Need help computing the double integral $int_{0}^{infty} int_{0}^{infty} frac{f(x + y)}{x + y} mathop{dy}...












3












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Need help computing the double integral $int_{0}^{infty} int_{0}^{infty} frac{f(x + y)}{x + y} mathop{dy} mathop{dx}.$



I know that $int_{0}^{infty} f(u) mathop{du}$ equals $1$. The entire integral should come out to be $1$.



I am new to multivariable calculus and would appreciate some help.



Making the substitution $u = x + y$ and $v = y$, we get $mathop{du} = mathop{dx} + mathop{dy}$ and $mathop{dv} = mathop{dy}$. So,



$$int_{0}^{infty} int_{0}^{infty} frac{f(u)}{u} mathop{dy} mathop{dx}$$



I don't know what to change the $dy$ and $dx$ to since $du$ has both $dx$ and $dy$ in it. Also, for the Jacobian, I know that it will equal $1$, but I don't know what I'm taking partial derivatives of in the determinant. Can someone please help me with this integral?






integration multivariable-calculus definite-integrals improper-integrals substitution







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  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
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    – Aloizio Macedo♦
    Dec 16 '18 at 6:17
















3












$begingroup$


Need help computing the double integral $int_{0}^{infty} int_{0}^{infty} frac{f(x + y)}{x + y} mathop{dy} mathop{dx}.$



I know that $int_{0}^{infty} f(u) mathop{du}$ equals $1$. The entire integral should come out to be $1$.



I am new to multivariable calculus and would appreciate some help.



Making the substitution $u = x + y$ and $v = y$, we get $mathop{du} = mathop{dx} + mathop{dy}$ and $mathop{dv} = mathop{dy}$. So,



$$int_{0}^{infty} int_{0}^{infty} frac{f(u)}{u} mathop{dy} mathop{dx}$$



I don't know what to change the $dy$ and $dx$ to since $du$ has both $dx$ and $dy$ in it. Also, for the Jacobian, I know that it will equal $1$, but I don't know what I'm taking partial derivatives of in the determinant. Can someone please help me with this integral?






integration multivariable-calculus definite-integrals improper-integrals substitution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 16 '18 at 6:17














3












3








3





$begingroup$


Need help computing the double integral $int_{0}^{infty} int_{0}^{infty} frac{f(x + y)}{x + y} mathop{dy} mathop{dx}.$



I know that $int_{0}^{infty} f(u) mathop{du}$ equals $1$. The entire integral should come out to be $1$.



I am new to multivariable calculus and would appreciate some help.



Making the substitution $u = x + y$ and $v = y$, we get $mathop{du} = mathop{dx} + mathop{dy}$ and $mathop{dv} = mathop{dy}$. So,



$$int_{0}^{infty} int_{0}^{infty} frac{f(u)}{u} mathop{dy} mathop{dx}$$



I don't know what to change the $dy$ and $dx$ to since $du$ has both $dx$ and $dy$ in it. Also, for the Jacobian, I know that it will equal $1$, but I don't know what I'm taking partial derivatives of in the determinant. Can someone please help me with this integral?






integration multivariable-calculus definite-integrals improper-integrals substitution







share|cite|improve this question











$endgroup$




Need help computing the double integral $int_{0}^{infty} int_{0}^{infty} frac{f(x + y)}{x + y} mathop{dy} mathop{dx}.$



I know that $int_{0}^{infty} f(u) mathop{du}$ equals $1$. The entire integral should come out to be $1$.



I am new to multivariable calculus and would appreciate some help.



Making the substitution $u = x + y$ and $v = y$, we get $mathop{du} = mathop{dx} + mathop{dy}$ and $mathop{dv} = mathop{dy}$. So,



$$int_{0}^{infty} int_{0}^{infty} frac{f(u)}{u} mathop{dy} mathop{dx}$$



I don't know what to change the $dy$ and $dx$ to since $du$ has both $dx$ and $dy$ in it. Also, for the Jacobian, I know that it will equal $1$, but I don't know what I'm taking partial derivatives of in the determinant. Can someone please help me with this integral?






integration multivariable-calculus definite-integrals improper-integrals substitution




integration multivariable-calculus definite-integrals improper-integrals substitution






share|cite|improve this question















share|cite|improve this question













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edited Dec 12 '18 at 22:39









Batominovski

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asked Dec 12 '18 at 21:11









josephjoseph

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  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 16 '18 at 6:17


















