Computing the Fourier transform of $cos(x/2)$
$begingroup$
How can I prove that $int_{-pi}^{pi} cos(x/2)e^{ixt} dx=frac{4cos(pi t)}{1-4t^2}$?
fourier-transform
edited Jan 13 at 20:34
Robert Z
101k1070143
asked Jan 13 at 20:22
eraldcoileraldcoil
393211
$endgroup$
add a comment |
$begingroup$
How can I prove that $int_{-pi}^{pi} cos(x/2)e^{ixt} dx=frac{4cos(pi t)}{1-4t^2}$?
fourier-transform
edited Jan 13 at 20:34
Robert Z
101k1070143
asked Jan 13 at 20:22
eraldcoileraldcoil
393211
$endgroup$
2
$begingroup$
I think integration by parts twice will work.
$endgroup$
– projectilemotion
Jan 13 at 20:25
add a comment |
$begingroup$
How can I prove that $int_{-pi}^{pi} cos(x/2)e^{ixt} dx=frac{4cos(pi t)}{1-4t^2}$?
fourier-transform
edited Jan 13 at 20:34
Robert Z
101k1070143
asked Jan 13 at 20:22
eraldcoileraldcoil
393211
$endgroup$
How can I prove that $int_{-pi}^{pi} cos(x/2)e^{ixt} dx=frac{4cos(pi t)}{1-4t^2}$?
fourier-transform
fourier-transform
edited Jan 13 at 20:34
Robert Z
101k1070143
asked Jan 13 at 20:22
eraldcoileraldcoil
393211
edited Jan 13 at 20:34
Robert Z
101k1070143
asked Jan 13 at 20:22
eraldcoileraldcoil
393211
edited Jan 13 at 20:34
Robert Z
101k1070143
edited Jan 13 at 20:34
Robert Z
101k1070143
edited Jan 13 at 20:34
Robert Z
101k1070143
101k1070143
asked Jan 13 at 20:22
eraldcoileraldcoil
393211
asked Jan 13 at 20:22
eraldcoileraldcoil
393211
asked Jan 13 at 20:22
eraldcoileraldcoil
393211
393211
2
$begingroup$
I think integration by parts twice will work.
$endgroup$
– projectilemotion
Jan 13 at 20:25
add a comment |
2
$begingroup$
I think integration by parts twice will work.
$endgroup$
– projectilemotion
Jan 13 at 20:25
2
2
$begingroup$
I think integration by parts twice will work.
$endgroup$
– projectilemotion
Jan 13 at 20:25
$begingroup$
I think integration by parts twice will work.
$endgroup$
– projectilemotion
Jan 13 at 20:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can utilize the exponential form of the cosine to combine exponentials
$$begin{align}int_{-pi}^pi cos(frac{x}{2})e^{ixt}dx&=frac{1}{2}int_{-pi}^pi (e^{frac{ix}{2}}+e^{-frac{ix}{2}})e^{ixt}dx\ \
&=frac{1}{2}int_{-pi}^pi e^{ix(t+frac{1}{2})}dx+frac{1}{2}int_{-pi}^pi e^{ix(t-frac{1}{2})}dx\ \
&=frac{1}{2}left(frac{e^{ipi(t+frac{1}{2})}-e^{-ipi(t+frac{1}{2})}}{i(t+frac{1}{2})}+frac{e^{ipi(t-frac{1}{2})}-e^{-ipi(t-frac{1}{2})}}{i(t-frac{1}{2})}right)end{align}$$ Adding the two terms together inside and using the exponential definition of the cosine again yields the answer.
edited Jan 13 at 20:43
Namaste
1
answered Jan 13 at 20:33
aledenaleden
2,5051511
$endgroup$
add a comment |
$begingroup$
By parts, as hinted by @projectilemotion.
