Mixing
Integral of Reciprocal Functions: Why The Modulus Sign?
$begingroup$
I've been taught that $intfrac{1}{x}dx=ln|x| + C$ rather than $ln(x) + C$ without the modulus sign and been told that it is incorrect to not have the modulus sign. Why is that?
I've been investigating the natural logarithms of complex numbers and derived the equation $ln(z)=ln|z|+iarg(z)$ from $z=|z|exp(iθ)$.
So for negative real numbers this simplifies to $ln(x)=ln|x|+iπ$ so clearly both logarithm functions only differ by a constant so should both be valid functions $intfrac{1}{x}dx$, due to the integration constant being arbitrary.
integration functions
asked Jan 29 at 3:34
TheTroll73TheTroll73
83
$endgroup$
add a comment |
$begingroup$
I've been taught that $intfrac{1}{x}dx=ln|x| + C$ rather than $ln(x) + C$ without the modulus sign and been told that it is incorrect to not have the modulus sign. Why is that?
I've been investigating the natural logarithms of complex numbers and derived the equation $ln(z)=ln|z|+iarg(z)$ from $z=|z|exp(iθ)$.
So for negative real numbers this simplifies to $ln(x)=ln|x|+iπ$ so clearly both logarithm functions only differ by a constant so should both be valid functions $intfrac{1}{x}dx$, due to the integration constant being arbitrary.
integration functions
asked Jan 29 at 3:34
TheTroll73TheTroll73
83
$endgroup$
add a comment |
$begingroup$
I've been taught that $intfrac{1}{x}dx=ln|x| + C$ rather than $ln(x) + C$ without the modulus sign and been told that it is incorrect to not have the modulus sign. Why is that?
I've been investigating the natural logarithms of complex numbers and derived the equation $ln(z)=ln|z|+iarg(z)$ from $z=|z|exp(iθ)$.
So for negative real numbers this simplifies to $ln(x)=ln|x|+iπ$ so clearly both logarithm functions only differ by a constant so should both be valid functions $intfrac{1}{x}dx$, due to the integration constant being arbitrary.
integration functions
asked Jan 29 at 3:34
TheTroll73TheTroll73
83
$endgroup$
I've been taught that $intfrac{1}{x}dx=ln|x| + C$ rather than $ln(x) + C$ without the modulus sign and been told that it is incorrect to not have the modulus sign. Why is that?
I've been investigating the natural logarithms of complex numbers and derived the equation $ln(z)=ln|z|+iarg(z)$ from $z=|z|exp(iθ)$.
So for negative real numbers this simplifies to $ln(x)=ln|x|+iπ$ so clearly both logarithm functions only differ by a constant so should both be valid functions $intfrac{1}{x}dx$, due to the integration constant being arbitrary.
integration functions
integration functions
asked Jan 29 at 3:34
TheTroll73TheTroll73
83
asked Jan 29 at 3:34
TheTroll73TheTroll73
83
asked Jan 29 at 3:34
TheTroll73TheTroll73
83
asked Jan 29 at 3:34
TheTroll73TheTroll73
83
asked Jan 29 at 3:34
TheTroll73TheTroll73
83
83
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.
(All of what follows assumes a real variable. The complex logarithm will not appear.)
On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.
In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.
answered Jan 29 at 3:57
jmerryjmerry
16.9k11633
$endgroup$
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%2f3091704%2fintegral-of-reciprocal-functions-why-the-modulus-sign%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
$begingroup$
This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.
(All of what follows assumes a real variable. The complex logarithm will not appear.)
On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.
In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.
answered Jan 29 at 3:57
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.
(All of what follows assumes a real variable. The complex logarithm will not appear.)
On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.
In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.
answered Jan 29 at 3:57
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.
(All of what follows assumes a real variable. The complex logarithm will not appear.)
On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.
In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.
answered Jan 29 at 3:57
jmerryjmerry
16.9k11633
$endgroup$
This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.
(All of what follows assumes a real variable. The complex logarithm will not appear.)
On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.
In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.
answered Jan 29 at 3:57
jmerryjmerry
16.9k11633
answered Jan 29 at 3:57
jmerryjmerry
16.9k11633
answered Jan 29 at 3:57
jmerryjmerry
16.9k11633
answered Jan 29 at 3:57
jmerryjmerry
16.9k11633
16.9k11633
add a comment |
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.
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%2f3091704%2fintegral-of-reciprocal-functions-why-the-modulus-sign%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
