Limits and Infinite Integration by PartsDelta distribution - integration by parts of its differentiationMethods of constructing rapidly convergent seriesGauss Hermite Integration of 1/(1+x^2)Infinite sum: $sum_n=-infty^infty frac110^(n/100)^2$Infinite Integration by PartsIntegral of $sin(1/x)$ from $0$ to $infty$Evaluating $limlimits_btoinftybint_0^1cos(b x) cosh^-1(frac1x)dx$Formula obtained by repeated integration by parts.When do Taylor series converge quickly?$intfracsin(x)arcsin(-x)dx$
Terse Method to Swap Lowest for Highest?
Quoting Keynes in a lecture
Why did the EU agree to delay the Brexit deadline?
Why "had" in "[something] we would have made had we used [something]"?
Mimic lecturing on blackboard, facing audience
How to say when an application is taking the half of your screen on a computer
Can a stoichiometric mixture of oxygen and methane exist as a liquid at standard pressure and some (low) temperature?
How can mimic phobia be cured?
Can disgust be a key component of horror?
What should you do if you miss a job interview (deliberately)?
Why is the "ls" command showing permissions of files in a FAT32 partition?
How to cover method return statement in Apex Class?
What are the advantages of simplicial model categories over non-simplicial ones?
Do we have to expect a queue for the shuttle from Watford Junction to Harry Potter Studio?
Is there a RAID 0 Equivalent for RAM?
Recommended PCB layout understanding - ADM2572 datasheet
How should I respond when I lied about my education and the company finds out through background check?
How much character growth crosses the line into breaking the character
What are the balance implications behind making invisible things auto-hide?
What does "Scientists rise up against statistical significance" mean? (Comment in Nature)
What if you are holding an Iron Flask with a demon inside and walk into Antimagic Field?
Are Captain Marvel's powers affected by Thanos' actions in Infinity War
Why should universal income be universal?
Why does AES have exactly 10 rounds for a 128-bit key, 12 for 192 bits and 14 for a 256-bit key size?
Limits and Infinite Integration by Parts
Delta distribution - integration by parts of its differentiationMethods of constructing rapidly convergent seriesGauss Hermite Integration of 1/(1+x^2)Infinite sum: $sum_n=-infty^infty frac110^(n/100)^2$Infinite Integration by PartsIntegral of $sin(1/x)$ from $0$ to $infty$Evaluating $limlimits_btoinftybint_0^1cos(b x) cosh^-1(frac1x)dx$Formula obtained by repeated integration by parts.When do Taylor series converge quickly?$intfracsin(x)arcsin(-x)dx$
$begingroup$
It is well known that
$$int fracsin(x)x ,dx$$
cannot be expressed in terms of elementary functions. However, if we repeatedly use integration by parts, we seem to be able to at least approximate the integral through the formula
$$int f(x) ,dx approx sum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
where $a in mathbbN$. When plugging this in to a graphing calculator, it converges, but very slowly. It also tends to converge more quickly for functions that tend to $0$ as $x to infty$. My guess is that
$$int f(x) ,dx = lim_atoinftysum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
at least on a certain interval, but I am uncertain where to look to learn more about these series. Any ideas? Thanks!
real-analysis calculus integration sequences-and-series
asked 57 mins ago
HyperionHyperion
653110
$endgroup$
add a comment |
$begingroup$
It is well known that
$$int fracsin(x)x ,dx$$
cannot be expressed in terms of elementary functions. However, if we repeatedly use integration by parts, we seem to be able to at least approximate the integral through the formula
$$int f(x) ,dx approx sum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
where $a in mathbbN$. When plugging this in to a graphing calculator, it converges, but very slowly. It also tends to converge more quickly for functions that tend to $0$ as $x to infty$. My guess is that
$$int f(x) ,dx = lim_atoinftysum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
at least on a certain interval, but I am uncertain where to look to learn more about these series. Any ideas? Thanks!
real-analysis calculus integration sequences-and-series
asked 57 mins ago
HyperionHyperion
653110
$endgroup$
add a comment |
$begingroup$
It is well known that
$$int fracsin(x)x ,dx$$
cannot be expressed in terms of elementary functions. However, if we repeatedly use integration by parts, we seem to be able to at least approximate the integral through the formula
$$int f(x) ,dx approx sum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
where $a in mathbbN$. When plugging this in to a graphing calculator, it converges, but very slowly. It also tends to converge more quickly for functions that tend to $0$ as $x to infty$. My guess is that
$$int f(x) ,dx = lim_atoinftysum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
at least on a certain interval, but I am uncertain where to look to learn more about these series. Any ideas? Thanks!
real-analysis calculus integration sequences-and-series
asked 57 mins ago
HyperionHyperion
653110
$endgroup$
It is well known that
$$int fracsin(x)x ,dx$$
cannot be expressed in terms of elementary functions. However, if we repeatedly use integration by parts, we seem to be able to at least approximate the integral through the formula
$$int f(x) ,dx approx sum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
where $a in mathbbN$. When plugging this in to a graphing calculator, it converges, but very slowly. It also tends to converge more quickly for functions that tend to $0$ as $x to infty$. My guess is that
$$int f(x) ,dx = lim_atoinftysum_n=1^a frac(-1)^n-1cdot f^(n-1)(x)cdot x^nn!$$
at least on a certain interval, but I am uncertain where to look to learn more about these series. Any ideas? Thanks!
real-analysis calculus integration sequences-and-series
real-analysis calculus integration sequences-and-series
asked 57 mins ago
HyperionHyperion
653110
asked 57 mins ago
HyperionHyperion
653110
asked 57 mins ago
HyperionHyperion
653110
asked 57 mins ago
HyperionHyperion
653110
asked 57 mins ago
HyperionHyperion
653110
653110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You've essentially rediscovered Taylor series.
Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,
$$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$
i.e.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
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%2f3158687%2flimits-and-infinite-integration-by-parts%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
$begingroup$
You've essentially rediscovered Taylor series.
Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,
$$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$
i.e.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
add a comment |
$begingroup$
You've essentially rediscovered Taylor series.
Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,
$$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$
i.e.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
add a comment |
$begingroup$
You've essentially rediscovered Taylor series.
Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,
$$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$
i.e.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
$endgroup$
You've essentially rediscovered Taylor series.
Let $G(x)$ be an antiderivative of $f(x)$, so $f^(k)(x) = G^(k+1)(x)$ for $k ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,
$$ G(0) = sum_k=0^infty G^(k)(x) frac(-x)^kk! = G(x) + sum_k=1^infty f^(k-1)(x) frac(-x)^kk!$$
i.e.
$$ int_0^x f(t); dt = G(x)- G(0) = sum_k=1^infty f^(k-1)(x) frac(-1)^k-1 x^kk! $$
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
answered 46 mins ago
Robert IsraelRobert Israel
328k23216469
328k23216469
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%2f3158687%2flimits-and-infinite-integration-by-parts%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