$$I=intcosfrac x2e^{ixt}dx=2sinfrac x2e^{ixt}-i2tintsinfrac x2e^{ixt}dx
\=2sinfrac x2e^{ixt}+i2tcosfrac x2e^{ixt}dx+4t^2I.$$
When integrating from $-pi$ to $pi$, the first term equals
$$2(e^{ixt}+e^{-ixt})$$ and the second vanishes.
answered Jan 13 at 20:43
Yves DaoustYves Daoust
131k676229
$endgroup$
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%2f3072472%2fcomputing-the-fourier-transform-of-cosx-2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can utilize the exponential form of the cosine to combine exponentials
$$begin{align}int_{-pi}^pi cos(frac{x}{2})e^{ixt}dx&=frac{1}{2}int_{-pi}^pi (e^{frac{ix}{2}}+e^{-frac{ix}{2}})e^{ixt}dx\ \
&=frac{1}{2}int_{-pi}^pi e^{ix(t+frac{1}{2})}dx+frac{1}{2}int_{-pi}^pi e^{ix(t-frac{1}{2})}dx\ \
&=frac{1}{2}left(frac{e^{ipi(t+frac{1}{2})}-e^{-ipi(t+frac{1}{2})}}{i(t+frac{1}{2})}+frac{e^{ipi(t-frac{1}{2})}-e^{-ipi(t-frac{1}{2})}}{i(t-frac{1}{2})}right)end{align}$$ Adding the two terms together inside and using the exponential definition of the cosine again yields the answer.
edited Jan 13 at 20:43
Namaste
1
answered Jan 13 at 20:33
aledenaleden
2,5051511
$endgroup$
add a comment |
$begingroup$
You can utilize the exponential form of the cosine to combine exponentials
$$begin{align}int_{-pi}^pi cos(frac{x}{2})e^{ixt}dx&=frac{1}{2}int_{-pi}^pi (e^{frac{ix}{2}}+e^{-frac{ix}{2}})e^{ixt}dx\ \
&=frac{1}{2}int_{-pi}^pi e^{ix(t+frac{1}{2})}dx+frac{1}{2}int_{-pi}^pi e^{ix(t-frac{1}{2})}dx\ \
&=frac{1}{2}left(frac{e^{ipi(t+frac{1}{2})}-e^{-ipi(t+frac{1}{2})}}{i(t+frac{1}{2})}+frac{e^{ipi(t-frac{1}{2})}-e^{-ipi(t-frac{1}{2})}}{i(t-frac{1}{2})}right)end{align}$$ Adding the two terms together inside and using the exponential definition of the cosine again yields the answer.
edited Jan 13 at 20:43
Namaste
1
answered Jan 13 at 20:33
aledenaleden
2,5051511
$endgroup$
add a comment |
$begingroup$
You can utilize the exponential form of the cosine to combine exponentials
$$begin{align}int_{-pi}^pi cos(frac{x}{2})e^{ixt}dx&=frac{1}{2}int_{-pi}^pi (e^{frac{ix}{2}}+e^{-frac{ix}{2}})e^{ixt}dx\ \
&=frac{1}{2}int_{-pi}^pi e^{ix(t+frac{1}{2})}dx+frac{1}{2}int_{-pi}^pi e^{ix(t-frac{1}{2})}dx\ \
&=frac{1}{2}left(frac{e^{ipi(t+frac{1}{2})}-e^{-ipi(t+frac{1}{2})}}{i(t+frac{1}{2})}+frac{e^{ipi(t-frac{1}{2})}-e^{-ipi(t-frac{1}{2})}}{i(t-frac{1}{2})}right)end{align}$$ Adding the two terms together inside and using the exponential definition of the cosine again yields the answer.
