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Small question on improper complex integral intuition
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Recently, I am dealing with an integral
$$int^infty_a+bif(z)dz$$ where $f(z)$ is a meromorphic function on $mathbb C$.
The $infty$ shall be understood as the real positive infinity, and conventionally the integral is along a straight line connecting the two endpoints.
When I draw out the integral path, I got 
The path is the hypotenuse of a right angled triangle with an infinitely long base and a finite height. Thus, the slope of the integral path is zero.
So, intuitively,
$$int^infty_a+bif(z)dz=int^infty+bi_a+bif(z)dz$$ and the integral path can be described as $Im(z)=b, Re(z)ge a$.
Is the equality (trivially) true? If yes, how can one prove it mathematically?
complex-integration
asked Jul 30 at 2:28
Szeto
3,8431421
add a comment |Â
up vote
0
down vote
favorite
Recently, I am dealing with an integral
$$int^infty_a+bif(z)dz$$ where $f(z)$ is a meromorphic function on $mathbb C$.
The $infty$ shall be understood as the real positive infinity, and conventionally the integral is along a straight line connecting the two endpoints.
When I draw out the integral path, I got
The path is the hypotenuse of a right angled triangle with an infinitely long base and a finite height. Thus, the slope of the integral path is zero.
So, intuitively,
$$int^infty_a+bif(z)dz=int^infty+bi_a+bif(z)dz$$ and the integral path can be described as $Im(z)=b, Re(z)ge a$.
Is the equality (trivially) true? If yes, how can one prove it mathematically?
complex-integration
asked Jul 30 at 2:28
Szeto
3,8431421
It depends on what you need to prove (you gave the definition of the r.h.s. but left the l.h.s. undefined). Anyway, questions of this kind are (usually) resolved by this theorem.
– metamorphy
Jul 30 at 8:46
@metamorphy Okay...So I essentially think that no poles can lie within the triangle with vertices $(a,b)$ $(R,0)$ $(R,b)$ where $Rto+infty$. Is that always true?
– Szeto
Jul 30 at 9:36
If the l.h.s. is defined to be $displaystylelim_Rto+inftyint_a+bi^R f(z) dz$ with straight line as the path of integration, then yes, under these conditions it is true (provided that the r.h.s. exists at all).
– metamorphy
Jul 30 at 10:01
@metamorphy Thank you. Can you provide a proof?
– Szeto
Jul 30 at 10:20
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Recently, I am dealing with an integral
$$int^infty_a+bif(z)dz$$ where $f(z)$ is a meromorphic function on $mathbb C$.
The $infty$ shall be understood as the real positive infinity, and conventionally the integral is along a straight line connecting the two endpoints.
When I draw out the integral path, I got
The path is the hypotenuse of a right angled triangle with an infinitely long base and a finite height. Thus, the slope of the integral path is zero.
So, intuitively,
$$int^infty_a+bif(z)dz=int^infty+bi_a+bif(z)dz$$ and the integral path can be described as $Im(z)=b, Re(z)ge a$.
Is the equality (trivially) true? If yes, how can one prove it mathematically?
complex-integration
asked Jul 30 at 2:28
Szeto
3,8431421
Recently, I am dealing with an integral
$$int^infty_a+bif(z)dz$$ where $f(z)$ is a meromorphic function on $mathbb C$.
The $infty$ shall be understood as the real positive infinity, and conventionally the integral is along a straight line connecting the two endpoints.
When I draw out the integral path, I got
The path is the hypotenuse of a right angled triangle with an infinitely long base and a finite height. Thus, the slope of the integral path is zero.
So, intuitively,
$$int^infty_a+bif(z)dz=int^infty+bi_a+bif(z)dz$$ and the integral path can be described as $Im(z)=b, Re(z)ge a$.
Is the equality (trivially) true? If yes, how can one prove it mathematically?
complex-integration
asked Jul 30 at 2:28
Szeto
3,8431421
asked Jul 30 at 2:28
Szeto
3,8431421
asked Jul 30 at 2:28
Szeto
3,8431421
asked Jul 30 at 2:28
Szeto
3,8431421
3,8431421
It depends on what you need to prove (you gave the definition of the r.h.s. but left the l.h.s. undefined). Anyway, questions of this kind are (usually) resolved by this theorem.