  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 16 '18 at 6:17
















$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 16 '18 at 6:17




$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 16 '18 at 6:17










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You almost nailed it with your substitution $(x,y)mapsto (u,v):=(x+y,y)$. That said, you made a mistake when you thought that the region of integration is still $mathbb{R}_{> 0}^2$. The correct region is
$$Omega:=big{(u,v)inmathbb{R}_{> 0}^2,big|,u> vbig},.$$
Hence,
$$I:=int_0^infty,int_0^infty,frac{f(x+y)}{x+y},text{d}x,text{d}y=iint_Omega,frac{f(u)}{u},text{d}u,text{d}v,.$$
Assuming that $f$ is sufficiently nice so that Fubini's Theorem applies, we get
$$I=int_0^infty,int_0^u,frac{f(u)}{u},text{d}v,text{d}u=int_0^infty,u,left(frac{f(u)}{u}right),text{d}u,.$$
Thus,
$$I=int_0^infty,f(u),text{d}u=1,.$$






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    $begingroup$

    You almost nailed it with your substitution $(x,y)mapsto (u,v):=(x+y,y)$. That said, you made a mistake when you thought that the region of integration is still $mathbb{R}_{> 0}^2$. The correct region is
    $$Omega:=big{(u,v)inmathbb{R}_{> 0}^2,big|,u> vbig},.$$
    Hence,
    $$I:=int_0^infty,int_0^infty,frac{f(x+y)}{x+y},text{d}x,text{d}y=iint_Omega,frac{f(u)}{u},text{d}u,text{d}v,.$$
    Assuming that $f$ is sufficiently nice so that Fubini's Theorem applies, we get
    $$I=int_0^infty,int_0^u,frac{f(u)}{u},text{d}v,text{d}u=int_0^infty,u,left(frac{f(u)}{u}right),text{d}u,.$$
    Thus,
    $$I=int_0^infty,f(u),text{d}u=1,.$$






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      3












      $begingroup$

      You almost nailed it with your substitution $(x,y)mapsto (u,v):=(x+y,y)$. That said, you made a mistake when you thought that the region of integration is still $mathbb{R}_{> 0}^2$. The correct region is
      $$Omega:=big{(u,v)inmathbb{R}_{> 0}^2,big|,u> vbig},.$$
      Hence,
      $$I:=int_0^infty,int_0^infty,frac{f(x+y)}{x+y},text{d}x,text{d}y=iint_Omega,frac{f(u)}{u},text{d}u,text{d}v,.$$
      Assuming that $f$ is sufficiently nice so that Fubini's Theorem applies, we get
      $$I=int_0^infty,int_0^u,frac{f(u)}{u},text{d}v,text{d}u=int_0^infty,u,left(frac{f(u)}{u}right),text{d}u,.$$
      Thus,
      $$I=int_0^infty,f(u),text{d}u=1,.$$






      share|cite|improve this answer









      $endgroup$
















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        3








        3





        $begingroup$

        You almost nailed it with your substitution $(x,y)mapsto (u,v):=(x+y,y)$. That said, you made a mistake when you thought that the region of integration is still $mathbb{R}_{> 0}^2$. The correct region is
        $$Omega:=big{(u,v)inmathbb{R}_{> 0}^2,big|,u> vbig},.$$
        Hence,
        $$I:=int_0^infty,int_0^infty,frac{f(x+y)}{x+y},text{d}x,text{d}y=iint_Omega,frac{f(u)}{u},text{d}u,text{d}v,.$$
        Assuming that $f$ is sufficiently nice so that Fubini's Theorem applies, we get
        $$I=int_0^infty,int_0^u,frac{f(u)}{u},text{d}v,text{d}u=int_0^infty,u,left(frac{f(u)}{u}right),text{d}u,.$$
        Thus,
        $$I=int_0^infty,f(u),text{d}u=1,.$$






        share|cite|improve this answer









        $endgroup$



        You almost nailed it with your substitution $(x,y)mapsto (u,v):=(x+y,y)$. That said, you made a mistake when you thought that the region of integration is still $mathbb{R}_{> 0}^2$. The correct region is
        $$Omega:=big{(u,v)inmathbb{R}_{> 0}^2,big|,u> vbig},.$$
        Hence,
        $$I:=int_0^infty,int_0^infty,frac{f(x+y)}{x+y},text{d}x,text{d}y=iint_Omega,frac{f(u)}{u},text{d}u,text{d}v,.$$
        Assuming that $f$ is sufficiently nice so that Fubini's Theorem applies, we get
        $$I=int_0^infty,int_0^u,frac{f(u)}{u},text{d}v,text{d}u=int_0^infty,u,left(frac{f(u)}{u}right),text{d}u,.$$
        Thus,
        $$I=int_0^infty,f(u),text{d}u=1,.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 22:37









        BatominovskiBatominovski

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