edited Jan 13 at 20:43
Namaste
1
answered Jan 13 at 20:33
aledenaleden
2,5051511
$endgroup$
You can utilize the exponential form of the cosine to combine exponentials
$$begin{align}int_{-pi}^pi cos(frac{x}{2})e^{ixt}dx&=frac{1}{2}int_{-pi}^pi (e^{frac{ix}{2}}+e^{-frac{ix}{2}})e^{ixt}dx\ \
&=frac{1}{2}int_{-pi}^pi e^{ix(t+frac{1}{2})}dx+frac{1}{2}int_{-pi}^pi e^{ix(t-frac{1}{2})}dx\ \
&=frac{1}{2}left(frac{e^{ipi(t+frac{1}{2})}-e^{-ipi(t+frac{1}{2})}}{i(t+frac{1}{2})}+frac{e^{ipi(t-frac{1}{2})}-e^{-ipi(t-frac{1}{2})}}{i(t-frac{1}{2})}right)end{align}$$ Adding the two terms together inside and using the exponential definition of the cosine again yields the answer.
edited Jan 13 at 20:43
Namaste
1
answered Jan 13 at 20:33
aledenaleden
2,5051511
edited Jan 13 at 20:43
Namaste
1
edited Jan 13 at 20:43
Namaste
1
edited Jan 13 at 20:43
Namaste
1
1
answered Jan 13 at 20:33
aledenaleden
2,5051511
answered Jan 13 at 20:33
aledenaleden
2,5051511
answered Jan 13 at 20:33
aledenaleden
2,5051511
2,5051511
add a comment |
add a comment |
$begingroup$
By parts, as hinted by @projectilemotion.
$$I=intcosfrac x2e^{ixt}dx=2sinfrac x2e^{ixt}-i2tintsinfrac x2e^{ixt}dx
\=2sinfrac x2e^{ixt}+i2tcosfrac x2e^{ixt}dx+4t^2I.$$
When integrating from $-pi$ to $pi$, the first term equals
$$2(e^{ixt}+e^{-ixt})$$ and the second vanishes.
answered Jan 13 at 20:43
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
By parts, as hinted by @projectilemotion.
$$I=intcosfrac x2e^{ixt}dx=2sinfrac x2e^{ixt}-i2tintsinfrac x2e^{ixt}dx
\=2sinfrac x2e^{ixt}+i2tcosfrac x2e^{ixt}dx+4t^2I.$$
When integrating from $-pi$ to $pi$, the first term equals
$$2(e^{ixt}+e^{-ixt})$$ and the second vanishes.
answered Jan 13 at 20:43
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
By parts, as hinted by @projectilemotion.
$$I=intcosfrac x2e^{ixt}dx=2sinfrac x2e^{ixt}-i2tintsinfrac x2e^{ixt}dx
\=2sinfrac x2e^{ixt}+i2tcosfrac x2e^{ixt}dx+4t^2I.$$
When integrating from $-pi$ to $pi$, the first term equals
$$2(e^{ixt}+e^{-ixt})$$ and the second vanishes.
answered Jan 13 at 20:43
Yves DaoustYves Daoust
131k676229
$endgroup$
By parts, as hinted by @projectilemotion.
$$I=intcosfrac x2e^{ixt}dx=2sinfrac x2e^{ixt}-i2tintsinfrac x2e^{ixt}dx
\=2sinfrac x2e^{ixt}+i2tcosfrac x2e^{ixt}dx+4t^2I.$$
When integrating from $-pi$ to $pi$, the first term equals
$$2(e^{ixt}+e^{-ixt})$$ and the second vanishes.
answered Jan 13 at 20:43
Yves DaoustYves Daoust
131k676229
answered Jan 13 at 20:43
Yves DaoustYves Daoust
131k676229
answered Jan 13 at 20:43
Yves DaoustYves Daoust
131k676229
answered Jan 13 at 20:43
Yves DaoustYves Daoust
131k676229
131k676229
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%2f3072472%2fcomputing-the-fourier-transform-of-cosx-2%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
2
$begingroup$
I think integration by parts twice will work.
$endgroup$
– projectilemotion
Jan 13 at 20:25