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 17 hours ago
Cute Brownie
936316
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 17 hours ago
Cute Brownie
936316
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
17 hours ago
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
17 hours ago
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
17 hours ago
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 17 hours ago
Cute Brownie
936316
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 17 hours ago
Cute Brownie
936316
asked 17 hours ago
Cute Brownie
936316
edited 17 hours ago
asked 17 hours ago
Cute Brownie
936316
asked 17 hours ago
Cute Brownie
936316
asked 17 hours ago
Cute Brownie
936316
936316
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
17 hours ago
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
17 hours ago
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
17 hours ago
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
17 hours ago
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
17 hours ago
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
17 hours ago
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
17 hours ago
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
17 hours ago
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
17 hours ago
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
17 hours ago
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
17 hours ago
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
17 hours ago
add a comment |
1 Answer
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up vote
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 17 hours ago
Kavi Rama Murthy
40k31750
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 17 hours ago
Kavi Rama Murthy
40k31750
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 17 hours ago
Kavi Rama Murthy
40k31750
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 17 hours ago
Kavi Rama Murthy
40k31750
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 17 hours ago
Kavi Rama Murthy
40k31750
answered 17 hours ago
Kavi Rama Murthy
40k31750
answered 17 hours ago
Kavi Rama Murthy
40k31750
answered 17 hours ago
Kavi Rama Murthy
40k31750
40k31750
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
17 hours ago
$int_{-infty}^inftyfrac1{x^2}dx$ doesn't converge, though. However, that's because of the pole in the middle, not because of the tails at $pminfty$.
– Arthur
17 hours ago
@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
17 hours ago