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







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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
    9 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
    9 hours ago












  • @Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
    – Cute Brownie
    9 hours ago















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







share|cite|improve this question
























  • 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
    9 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
    9 hours ago












  • @Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
    – Cute Brownie
    9 hours ago













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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago

























asked 9 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
    9 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
    9 hours ago












  • @Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
    – Cute Brownie
    9 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
    9 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
    9 hours ago












  • @Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
    – Cute Brownie
    9 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
9 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
9 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
9 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
9 hours ago














@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
9 hours ago




@Ennar thanks, would you elaborate using epsilon-delta approximations and I'll select yours as an answer?
– Cute Brownie
9 hours ago










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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.






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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.






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      up vote
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      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.






      share|cite|improve this answer























        up vote
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        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.






        share|cite|improve this answer












        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.







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        share|cite|improve this answer



        share|cite|improve this answer










        answered 9 hours ago









        Kavi Rama Murthy

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