Why does $int_{-infty}^infty R(x) dx$ converge iff the rational function $R(x)$ has degree of denom. at least...
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I am readinf Ahlfors and came across the fact that:
$int_{-infty}^infty R(x) dx$, where $R(x)$ is a rational function, converges if and only if in the rational function $R(x)$ the degree of the denominator is at least two units higher than the degree of the numerator.
I am unsure of how to prove this fact rigorously. I do get that the condition on the degrees means roughly that $R(x)$ is approximately $frac cx$ (whose integral diverges) with $c$ a constant, but could anyone post a rigorus proof of this?
real-analysis complex-analysis rational-functions
asked Nov 19 at 9:38
Cute Brownie
980316
add a comment |
up vote
0
down vote
favorite
I am readinf Ahlfors and came across the fact that:
$int_{-infty}^infty R(x) dx$, where $R(x)$ is a rational function, converges if and only if in the rational function $R(x)$ the degree of the denominator is at least two units higher than the degree of the numerator.
I am unsure of how to prove this fact rigorously. I do get that the condition on the degrees means roughly that $R(x)$ is approximately $frac cx$ (whose integral diverges) with $c$ a constant, but could anyone post a rigorus proof of this?
real-analysis complex-analysis rational-functions
asked Nov 19 at 9:38
Cute Brownie
980316
You mean $int_1^infty$, right? It's basically because $int_1^xfrac{dx} x$ doesn't converge, yet $int_1^xfrac{dx} {x^2}$ does.
– Ennar
Nov 19 at 9:44
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
Nov 19 at 9:45
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
Nov 19 at 9:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am readinf Ahlfors and came across the fact that:
$int_{-infty}^infty R(x) dx$, where $R(x)$ is a rational function, converges if and only if in the rational function $R(x)$ the degree of the denominator is at least two units higher than the degree of the numerator.
I am unsure of how to prove this fact rigorously. I do get that the condition on the degrees means roughly that $R(x)$ is approximately $frac cx$ (whose integral diverges) with $c$ a constant, but could anyone post a rigorus proof of this?
real-analysis complex-analysis rational-functions
asked Nov 19 at 9:38
Cute Brownie
980316
I am readinf Ahlfors and came across the fact that:
$int_{-infty}^infty R(x) dx$, where $R(x)$ is a rational function, converges if and only if in the rational function $R(x)$ the degree of the denominator is at least two units higher than the degree of the numerator.
I am unsure of how to prove this fact rigorously. I do get that the condition on the degrees means roughly that $R(x)$ is approximately $frac cx$ (whose integral diverges) with $c$ a constant, but could anyone post a rigorus proof of this?
real-analysis complex-analysis rational-functions
real-analysis complex-analysis rational-functions
asked Nov 19 at 9:38
Cute Brownie
980316
asked Nov 19 at 9:38
Cute Brownie
980316
edited Nov 19 at 9:49
asked Nov 19 at 9:38
Cute Brownie
980316
asked Nov 19 at 9:38
Cute Brownie
980316
asked Nov 19 at 9:38
Cute Brownie
980316
980316
You mean $int_1^infty$, right? It's basically because $int_1^xfrac{dx} x$ doesn't converge, yet $int_1^xfrac{dx} {x^2}$ does.
– Ennar
Nov 19 at 9:44
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
Nov 19 at 9:45
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
Nov 19 at 9:45
add a comment |
You mean $int_1^infty$, right? It's basically because $int_1^xfrac{dx} x$ doesn't converge, yet $int_1^xfrac{dx} {x^2}$ does.
– Ennar
Nov 19 at 9:44
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
Nov 19 at 9:45
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
Nov 19 at 9:45
You mean $int_1^infty$, right? It's basically because $int_1^xfrac{dx} x$ doesn't converge, yet $int_1^xfrac{dx} {x^2}$ does.
– Ennar
Nov 19 at 9:44
You mean $int_1^infty$, right? It's basically because $int_1^xfrac{dx} x$ doesn't converge, yet $int_1^xfrac{dx} {x^2}$ does.
– Ennar
Nov 19 at 9:44
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
Nov 19 at 9:45
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
Nov 19 at 9:45
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
Nov 19 at 9:45
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
Nov 19 at 9:45
add a comment |
1 Answer
1
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3
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You need an extra assumption on zeros of the denominator. For example $frac 1 {x^{2}}$ is not integrable. Let $p$ and $q$ are polynomials and degree of $q$ is at least $2$ more than the degree of $p$ and let us assume that the denominator has no zeros on $mathbb R$. Then there is a constant $C$ such that $|frac p q |leq frac C {x^{2}}$ for $|x|$ sufficiently large and this makes $R=frac p q$ integrable.
answered Nov 19 at 9:45
Kavi Rama Murthy
45.8k31853
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You need an extra assumption on zeros of the denominator. For example $frac 1 {x^{2}}$ is not integrable. Let $p$ and $q$ are polynomials and degree of $q$ is at least $2$ more than the degree of $p$ and let us assume that the denominator has no zeros on $mathbb R$. Then there is a constant $C$ such that $|frac p q |leq frac C {x^{2}}$ for $|x|$ sufficiently large and this makes $R=frac p q$ integrable.
answered Nov 19 at 9:45
Kavi Rama Murthy
45.8k31853
add a comment |
up vote
3
down vote
You need an extra assumption on zeros of the denominator. For example $frac 1 {x^{2}}$ is not integrable. Let $p$ and $q$ are polynomials and degree of $q$ is at least $2$ more than the degree of $p$ and let us assume that the denominator has no zeros on $mathbb R$. Then there is a constant $C$ such that $|frac p q |leq frac C {x^{2}}$ for $|x|$ sufficiently large and this makes $R=frac p q$ integrable.
answered Nov 19 at 9:45
Kavi Rama Murthy
45.8k31853
add a comment |
up vote
3
down vote
up vote
3
down vote
You need an extra assumption on zeros of the denominator. For example $frac 1 {x^{2}}$ is not integrable. Let $p$ and $q$ are polynomials and degree of $q$ is at least $2$ more than the degree of $p$ and let us assume that the denominator has no zeros on $mathbb R$. Then there is a constant $C$ such that $|frac p q |leq frac C {x^{2}}$ for $|x|$ sufficiently large and this makes $R=frac p q$ integrable.
answered Nov 19 at 9:45
Kavi Rama Murthy
45.8k31853
You need an extra assumption on zeros of the denominator. For example $frac 1 {x^{2}}$ is not integrable. Let $p$ and $q$ are polynomials and degree of $q$ is at least $2$ more than the degree of $p$ and let us assume that the denominator has no zeros on $mathbb R$. Then there is a constant $C$ such that $|frac p q |leq frac C {x^{2}}$ for $|x|$ sufficiently large and this makes $R=frac p q$ integrable.
answered Nov 19 at 9:45
Kavi Rama Murthy
45.8k31853
answered Nov 19 at 9:45
Kavi Rama Murthy
45.8k31853
answered Nov 19 at 9:45
Kavi Rama Murthy
45.8k31853
answered Nov 19 at 9:45
Kavi Rama Murthy
45.8k31853
45.8k31853
add a comment |
add a comment |
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You mean $int_1^infty$, right? It's basically because $int_1^xfrac{dx} x$ doesn't converge, yet $int_1^xfrac{dx} {x^2}$ does.
– Ennar
Nov 19 at 9:44
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
Nov 19 at 9:45
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
Nov 19 at 9:45