General procedure to switch order of the integrals
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I am having problems in understanding how the limits of the integral change when we switch the order of integration. In particular, let $f,g : mathbb{R} rightarrow mathbb{R}$ be integrable functions. How do I show that for $- infty < a < b < + infty$ the following is true
begin{align}
int_a^b int_a^t f(s) g(t) ds dt = int_a^b int_s^b f(s) g(t) dt ds
end{align}
integration definite-integrals order-of-integration
asked Nov 23 at 17:16
F.Vitiello
111
add a comment |
up vote
0
down vote
favorite
I am having problems in understanding how the limits of the integral change when we switch the order of integration. In particular, let $f,g : mathbb{R} rightarrow mathbb{R}$ be integrable functions. How do I show that for $- infty < a < b < + infty$ the following is true
begin{align}
int_a^b int_a^t f(s) g(t) ds dt = int_a^b int_s^b f(s) g(t) dt ds
end{align}
integration definite-integrals order-of-integration
asked Nov 23 at 17:16
F.Vitiello
111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am having problems in understanding how the limits of the integral change when we switch the order of integration. In particular, let $f,g : mathbb{R} rightarrow mathbb{R}$ be integrable functions. How do I show that for $- infty < a < b < + infty$ the following is true
begin{align}
int_a^b int_a^t f(s) g(t) ds dt = int_a^b int_s^b f(s) g(t) dt ds
end{align}
integration definite-integrals order-of-integration
asked Nov 23 at 17:16
F.Vitiello
111
I am having problems in understanding how the limits of the integral change when we switch the order of integration. In particular, let $f,g : mathbb{R} rightarrow mathbb{R}$ be integrable functions. How do I show that for $- infty < a < b < + infty$ the following is true
begin{align}
int_a^b int_a^t f(s) g(t) ds dt = int_a^b int_s^b f(s) g(t) dt ds
end{align}
integration definite-integrals order-of-integration
integration definite-integrals order-of-integration
asked Nov 23 at 17:16
F.Vitiello
111
asked Nov 23 at 17:16
F.Vitiello
111
asked Nov 23 at 17:16
F.Vitiello
111
asked Nov 23 at 17:16
F.Vitiello
111
asked Nov 23 at 17:16
F.Vitiello
111
111
add a comment |
add a comment |
1 Answer
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Your integration domain is $D := lbrace (s,t) in (a,b)^2: s < trbrace$.
Therefore, $$int_a^b int_a^t f(s)g(t) ,ds,dt = iint_D f(s)g(t) ,d(s,t) = int_a^b int_s^b f(s)g(t) ,dt,ds.$$
If you let $t in(a,b)$ arbitrary, you want $a < s < t$ to hold and if $sin (a,b)$ is some real number, you want $s < t <b$.
answered Nov 23 at 18:10
maliesen
18111
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your integration domain is $D := lbrace (s,t) in (a,b)^2: s < trbrace$.
Therefore, $$int_a^b int_a^t f(s)g(t) ,ds,dt = iint_D f(s)g(t) ,d(s,t) = int_a^b int_s^b f(s)g(t) ,dt,ds.$$
If you let $t in(a,b)$ arbitrary, you want $a < s < t$ to hold and if $sin (a,b)$ is some real number, you want $s < t <b$.
answered Nov 23 at 18:10
maliesen
18111
add a comment |
up vote
1
down vote
accepted
Your integration domain is $D := lbrace (s,t) in (a,b)^2: s < trbrace$.
Therefore, $$int_a^b int_a^t f(s)g(t) ,ds,dt = iint_D f(s)g(t) ,d(s,t) = int_a^b int_s^b f(s)g(t) ,dt,ds.$$
If you let $t in(a,b)$ arbitrary, you want $a < s < t$ to hold and if $sin (a,b)$ is some real number, you want $s < t <b$.
answered Nov 23 at 18:10
maliesen
18111
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your integration domain is $D := lbrace (s,t) in (a,b)^2: s < trbrace$.
Therefore, $$int_a^b int_a^t f(s)g(t) ,ds,dt = iint_D f(s)g(t) ,d(s,t) = int_a^b int_s^b f(s)g(t) ,dt,ds.$$
If you let $t in(a,b)$ arbitrary, you want $a < s < t$ to hold and if $sin (a,b)$ is some real number, you want $s < t <b$.
answered Nov 23 at 18:10
maliesen
18111
Your integration domain is $D := lbrace (s,t) in (a,b)^2: s < trbrace$.
Therefore, $$int_a^b int_a^t f(s)g(t) ,ds,dt = iint_D f(s)g(t) ,d(s,t) = int_a^b int_s^b f(s)g(t) ,dt,ds.$$
If you let $t in(a,b)$ arbitrary, you want $a < s < t$ to hold and if $sin (a,b)$ is some real number, you want $s < t <b$.
answered Nov 23 at 18:10
maliesen
18111
answered Nov 23 at 18:10
maliesen
18111
answered Nov 23 at 18:10
maliesen
18111
answered Nov 23 at 18:10
maliesen
18111
18111
add a comment |
add a comment |
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