Mixing
Integrals: can we take limits ‘partially’?
Clash Royale CLAN TAG#URR8PPP
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Suppose I have an integral that converges properly
$$int^b_a f(t)dt$$ where $f(t)$ is a complex analytic function.
($a,binmathbb R, b>a$).
The integral is taken on a straight path.
I applied a substitution $u=frac1t-a+frac1R$, then we have the original integral equals to
$$int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du$$
My question is, under what conditions is this true?
$$int^b_a f(t)dt=lim_Rtoinfty int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du =lim_Rtoinfty int^R_frac1b-afracf(frac1u+a)u^2du$$
I apologize for not being able to provide much context, as I have no idea how to tackle this problem.
integration limits
asked 2 days ago
Szeto
3,8431421
add a comment |Â
up vote
1
down vote
favorite
Suppose I have an integral that converges properly
$$int^b_a f(t)dt$$ where $f(t)$ is a complex analytic function.
($a,binmathbb R, b>a$).
The integral is taken on a straight path.
I applied a substitution $u=frac1t-a+frac1R$, then we have the original integral equals to
$$int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du$$
My question is, under what conditions is this true?
$$int^b_a f(t)dt=lim_Rtoinfty int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du =lim_Rtoinfty int^R_frac1b-afracf(frac1u+a)u^2du$$
I apologize for not being able to provide much context, as I have no idea how to tackle this problem.
integration limits
asked 2 days ago
Szeto
3,8431421
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose I have an integral that converges properly
$$int^b_a f(t)dt$$ where $f(t)$ is a complex analytic function.
($a,binmathbb R, b>a$).
The integral is taken on a straight path.
I applied a substitution $u=frac1t-a+frac1R$, then we have the original integral equals to
$$int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du$$
My question is, under what conditions is this true?
$$int^b_a f(t)dt=lim_Rtoinfty int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du =lim_Rtoinfty int^R_frac1b-afracf(frac1u+a)u^2du$$
I apologize for not being able to provide much context, as I have no idea how to tackle this problem.
integration limits
asked 2 days ago
Szeto
3,8431421
Suppose I have an integral that converges properly
$$int^b_a f(t)dt$$ where $f(t)$ is a complex analytic function.
($a,binmathbb R, b>a$).
The integral is taken on a straight path.
I applied a substitution $u=frac1t-a+frac1R$, then we have the original integral equals to
$$int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du$$
My question is, under what conditions is this true?
$$int^b_a f(t)dt=lim_Rtoinfty int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du =lim_Rtoinfty int^R_frac1b-afracf(frac1u+a)u^2du$$
I apologize for not being able to provide much context, as I have no idea how to tackle this problem.
integration limits
asked 2 days ago
Szeto
3,8431421
asked 2 days ago
Szeto
3,8431421
asked 2 days ago
Szeto
3,8431421
asked 2 days ago
Szeto
3,8431421
3,8431421
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
After the change of variables we already have for any $R in (0,infty)$
$$int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
so
$$lim_R to inftyint^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
since for all $epsilon > 0$, regardless of $R$,
$$left|int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du - int^b_a f(t)dtright| = 0 < epsilon$$
This is not entirely trivial in that the validity of change-of-variables has some conditions. However, integrability of $f$ and monotonicity of $u mapsto 1/u +a - 1/R$ is sufficient.
For the second part apply the variable change $t = 1/u +a$ first to the integral over $[a + 1/R,b]$ to get
$$int_a + 1/R^b f(t) , dt = int^R_frac1b-afracf(frac1u+a)u^2du.$$
We always have $lim_c to a+int_c^b f(t) , dt = int_a^bf(t) , dt$ if $f$ is properly integrable. Hence,
$$int_a^b f(t) , dt = lim_R to infty int_a + 1/R^b f(t) , dt = lim_R to infty int^R_frac1b-afracf(frac1u+a)u^2du$$
answered 2 days ago
RRL
43.4k42160
Thank you for your answer, but it is not complete yet because there are two equalities in the yellow box. Could you please also prove the second equality?
– Szeto
2 days ago
@Szeto: Sure -- that can be done.
– RRL
2 days ago
Tons of thanks to you!
– Szeto
2 days ago
@Szeto: You're welcome. A most unusual question!
– RRL
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
After the change of variables we already have for any $R in (0,infty)$
$$int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
so
$$lim_R to inftyint^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
since for all $epsilon > 0$, regardless of $R$,
$$left|int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du - int^b_a f(t)dtright| = 0 < epsilon$$
This is not entirely trivial in that the validity of change-of-variables has some conditions. However, integrability of $f$ and monotonicity of $u mapsto 1/u +a - 1/R$ is sufficient.
For the second part apply the variable change $t = 1/u +a$ first to the integral over $[a + 1/R,b]$ to get
$$int_a + 1/R^b f(t) , dt = int^R_frac1b-afracf(frac1u+a)u^2du.$$
We always have $lim_c to a+int_c^b f(t) , dt = int_a^bf(t) , dt$ if $f$ is properly integrable. Hence,
$$int_a^b f(t) , dt = lim_R to infty int_a + 1/R^b f(t) , dt = lim_R to infty int^R_frac1b-afracf(frac1u+a)u^2du$$
answered 2 days ago
RRL
43.4k42160
Thank you for your answer, but it is not complete yet because there are two equalities in the yellow box. Could you please also prove the second equality?
