Gamma(2-n, -xt)right)+c_1$$
Which reduces to
$$boxed{displaystyle I(x)=-(-t)^{n-1}Gamma(2-n, -xt)+c_2}$$






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    1 Answer
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    1 Answer
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    up vote
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    $$I(x)=int-x^{1-n}cdotexp(xt),mathrm dx$$
    Let $u=-xt$
    $$I(x)=intleft(frac {-u}tright)^{1-n}exp(-u),mathrm du$$
    $$I(x)=(-t)^{n-1}int u^{1-n}exp(-u),mathrm du$$
    The lower incomplete gamma function $gamma(a,x)$ is defined by
    $$gamma(a,x)=int_0^xt^{a-1}exp(-t),mathrm dt$$
    If $a=2-n$ then the above integral is equal to the integral in our problem, up to a constant. Therefore
    $$I(x)=(-t)^{n-1}gamma(2-n, -xt)+c_1$$
    However the lower incomplete gamma function is very unknown to most people, however there is another function called the incomplete gamma function, which is more known, and defined as the following:
    $$Gamma(a,x)=int_x^infty t^{a-1}exp(-t),mathrm dt$$
    By definition, the following is true, where $Gamma(x)$ is the gamma function
    $$gamma(a,x)+Gamma(a,x)=Gamma(a)$$
    $$gamma(a,x)=Gamma(a)-Gamma(a,x)$$
    Therefore we can say that
    $$I(x)=(-t)^{n-1}left(Gamma(2-n)-Gamma(2-n, -xt)right)+c_1$$
    Which reduces to
    $$boxed{displaystyle I(x)=-(-t)^{n-1}Gamma(2-n, -xt)+c_2}$$






    share|cite|improve this answer

























      up vote
      0
      down vote













      $$I(x)=int-x^{1-n}cdotexp(xt),mathrm dx$$
      Let $u=-xt$
      $$I(x)=intleft(frac {-u}tright)^{1-n}exp(-u),mathrm du$$
      $$I(x)=(-t)^{n-1}int u^{1-n}exp(-u),mathrm du$$
      The lower incomplete gamma function $gamma(a,x)$ is defined by
      $$gamma(a,x)=int_0^xt^{a-1}exp(-t),mathrm dt$$
      If $a=2-n$ then the above integral is equal to the integral in our problem, up to a constant. Therefore
      $$I(x)=(-t)^{n-1}gamma(2-n, -xt)+c_1$$
      However the lower incomplete gamma function is very unknown to most people, however there is another function called the incomplete gamma function, which is more known, and defined as the following:
      $$Gamma(a,x)=int_x^infty t^{a-1}exp(-t),mathrm dt$$
      By definition, the following is true, where $Gamma(x)$ is the gamma function
      $$gamma(a,x)+Gamma(a,x)=Gamma(a)$$
      $$gamma(a,x)=Gamma(a)-Gamma(a,x)$$
      Therefore we can say that
      $$I(x)=(-t)^{n-1}left(Gamma(2-n)-Gamma(2-n, -xt)right)+c_1$$
      Which reduces to
      $$boxed{displaystyle I(x)=-(-t)^{n-1}Gamma(2-n, -xt)+c_2}$$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        $$I(x)=int-x^{1-n}cdotexp(xt),mathrm dx$$
        Let $u=-xt$
        $$I(x)=intleft(frac {-u}tright)^{1-n}exp(-u),mathrm du$$
        $$I(x)=(-t)^{n-1}int u^{1-n}exp(-u),mathrm du$$
        The lower incomplete gamma function $gamma(a,x)$ is defined by
        $$gamma(a,x)=int_0^xt^{a-1}exp(-t),mathrm dt$$
        If $a=2-n$ then the above integral is equal to the integral in our problem, up to a constant. Therefore
        $$I(x)=(-t)^{n-1}gamma(2-n, -xt)+c_1$$
        However the lower incomplete gamma function is very unknown to most people, however there is another function called the incomplete gamma function, which is more known, and defined as the following:
        $$Gamma(a,x)=int_x^infty t^{a-1}exp(-t),mathrm dt$$
        By definition, the following is true, where $Gamma(x)$ is the gamma function
        $$gamma(a,x)+Gamma(a,x)=Gamma(a)$$
        $$gamma(a,x)=Gamma(a)-Gamma(a,x)$$
        Therefore we can say that
        $$I(x)=(-t)^{n-1}left(Gamma(2-n)-Gamma(2-n, -xt)right)+c_1$$
        Which reduces to
        $$boxed{displaystyle I(x)=-(-t)^{n-1}Gamma(2-n, -xt)+c_2}$$






        share|cite|improve this answer












        $$I(x)=int-x^{1-n}cdotexp(xt),mathrm dx$$
        Let $u=-xt$
        $$I(x)=intleft(frac {-u}tright)^{1-n}exp(-u),mathrm du$$
        $$I(x)=(-t)^{n-1}int u^{1-n}exp(-u),mathrm du$$
        The lower incomplete gamma function $gamma(a,x)$ is defined by
        $$gamma(a,x)=int_0^xt^{a-1}exp(-t),mathrm dt$$
        If $a=2-n$ then the above integral is equal to the integral in our problem, up to a constant. Therefore
        $$I(x)=(-t)^{n-1}gamma(2-n, -xt)+c_1$$
        However the lower incomplete gamma function is very unknown to most people, however there is another function called the incomplete gamma function, which is more known, and defined as the following:
        $$Gamma(a,x)=int_x^infty t^{a-1}exp(-t),mathrm dt$$
        By definition, the following is true, where $Gamma(x)$ is the gamma function
        $$gamma(a,x)+Gamma(a,x)=Gamma(a)$$
        $$gamma(a,x)=Gamma(a)-Gamma(a,x)$$
        Therefore we can say that
        $$I(x)=(-t)^{n-1}left(Gamma(2-n)-Gamma(2-n, -xt)right)+c_1$$
        Which reduces to
        $$boxed{displaystyle I(x)=-(-t)^{n-1}Gamma(2-n, -xt)+c_2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '14 at 20:03









        Alice Ryhl

        5,89011235




        5,89011235






























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