Need help computing the double integral $int_{0}^{infty} int_{0}^{infty} frac{f(x + y)}{x + y} mathop{dy}...
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
edited Dec 12 '18 at 22:39
Batominovski
1
asked Dec 12 '18 at 21:11
josephjoseph
46210
add a comment |
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
edited Dec 12 '18 at 22:39
Batominovski
1
asked Dec 12 '18 at 21:11
josephjoseph
46210
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Dec 16 '18 at 6:17
add a comment |
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
edited Dec 12 '18 at 22:39
Batominovski
1
asked Dec 12 '18 at 21:11
josephjoseph
46210
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
edited Dec 12 '18 at 22:39
Batominovski
1
asked Dec 12 '18 at 21:11
josephjoseph
46210
edited Dec 12 '18 at 22:39
Batominovski
1
asked Dec 12 '18 at 21:11
josephjoseph
46210
edited Dec 12 '18 at 22:39
Batominovski
1
edited Dec 12 '18 at 22:39
Batominovski
1
edited Dec 12 '18 at 22:39
Batominovski
1
1
asked Dec 12 '18 at 21:11
josephjoseph
46210
asked Dec 12 '18 at 21:11
josephjoseph
46210
asked Dec 12 '18 at 21:11
josephjoseph
46210
46210
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Dec 16 '18 at 6:17
add a comment |
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Dec 16 '18 at 6:17
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Dec 16 '18 at 6:17
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Dec 16 '18 at 6:17
add a comment |
1 Answer
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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,
$$I=int_0^infty,f(u),text{d}u=1,.$$
answered Dec 12 '18 at 22:37
BatominovskiBatominovski
1
add a comment |
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1 Answer
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1 Answer
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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,
$$I=int_0^infty,f(u),text{d}u=1,.$$
answered Dec 12 '18 at 22:37
BatominovskiBatominovski
1
add a comment |
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,
$$I=int_0^infty,f(u),text{d}u=1,.$$
answered Dec 12 '18 at 22:37
BatominovskiBatominovski
1
add a comment |
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,
$$I=int_0^infty,f(u),text{d}u=1,.$$
answered Dec 12 '18 at 22:37
BatominovskiBatominovski
1
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,
$$I=int_0^infty,f(u),text{d}u=1,.$$
answered Dec 12 '18 at 22:37
BatominovskiBatominovski
1
answered Dec 12 '18 at 22:37
BatominovskiBatominovski
1
answered Dec 12 '18 at 22:37
BatominovskiBatominovski
1
answered Dec 12 '18 at 22:37
BatominovskiBatominovski
1
1
add a comment |
add a comment |
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– Aloizio Macedo♦
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