– Szeto
2 days ago
@Szeto: Sure -- that can be done.
– RRL
2 days ago
Tons of thanks to you!
– Szeto
2 days ago
@Szeto: You're welcome. A most unusual question!
– RRL
2 days ago
add a comment |Â
up vote
1
down vote
accepted
After the change of variables we already have for any $R in (0,infty)$
$$int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
so
$$lim_R to inftyint^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
since for all $epsilon > 0$, regardless of $R$,
$$left|int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du - int^b_a f(t)dtright| = 0 < epsilon$$
This is not entirely trivial in that the validity of change-of-variables has some conditions. However, integrability of $f$ and monotonicity of $u mapsto 1/u +a - 1/R$ is sufficient.
For the second part apply the variable change $t = 1/u +a$ first to the integral over $[a + 1/R,b]$ to get
$$int_a + 1/R^b f(t) , dt = int^R_frac1b-afracf(frac1u+a)u^2du.$$
We always have $lim_c to a+int_c^b f(t) , dt = int_a^bf(t) , dt$ if $f$ is properly integrable. Hence,
$$int_a^b f(t) , dt = lim_R to infty int_a + 1/R^b f(t) , dt = lim_R to infty int^R_frac1b-afracf(frac1u+a)u^2du$$
answered 2 days ago
RRL
43.4k42160
Thank you for your answer, but it is not complete yet because there are two equalities in the yellow box. Could you please also prove the second equality?
– Szeto
2 days ago
@Szeto: Sure -- that can be done.
– RRL
2 days ago
Tons of thanks to you!
– Szeto
2 days ago
@Szeto: You're welcome. A most unusual question!
– RRL
2 days ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
After the change of variables we already have for any $R in (0,infty)$
$$int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
so
$$lim_R to inftyint^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
since for all $epsilon > 0$, regardless of $R$,
$$left|int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du - int^b_a f(t)dtright| = 0 < epsilon$$
This is not entirely trivial in that the validity of change-of-variables has some conditions. However, integrability of $f$ and monotonicity of $u mapsto 1/u +a - 1/R$ is sufficient.
For the second part apply the variable change $t = 1/u +a$ first to the integral over $[a + 1/R,b]$ to get
$$int_a + 1/R^b f(t) , dt = int^R_frac1b-afracf(frac1u+a)u^2du.$$
We always have $lim_c to a+int_c^b f(t) , dt = int_a^bf(t) , dt$ if $f$ is properly integrable. Hence,
$$int_a^b f(t) , dt = lim_R to infty int_a + 1/R^b f(t) , dt = lim_R to infty int^R_frac1b-afracf(frac1u+a)u^2du$$
answered 2 days ago
RRL
43.4k42160
After the change of variables we already have for any $R in (0,infty)$
$$int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
so
$$lim_R to inftyint^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du = int^b_a f(t)dt,$$
since for all $epsilon > 0$, regardless of $R$,
$$left|int^R_frac1b-a+frac1Rfracf(frac1u+a-frac1R)u^2du - int^b_a f(t)dtright| = 0 < epsilon$$
This is not entirely trivial in that the validity of change-of-variables has some conditions. However, integrability of $f$ and monotonicity of $u mapsto 1/u +a - 1/R$ is sufficient.
For the second part apply the variable change $t = 1/u +a$ first to the integral over $[a + 1/R,b]$ to get
$$int_a + 1/R^b f(t) , dt = int^R_frac1b-afracf(frac1u+a)u^2du.$$
We always have $lim_c to a+int_c^b f(t) , dt = int_a^bf(t) , dt$ if $f$ is properly integrable. Hence,
$$int_a^b f(t) , dt = lim_R to infty int_a + 1/R^b f(t) , dt = lim_R to infty int^R_frac1b-afracf(frac1u+a)u^2du$$
answered 2 days ago
RRL
43.4k42160
edited 2 days ago
answered 2 days ago
RRL
43.4k42160
answered 2 days ago
RRL
43.4k42160
answered 2 days ago
RRL
43.4k42160
43.4k42160
Thank you for your answer, but it is not complete yet because there are two equalities in the yellow box. Could you please also prove the second equality?
– Szeto
2 days ago
@Szeto: Sure -- that can be done.
– RRL
2 days ago
Tons of thanks to you!
– Szeto
2 days ago
@Szeto: You're welcome. A most unusual question!
– RRL
2 days ago
add a comment |Â
Thank you for your answer, but it is not complete yet because there are two equalities in the yellow box. Could you please also prove the second equality?
– Szeto
2 days ago
@Szeto: Sure -- that can be done.
– RRL
2 days ago
Tons of thanks to you!
– Szeto
2 days ago
@Szeto: You're welcome. A most unusual question!
– RRL
2 days ago
Thank you for your answer, but it is not complete yet because there are two equalities in the yellow box. Could you please also prove the second equality?
– Szeto
2 days ago
Thank you for your answer, but it is not complete yet because there are two equalities in the yellow box. Could you please also prove the second equality?
– Szeto
2 days ago
@Szeto: Sure -- that can be done.
– RRL
2 days ago
@Szeto: Sure -- that can be done.
– RRL
2 days ago
Tons of thanks to you!
– Szeto
2 days ago
Tons of thanks to you!
– Szeto
2 days ago
@Szeto: You're welcome. A most unusual question!
– RRL
2 days ago
@Szeto: You're welcome. A most unusual question!
– RRL
2 days ago
add a comment |Â
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