Absolute convergence of an complex integral
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I would like to show that the integral $$I(t)=int_{B-iinfty}^{B+iinfty}left|at^2+frac{M}{4a}right|^{-(1+s)}|e^{2pi nt}|dt$$ converges for $Re(s)>-frac{1}{2},B>0, ain mathbb{Z}, n,Min mathbb{N}$ and satisfies an uniform estimate. Unfortunately, I do not know how to start, because I did not find anything about the convergence of a complex integral. Furthermore, I am not sure what is meant by uniform estimate meaning from what is the estimate independent. Note that I do not need the evaluation of the integral. I am thankful for every help and hint.
Using the tip I got begin{align*}I(t)&=iint_{-infty}^{infty}left|aB^2+2aBix-x^2+frac{M}{4a}right|^{-(1+s)}underbrace{|e^{2pi n ix}|}_{=1}|e^{2pi nB}|dx\&= i|e^{2pi nB}|int_{-infty}^{infty}left|-x^2+2aBix+aB^2+frac{M}{4a}right|^{-(1+s)}dx.end{align*} I would say that the integrand is of order $O(x^{-2-s})$. However, I am not sure how to formally show that this integral is now convergent since it is still over the whole real line.
integration complex-analysis
asked Nov 16 at 1:25
Masterexe
53
add a comment |
up vote
0
down vote
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I would like to show that the integral $$I(t)=int_{B-iinfty}^{B+iinfty}left|at^2+frac{M}{4a}right|^{-(1+s)}|e^{2pi nt}|dt$$ converges for $Re(s)>-frac{1}{2},B>0, ain mathbb{Z}, n,Min mathbb{N}$ and satisfies an uniform estimate. Unfortunately, I do not know how to start, because I did not find anything about the convergence of a complex integral. Furthermore, I am not sure what is meant by uniform estimate meaning from what is the estimate independent. Note that I do not need the evaluation of the integral. I am thankful for every help and hint.
Using the tip I got begin{align*}I(t)&=iint_{-infty}^{infty}left|aB^2+2aBix-x^2+frac{M}{4a}right|^{-(1+s)}underbrace{|e^{2pi n ix}|}_{=1}|e^{2pi nB}|dx\&= i|e^{2pi nB}|int_{-infty}^{infty}left|-x^2+2aBix+aB^2+frac{M}{4a}right|^{-(1+s)}dx.end{align*} I would say that the integrand is of order $O(x^{-2-s})$. However, I am not sure how to formally show that this integral is now convergent since it is still over the whole real line.
integration complex-analysis
asked Nov 16 at 1:25
Masterexe
53
Try changing variables $t = B + i x$ so it becomes an integral over the real line.
– eyeballfrog
Nov 16 at 1:39
Hi, I tried the changing of variables and edited it in the question. Was this the correct approach?
– Masterexe
Nov 16 at 2:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to show that the integral $$I(t)=int_{B-iinfty}^{B+iinfty}left|at^2+frac{M}{4a}right|^{-(1+s)}|e^{2pi nt}|dt$$ converges for $Re(s)>-frac{1}{2},B>0, ain mathbb{Z}, n,Min mathbb{N}$ and satisfies an uniform estimate. Unfortunately, I do not know how to start, because I did not find anything about the convergence of a complex integral. Furthermore, I am not sure what is meant by uniform estimate meaning from what is the estimate independent. Note that I do not need the evaluation of the integral. I am thankful for every help and hint.
Using the tip I got begin{align*}I(t)&=iint_{-infty}^{infty}left|aB^2+2aBix-x^2+frac{M}{4a}right|^{-(1+s)}underbrace{|e^{2pi n ix}|}_{=1}|e^{2pi nB}|dx\&= i|e^{2pi nB}|int_{-infty}^{infty}left|-x^2+2aBix+aB^2+frac{M}{4a}right|^{-(1+s)}dx.end{align*} I would say that the integrand is of order $O(x^{-2-s})$. However, I am not sure how to formally show that this integral is now convergent since it is still over the whole real line.
integration complex-analysis
asked Nov 16 at 1:25
Masterexe
53
I would like to show that the integral $$I(t)=int_{B-iinfty}^{B+iinfty}left|at^2+frac{M}{4a}right|^{-(1+s)}|e^{2pi nt}|dt$$ converges for $Re(s)>-frac{1}{2},B>0, ain mathbb{Z}, n,Min mathbb{N}$ and satisfies an uniform estimate. Unfortunately, I do not know how to start, because I did not find anything about the convergence of a complex integral. Furthermore, I am not sure what is meant by uniform estimate meaning from what is the estimate independent. Note that I do not need the evaluation of the integral. I am thankful for every help and hint.
