Mixing
Complex integration lemma: shorter proof?
The black line is the branch cut.
Lemma
$$lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt$$ where $arg(z-s)in[theta,theta+2pi)$, $f$ being holomorphic on the path of integration.
Many advanced users on this site use this lemma without stating, letting alone proving it. I wrote a proof here, but it is quite long.
Is there a shorter proof of this lemma?
complex-analysis complex-integration
asked Nov 19 '18 at 0:27
Szeto
6,4362926
add a comment |
The black line is the branch cut.
Lemma
$$lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt$$ where $arg(z-s)in[theta,theta+2pi)$, $f$ being holomorphic on the path of integration.
Many advanced users on this site use this lemma without stating, letting alone proving it. I wrote a proof here, but it is quite long.
Is there a shorter proof of this lemma?
complex-analysis complex-integration
asked Nov 19 '18 at 0:27
Szeto
6,4362926
add a comment |
The black line is the branch cut.
Lemma
$$lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt$$ where $arg(z-s)in[theta,theta+2pi)$, $f$ being holomorphic on the path of integration.
Many advanced users on this site use this lemma without stating, letting alone proving it. I wrote a proof here, but it is quite long.
Is there a shorter proof of this lemma?
complex-analysis complex-integration
asked Nov 19 '18 at 0:27
Szeto
6,4362926
The black line is the branch cut.
Lemma
$$lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt$$ where $arg(z-s)in[theta,theta+2pi)$, $f$ being holomorphic on the path of integration.
Many advanced users on this site use this lemma without stating, letting alone proving it. I wrote a proof here, but it is quite long.
Is there a shorter proof of this lemma?
complex-analysis complex-integration
complex-analysis complex-integration
asked Nov 19 '18 at 0:27
Szeto
6,4362926
asked Nov 19 '18 at 0:27
Szeto
6,4362926
asked Nov 19 '18 at 0:27
Szeto
6,4362926
asked Nov 19 '18 at 0:27
Szeto
6,4362926
asked Nov 19 '18 at 0:27
Szeto
6,4362926
6,4362926
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I just found a short proof using integration by parts:
Let $hat k=ifrac{s}{|s|}$.
Let $P=pe^{itheta}, Q=qe^{itheta}$.
Let $P^{pm}=Ppm Deltahat k,Q^{pm}=Qpm Deltahat k$.
Let $F$ be the local antiderivative of $f$. (A local antiderivative exists due to local continuity.)
Then,
$$
begin{align}
&~~~~~lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-} -lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)frac{F(z)}{z-s}dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-}+0 \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+}^{P^-}
+lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{P^+}^{P^-}
+F(Q)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)(2pi i)+F(Q)(-2pi i) \
&=-2pi ibigg(F(Q)-F(P)bigg) \
&=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt
end{align}
$$
Q.E.D.
Essentially the proof is only 9 lines long.
answered Nov 20 '18 at 10:17
Szeto
6,4362926
1
You may accept your own answer(?)
– Tianlalu
Nov 30 '18 at 7:36
add a comment |
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004332%2fcomplex-integration-lemma-shorter-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I just found a short proof using integration by parts:
Let $hat k=ifrac{s}{|s|}$.
Let $P=pe^{itheta}, Q=qe^{itheta}$.
Let $P^{pm}=Ppm Deltahat k,Q^{pm}=Qpm Deltahat k$.
Let $F$ be the local antiderivative of $f$. (A local antiderivative exists due to local continuity.)
Then,
$$
begin{align}
&~~~~~lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-} -lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)frac{F(z)}{z-s}dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-}+0 \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+}^{P^-}
+lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{P^+}^{P^-}
+F(Q)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)(2pi i)+F(Q)(-2pi i) \
&=-2pi ibigg(F(Q)-F(P)bigg) \
&=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt
end{align}
$$
Q.E.D.
Essentially the proof is only 9 lines long.
answered Nov 20 '18 at 10:17
Szeto
6,4362926
1
You may accept your own answer(?)
– Tianlalu
Nov 30 '18 at 7:36
add a comment |
I just found a short proof using integration by parts:
Let $hat k=ifrac{s}{|s|}$.
Let $P=pe^{itheta}, Q=qe^{itheta}$.
Let $P^{pm}=Ppm Deltahat k,Q^{pm}=Qpm Deltahat k$.
Let $F$ be the local antiderivative of $f$. (A local antiderivative exists due to local continuity.)
Then,
$$
begin{align}
&~~~~~lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-} -lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)frac{F(z)}{z-s}dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-}+0 \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+}^{P^-}
+lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{P^+}^{P^-}
+F(Q)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)(2pi i)+F(Q)(-2pi i) \
&=-2pi ibigg(F(Q)-F(P)bigg) \
&=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt
end{align}
$$
Q.E.D.
Essentially the proof is only 9 lines long.
answered Nov 20 '18 at 10:17
Szeto
6,4362926
1
You may accept your own answer(?)
– Tianlalu
Nov 30 '18 at 7:36
add a comment |
I just found a short proof using integration by parts:
Let $hat k=ifrac{s}{|s|}$.
Let $P=pe^{itheta}, Q=qe^{itheta}$.
Let $P^{pm}=Ppm Deltahat k,Q^{pm}=Qpm Deltahat k$.
Let $F$ be the local antiderivative of $f$. (A local antiderivative exists due to local continuity.)
Then,
$$
begin{align}
&~~~~~lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-} -lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)frac{F(z)}{z-s}dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-}+0 \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+}^{P^-}
+lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{P^+}^{P^-}
+F(Q)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)(2pi i)+F(Q)(-2pi i) \
&=-2pi ibigg(F(Q)-F(P)bigg) \
&=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt
end{align}
$$
Q.E.D.
Essentially the proof is only 9 lines long.
answered Nov 20 '18 at 10:17
Szeto
6,4362926
I just found a short proof using integration by parts:
Let $hat k=ifrac{s}{|s|}$.
Let $P=pe^{itheta}, Q=qe^{itheta}$.
Let $P^{pm}=Ppm Deltahat k,Q^{pm}=Qpm Deltahat k$.
Let $F$ be the local antiderivative of $f$. (A local antiderivative exists due to local continuity.)
Then,
$$
begin{align}
&~~~~~lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-} -lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)frac{F(z)}{z-s}dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-}+0 \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+}^{P^-}
+lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{P^+}^{P^-}
+F(Q)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)(2pi i)+F(Q)(-2pi i) \
&=-2pi ibigg(F(Q)-F(P)bigg) \
&=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt
end{align}
$$
Q.E.D.
Essentially the proof is only 9 lines long.
answered Nov 20 '18 at 10:17
Szeto
6,4362926
edited Nov 22 '18 at 5:30
answered Nov 20 '18 at 10:17
Szeto
6,4362926
answered Nov 20 '18 at 10:17
Szeto
6,4362926
answered Nov 20 '18 at 10:17
Szeto
6,4362926
6,4362926
1
You may accept your own answer(?)
– Tianlalu
Nov 30 '18 at 7:36
add a comment |
1
You may accept your own answer(?)
– Tianlalu
Nov 30 '18 at 7:36
1
1
You may accept your own answer(?)
– Tianlalu
Nov 30 '18 at 7:36
You may accept your own answer(?)
– Tianlalu
Nov 30 '18 at 7:36
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004332%2fcomplex-integration-lemma-shorter-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
