How to show that the following integral is infinite
For an application I want to show the following claim:
Let $xin(0,1)$ and $yin[0,1]$ be two parameters, then
$$int_{mathbb{R}}frac{1+cos(pi x (y-0.5))|z|^{x}}{1+2cos(pi x (y-0.5))|z|^x+|z|^{2x}}, dz=infty$$
holds true. I have tried to use the facts, that $lambda:=pi x (y-0.5)in(-pi/2,pi/2)(rightarrow cos(lambda)in(0,1))$ and
$$int_{1}^{infty}z^beta,dz=inftyquadtext{iff}quad beta>-1$$ but I didn't get it.
real-analysis estimation
asked Dec 4 '18 at 20:58
Math95Math95
113
add a comment |
For an application I want to show the following claim:
Let $xin(0,1)$ and $yin[0,1]$ be two parameters, then
$$int_{mathbb{R}}frac{1+cos(pi x (y-0.5))|z|^{x}}{1+2cos(pi x (y-0.5))|z|^x+|z|^{2x}}, dz=infty$$
holds true. I have tried to use the facts, that $lambda:=pi x (y-0.5)in(-pi/2,pi/2)(rightarrow cos(lambda)in(0,1))$ and
$$int_{1}^{infty}z^beta,dz=inftyquadtext{iff}quad beta>-1$$ but I didn't get it.
real-analysis estimation
asked Dec 4 '18 at 20:58
Math95Math95
113
What do we know about $x,y?$ For large $z$, the integrand is about $|z|^{-x}$, so if $x lt 1$ you are done.
– Ross Millikan
Dec 4 '18 at 22:10
Thank you. Indeed $x<1$. This was also the plan I wanted to show. But how to show formally that the integrand is about $|z|^{-x}$ for large $z$?
– Math95
Dec 5 '18 at 9:48
Just look at the leading terms. The numerator goes like $|z|^x$ with the cosine constant out front. The denominator goes like $|z|^{2x}$. For large $z$ the other terms do not matter. Convergence is determined by what happens at large $z$ unless there is a non-integrable pole somewhere.
– Ross Millikan
Dec 5 '18 at 14:40
add a comment |
For an application I want to show the following claim:
Let $xin(0,1)$ and $yin[0,1]$ be two parameters, then
$$int_{mathbb{R}}frac{1+cos(pi x (y-0.5))|z|^{x}}{1+2cos(pi x (y-0.5))|z|^x+|z|^{2x}}, dz=infty$$
holds true. I have tried to use the facts, that $lambda:=pi x (y-0.5)in(-pi/2,pi/2)(rightarrow cos(lambda)in(0,1))$ and
$$int_{1}^{infty}z^beta,dz=inftyquadtext{iff}quad beta>-1$$ but I didn't get it.
real-analysis estimation
asked Dec 4 '18 at 20:58
Math95Math95
113
For an application I want to show the following claim:
Let $xin(0,1)$ and $yin[0,1]$ be two parameters, then
$$int_{mathbb{R}}frac{1+cos(pi x (y-0.5))|z|^{x}}{1+2cos(pi x (y-0.5))|z|^x+|z|^{2x}}, dz=infty$$
holds true. I have tried to use the facts, that $lambda:=pi x (y-0.5)in(-pi/2,pi/2)(rightarrow cos(lambda)in(0,1))$ and
$$int_{1}^{infty}z^beta,dz=inftyquadtext{iff}quad beta>-1$$ but I didn't get it.
real-analysis estimation
real-analysis estimation
asked Dec 4 '18 at 20:58
Math95Math95
113
asked Dec 4 '18 at 20:58
Math95Math95
113
asked Dec 4 '18 at 20:58
Math95Math95
113
asked Dec 4 '18 at 20:58
Math95Math95
113
asked Dec 4 '18 at 20:58
Math95Math95
113
113
What do we know about $x,y?$ For large $z$, the integrand is about $|z|^{-x}$, so if $x lt 1$ you are done.
– Ross Millikan
Dec 4 '18 at 22:10
Thank you. Indeed $x<1$. This was also the plan I wanted to show. But how to show formally that the integrand is about $|z|^{-x}$ for large $z$?
– Math95
Dec 5 '18 at 9:48
Just look at the leading terms. The numerator goes like $|z|^x$ with the cosine constant out front. The denominator goes like $|z|^{2x}$. For large $z$ the other terms do not matter. Convergence is determined by what happens at large $z$ unless there is a non-integrable pole somewhere.
– Ross Millikan
Dec 5 '18 at 14:40
add a comment |
What do we know about $x,y?$ For large $z$, the integrand is about $|z|^{-x}$, so if $x lt 1$ you are done.
– Ross Millikan
Dec 4 '18 at 22:10
Thank you. Indeed $x<1$. This was also the plan I wanted to show. But how to show formally that the integrand is about $|z|^{-x}$ for large $z$?
– Math95
Dec 5 '18 at 9:48
Just look at the leading terms. The numerator goes like $|z|^x$ with the cosine constant out front. The denominator goes like $|z|^{2x}$. For large $z$ the other terms do not matter. Convergence is determined by what happens at large $z$ unless there is a non-integrable pole somewhere.
– Ross Millikan
Dec 5 '18 at 14:40
What do we know about $x,y?$ For large $z$, the integrand is about $|z|^{-x}$, so if $x lt 1$ you are done.
– Ross Millikan
Dec 4 '18 at 22:10
What do we know about $x,y?$ For large $z$, the integrand is about $|z|^{-x}$, so if $x lt 1$ you are done.
– Ross Millikan
Dec 4 '18 at 22:10
Thank you. Indeed $x<1$. This was also the plan I wanted to show. But how to show formally that the integrand is about $|z|^{-x}$ for large $z$?
– Math95
Dec 5 '18 at 9:48
Thank you. Indeed $x<1$. This was also the plan I wanted to show. But how to show formally that the integrand is about $|z|^{-x}$ for large $z$?
– Math95
Dec 5 '18 at 9:48
Just look at the leading terms. The numerator goes like $|z|^x$ with the cosine constant out front. The denominator goes like $|z|^{2x}$. For large $z$ the other terms do not matter. Convergence is determined by what happens at large $z$ unless there is a non-integrable pole somewhere.
– Ross Millikan
Dec 5 '18 at 14:40
Just look at the leading terms. The numerator goes like $|z|^x$ with the cosine constant out front. The denominator goes like $|z|^{2x}$. For large $z$ the other terms do not matter. Convergence is determined by what happens at large $z$ unless there is a non-integrable pole somewhere.
– Ross Millikan
Dec 5 '18 at 14:40
add a comment |
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What do we know about $x,y?$ For large $z$, the integrand is about $|z|^{-x}$, so if $x lt 1$ you are done.
– Ross Millikan
Dec 4 '18 at 22:10
Thank you. Indeed $x<1$. This was also the plan I wanted to show. But how to show formally that the integrand is about $|z|^{-x}$ for large $z$?
– Math95
Dec 5 '18 at 9:48
Just look at the leading terms. The numerator goes like $|z|^x$ with the cosine constant out front. The denominator goes like $|z|^{2x}$. For large $z$ the other terms do not matter. Convergence is determined by what happens at large $z$ unless there is a non-integrable pole somewhere.
– Ross Millikan
Dec 5 '18 at 14:40