Using the tip I got begin{align*}I(t)&=iint_{-infty}^{infty}left|aB^2+2aBix-x^2+frac{M}{4a}right|^{-(1+s)}underbrace{|e^{2pi n ix}|}_{=1}|e^{2pi nB}|dx\&= i|e^{2pi nB}|int_{-infty}^{infty}left|-x^2+2aBix+aB^2+frac{M}{4a}right|^{-(1+s)}dx.end{align*} I would say that the integrand is of order $O(x^{-2-s})$. However, I am not sure how to formally show that this integral is now convergent since it is still over the whole real line.
integration complex-analysis
integration complex-analysis
asked Nov 16 at 1:25
Masterexe
53
asked Nov 16 at 1:25
Masterexe
53
edited Nov 16 at 1:59
asked Nov 16 at 1:25
Masterexe
53
asked Nov 16 at 1:25
Masterexe
53
asked Nov 16 at 1:25
Masterexe
53
53
Try changing variables $t = B + i x$ so it becomes an integral over the real line.
– eyeballfrog
Nov 16 at 1:39
Hi, I tried the changing of variables and edited it in the question. Was this the correct approach?
– Masterexe
Nov 16 at 2:00
add a comment |
Try changing variables $t = B + i x$ so it becomes an integral over the real line.
– eyeballfrog
Nov 16 at 1:39
Hi, I tried the changing of variables and edited it in the question. Was this the correct approach?
– Masterexe
Nov 16 at 2:00
Try changing variables $t = B + i x$ so it becomes an integral over the real line.
– eyeballfrog
Nov 16 at 1:39
Try changing variables $t = B + i x$ so it becomes an integral over the real line.
– eyeballfrog
Nov 16 at 1:39
Hi, I tried the changing of variables and edited it in the question. Was this the correct approach?
– Masterexe
Nov 16 at 2:00
Hi, I tried the changing of variables and edited it in the question. Was this the correct approach?
– Masterexe
Nov 16 at 2:00
add a comment |
1 Answer
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The integrand is continuous on $mathbb{R}$ and $O(|x|^{-2(1-Re{s})})$ when $xgg 1$ which is integrable because the exponent is $>1$ this concludes your argument
answered Nov 16 at 4:59
Ezy
54429
The function is continuous. So on any finite domain the function is integrable. What matters is domination for large $|x|$
– Ezy
Nov 16 at 12:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The integrand is continuous on $mathbb{R}$ and $O(|x|^{-2(1-Re{s})})$ when $xgg 1$ which is integrable because the exponent is $>1$ this concludes your argument
answered Nov 16 at 4:59
Ezy
54429
The function is continuous. So on any finite domain the function is integrable. What matters is domination for large $|x|$
– Ezy
Nov 16 at 12:56
add a comment |
up vote
0
down vote
accepted
The integrand is continuous on $mathbb{R}$ and $O(|x|^{-2(1-Re{s})})$ when $xgg 1$ which is integrable because the exponent is $>1$ this concludes your argument
answered Nov 16 at 4:59
Ezy
54429
The function is continuous. So on any finite domain the function is integrable. What matters is domination for large $|x|$
– Ezy
Nov 16 at 12:56
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The integrand is continuous on $mathbb{R}$ and $O(|x|^{-2(1-Re{s})})$ when $xgg 1$ which is integrable because the exponent is $>1$ this concludes your argument
answered Nov 16 at 4:59
Ezy
54429
The integrand is continuous on $mathbb{R}$ and $O(|x|^{-2(1-Re{s})})$ when $xgg 1$ which is integrable because the exponent is $>1$ this concludes your argument
answered Nov 16 at 4:59
Ezy
54429
answered Nov 16 at 4:59
Ezy
54429
answered Nov 16 at 4:59
Ezy
54429
answered Nov 16 at 4:59
Ezy
54429
54429
The function is continuous. So on any finite domain the function is integrable. What matters is domination for large $|x|$
– Ezy
Nov 16 at 12:56
add a comment |
The function is continuous. So on any finite domain the function is integrable. What matters is domination for large $|x|$
– Ezy
Nov 16 at 12:56
The function is continuous. So on any finite domain the function is integrable. What matters is domination for large $|x|$
– Ezy
Nov 16 at 12:56
The function is continuous. So on any finite domain the function is integrable. What matters is domination for large $|x|$
– Ezy
Nov 16 at 12:56
add a comment |
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Try changing variables $t = B + i x$ so it becomes an integral over the real line.
– eyeballfrog
Nov 16 at 1:39
Hi, I tried the changing of variables and edited it in the question. Was this the correct approach?
– Masterexe
Nov 16 at 2:00