Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.












1















Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.




I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.






integration multivariable-calculus substitution jacobian







share|cite|improve this question


















  • 1




    Maybe try reversing order of integration?
    – Zachary Selk
    Jan 6 at 17:44






  • 2




    Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
    – Did
    Jan 6 at 17:46












  • @ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
    – J. Doe
    Jan 6 at 17:52






  • 1




    Draw a picture of the region.
    – Zachary Selk
    Jan 6 at 17:53






  • 1




    Yeah now it's totally doable.
    – Zachary Selk
    Jan 6 at 18:12
















1















Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.




I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.






integration multivariable-calculus substitution jacobian







share|cite|improve this question


















  • 1




    Maybe try reversing order of integration?
    – Zachary Selk
    Jan 6 at 17:44






  • 2




    Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
    – Did
    Jan 6 at 17:46












  • @ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
    – J. Doe
    Jan 6 at 17:52






  • 1




    Draw a picture of the region.
    – Zachary Selk
    Jan 6 at 17:53






  • 1




    Yeah now it's totally doable.
    – Zachary Selk
    Jan 6 at 18:12














1












1








1








Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.




I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.






integration multivariable-calculus substitution jacobian







share|cite|improve this question














Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.




I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.






integration multivariable-calculus substitution jacobian




integration multivariable-calculus substitution jacobian






share|cite|improve this question













share|cite|improve this question











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asked Jan 6 at 17:43









J. DoeJ. Doe

1286




1286








  • 1




    Maybe try reversing order of integration?
    – Zachary Selk
    Jan 6 at 17:44






  • 2




    Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
    – Did
    Jan 6 at 17:46












  • @ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
    – J. Doe
    Jan 6 at 17:52






  • 1




    Draw a picture of the region.
    – Zachary Selk
    Jan 6 at 17:53






  • 1




    Yeah now it's totally doable.
    – Zachary Selk
    Jan 6 at 18:12














  • 1




    Maybe try reversing order of integration?
    – Zachary Selk
    Jan 6 at 17:44






  • 2




    Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
    – Did
    Jan 6 at 17:46












  • @ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
    – J. Doe
    Jan 6 at 17:52






  • 1




    Draw a picture of the region.
    – Zachary Selk
    Jan 6 at 17:53






  • 1




    Yeah now it's totally doable.
    – Zachary Selk
    Jan 6 at 18:12








1




1




Maybe try reversing order of integration?
– Zachary Selk
Jan 6 at 17:44




Maybe try reversing order of integration?
– Zachary Selk
Jan 6 at 17:44




2




2




Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
– Did
Jan 6 at 17:46






Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
– Did
Jan 6 at 17:46














@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
– J. Doe
Jan 6 at 17:52




@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
– J. Doe
Jan 6 at 17:52




1




1




Draw a picture of the region.
– Zachary Selk
Jan 6 at 17:53




Draw a picture of the region.
– Zachary Selk
Jan 6 at 17:53




1




1




Yeah now it's totally doable.
– Zachary Selk
Jan 6 at 18:12




Yeah now it's totally doable.
– Zachary Selk
Jan 6 at 18:12










2 Answers
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oldest

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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$




$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.




begin{align}
&bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
\[5mm] = &
int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
bbx{expo{} - 2 over 3} approx 0.2394
end{align}






share|cite|improve this answer





























    2














    Well, solving a much more general problem:



    $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



    Using that (for all $x$):



    $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



    We can write:



    $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
    $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
    $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
    $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
    $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
    $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



    When $alpha=1$, we get:



    $$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$






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      2 Answers
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      2 Answers
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      $ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
      {LARGE ?}}$      .




      begin{align}
      &bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
      int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
      int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
      \[5mm] = &
      int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
      left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
      bbx{expo{} - 2 over 3} approx 0.2394
      end{align}






      share|cite|improve this answer


























        5














        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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        $ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
        {LARGE ?}}$        .




        begin{align}
        &bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
        int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
        int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
        \[5mm] = &
        int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
        left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
        bbx{expo{} - 2 over 3} approx 0.2394
        end{align}






        share|cite|improve this answer
























          5












          5








          5






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          $ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
          {LARGE ?}}$          .




          begin{align}
          &bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
          int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
          int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
          \[5mm] = &
          int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
          left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
          bbx{expo{} - 2 over 3} approx 0.2394
          end{align}






          share|cite|improve this answer












          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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          newcommand{verts}[1]{leftvert,{#1},rightvert}$




          $ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
          {LARGE ?}}$          .




          begin{align}
          &bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
          int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
          int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
          \[5mm] = &
          int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
          left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
          bbx{expo{} - 2 over 3} approx 0.2394
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 21:48









          Felix MarinFelix Marin

          67.2k7107141




          67.2k7107141























              2














              Well, solving a much more general problem:



              $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



              Using that (for all $x$):



              $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



              We can write:



              $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
              $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
              $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
              $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
              $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
              $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



              When $alpha=1$, we get:



              $$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$






              share|cite|improve this answer




























                2














                Well, solving a much more general problem:



                $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



                Using that (for all $x$):



                $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



                We can write:



                $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
                $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



                When $alpha=1$, we get:



                $$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$






                share|cite|improve this answer


























                  2












                  2








                  2






                  Well, solving a much more general problem:



                  $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



                  Using that (for all $x$):



                  $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



                  We can write:



                  $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                  $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



                  When $alpha=1$, we get:



                  $$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$






                  share|cite|improve this answer














                  Well, solving a much more general problem:



                  $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



                  Using that (for all $x$):



                  $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



                  We can write:



                  $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                  $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



                  When $alpha=1$, we get:



                  $$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 7 at 20:05

























                  answered Jan 6 at 20:52









                  JanJan

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                  21.7k31240






























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