Small question on improper complex integral intuition

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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




share|cite|improve this question



















  • 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














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




share|cite|improve this question



















  • 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












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




share|cite|improve this question











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





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









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
















  • 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















active

oldest

votes











Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866615%2fsmall-question-on-improper-complex-integral-intuition%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes










 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866615%2fsmall-question-on-improper-complex-integral-intuition%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more