– metamorphy
Jul 30 at 8:46
@metamorphy Okay...So I essentially think that no poles can lie within the triangle with vertices $(a,b)$ $(R,0)$ $(R,b)$ where $Rto+infty$. Is that always true?
– Szeto
Jul 30 at 9:36
If the l.h.s. is defined to be $displaystylelim_Rto+inftyint_a+bi^R f(z) dz$ with straight line as the path of integration, then yes, under these conditions it is true (provided that the r.h.s. exists at all).
– metamorphy
Jul 30 at 10:01
@metamorphy Thank you. Can you provide a proof?
– Szeto
Jul 30 at 10:20
add a comment |Â
It depends on what you need to prove (you gave the definition of the r.h.s. but left the l.h.s. undefined). Anyway, questions of this kind are (usually) resolved by this theorem.
– metamorphy
Jul 30 at 8:46
@metamorphy Okay...So I essentially think that no poles can lie within the triangle with vertices $(a,b)$ $(R,0)$ $(R,b)$ where $Rto+infty$. Is that always true?
– Szeto
Jul 30 at 9:36
If the l.h.s. is defined to be $displaystylelim_Rto+inftyint_a+bi^R f(z) dz$ with straight line as the path of integration, then yes, under these conditions it is true (provided that the r.h.s. exists at all).
– metamorphy
Jul 30 at 10:01
@metamorphy Thank you. Can you provide a proof?
– Szeto
Jul 30 at 10:20
It depends on what you need to prove (you gave the definition of the r.h.s. but left the l.h.s. undefined). Anyway, questions of this kind are (usually) resolved by this theorem.
– metamorphy
Jul 30 at 8:46
It depends on what you need to prove (you gave the definition of the r.h.s. but left the l.h.s. undefined). Anyway, questions of this kind are (usually) resolved by this theorem.
– metamorphy
Jul 30 at 8:46
@metamorphy Okay...So I essentially think that no poles can lie within the triangle with vertices $(a,b)$ $(R,0)$ $(R,b)$ where $Rto+infty$. Is that always true?
– Szeto
Jul 30 at 9:36
@metamorphy Okay...So I essentially think that no poles can lie within the triangle with vertices $(a,b)$ $(R,0)$ $(R,b)$ where $Rto+infty$. Is that always true?
– Szeto
Jul 30 at 9:36
If the l.h.s. is defined to be $displaystylelim_Rto+inftyint_a+bi^R f(z) dz$ with straight line as the path of integration, then yes, under these conditions it is true (provided that the r.h.s. exists at all).
– metamorphy
Jul 30 at 10:01
If the l.h.s. is defined to be $displaystylelim_Rto+inftyint_a+bi^R f(z) dz$ with straight line as the path of integration, then yes, under these conditions it is true (provided that the r.h.s. exists at all).
– metamorphy
Jul 30 at 10:01
@metamorphy Thank you. Can you provide a proof?
– Szeto
Jul 30 at 10:20
@metamorphy Thank you. Can you provide a proof?
– Szeto
Jul 30 at 10:20
add a comment |Â
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It depends on what you need to prove (you gave the definition of the r.h.s. but left the l.h.s. undefined). Anyway, questions of this kind are (usually) resolved by this theorem.
– metamorphy
Jul 30 at 8:46
@metamorphy Okay...So I essentially think that no poles can lie within the triangle with vertices $(a,b)$ $(R,0)$ $(R,b)$ where $Rto+infty$. Is that always true?
– Szeto
Jul 30 at 9:36
If the l.h.s. is defined to be $displaystylelim_Rto+inftyint_a+bi^R f(z) dz$ with straight line as the path of integration, then yes, under these conditions it is true (provided that the r.h.s. exists at all).
– metamorphy
Jul 30 at 10:01
@metamorphy Thank you. Can you provide a proof?
– Szeto
Jul 30 at 10:20