Imaginary part of integral of $exp(kx - x^2/2 + 2pi i x)$ vanishes over the real axis.
I want to show that
$$
int_0^infty x^k frac{1}{xsqrt{2pi}} e^{-log(x)^2 / 2} sin(2pi log(x)) = 0
$$
for $k in mathbb{N}$.
I thought, that it would be benificial to lift this integral into the complex plane and showed that the upper is equal to
$$
text{Im}left[ int_mathbb{R} exp(ky-y^2/2 + i2pi y)dyright] = 0.
$$
However, I don't know how to proceed from there.
complex-analysis analysis
edited Nov 29 '18 at 20:27
Bernard
118k639112
asked Nov 29 '18 at 20:27
fpmoofpmoo
368113
add a comment |
I want to show that
$$
int_0^infty x^k frac{1}{xsqrt{2pi}} e^{-log(x)^2 / 2} sin(2pi log(x)) = 0
$$
for $k in mathbb{N}$.
I thought, that it would be benificial to lift this integral into the complex plane and showed that the upper is equal to
$$
text{Im}left[ int_mathbb{R} exp(ky-y^2/2 + i2pi y)dyright] = 0.
$$
However, I don't know how to proceed from there.
complex-analysis analysis
edited Nov 29 '18 at 20:27
Bernard
118k639112
asked Nov 29 '18 at 20:27
fpmoofpmoo
368113
add a comment |
I want to show that
$$
int_0^infty x^k frac{1}{xsqrt{2pi}} e^{-log(x)^2 / 2} sin(2pi log(x)) = 0
$$
for $k in mathbb{N}$.
I thought, that it would be benificial to lift this integral into the complex plane and showed that the upper is equal to
$$
text{Im}left[ int_mathbb{R} exp(ky-y^2/2 + i2pi y)dyright] = 0.
$$
However, I don't know how to proceed from there.
complex-analysis analysis
edited Nov 29 '18 at 20:27
Bernard
118k639112
asked Nov 29 '18 at 20:27
fpmoofpmoo
368113
I want to show that
$$
int_0^infty x^k frac{1}{xsqrt{2pi}} e^{-log(x)^2 / 2} sin(2pi log(x)) = 0
$$
for $k in mathbb{N}$.
I thought, that it would be benificial to lift this integral into the complex plane and showed that the upper is equal to
$$
text{Im}left[ int_mathbb{R} exp(ky-y^2/2 + i2pi y)dyright] = 0.
$$
However, I don't know how to proceed from there.
complex-analysis analysis
complex-analysis analysis
edited Nov 29 '18 at 20:27
Bernard
118k639112
asked Nov 29 '18 at 20:27
fpmoofpmoo
368113
edited Nov 29 '18 at 20:27
Bernard
118k639112
asked Nov 29 '18 at 20:27
fpmoofpmoo
368113
edited Nov 29 '18 at 20:27
Bernard
118k639112
edited Nov 29 '18 at 20:27
Bernard
118k639112
edited Nov 29 '18 at 20:27
Bernard
118k639112
118k639112
asked Nov 29 '18 at 20:27
fpmoofpmoo
368113
asked Nov 29 '18 at 20:27
fpmoofpmoo
368113
asked Nov 29 '18 at 20:27
fpmoofpmoo
368113
368113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Introduce a new variable $x = y - k$:
begin{align*}int_mathbb{R} exp(ky-y^2/2 + i2pi y),dy &= int_mathbb{R} exp left( -frac{1}{2} (y-k)^2 + frac{k^2}{2} + i 2pi yright),dy \
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x + i 2pi kright),dx\
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x right),dx
end{align*}
We can drop the $i2pi k$ term because $k in mathbb{Z}$ and $exp(z) = exp(z+2pi i)$. Switching $x$ with $-x$ replaces the argument of $exp$ with its complex conjugate, and $e^overline{z} = overline{e^z}$, so the imaginary part of the integral over $(-infty, 0]$ cancels the imaginary part of the integral over $[0, infty)$, and we are done.
edited Nov 29 '18 at 21:26
fpmoo
368113
answered Nov 29 '18 at 21:02
Connor HarrisConnor Harris
4,350723
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%2f3019157%2fimaginary-part-of-integral-of-expkx-x2-2-2-pi-i-x-vanishes-over-the-re%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
Introduce a new variable $x = y - k$:
begin{align*}int_mathbb{R} exp(ky-y^2/2 + i2pi y),dy &= int_mathbb{R} exp left( -frac{1}{2} (y-k)^2 + frac{k^2}{2} + i 2pi yright),dy \
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x + i 2pi kright),dx\
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x right),dx
end{align*}
We can drop the $i2pi k$ term because $k in mathbb{Z}$ and $exp(z) = exp(z+2pi i)$. Switching $x$ with $-x$ replaces the argument of $exp$ with its complex conjugate, and $e^overline{z} = overline{e^z}$, so the imaginary part of the integral over $(-infty, 0]$ cancels the imaginary part of the integral over $[0, infty)$, and we are done.
edited Nov 29 '18 at 21:26
fpmoo
368113
answered Nov 29 '18 at 21:02
Connor HarrisConnor Harris
4,350723
add a comment |
Introduce a new variable $x = y - k$:
begin{align*}int_mathbb{R} exp(ky-y^2/2 + i2pi y),dy &= int_mathbb{R} exp left( -frac{1}{2} (y-k)^2 + frac{k^2}{2} + i 2pi yright),dy \
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x + i 2pi kright),dx\
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x right),dx
end{align*}
We can drop the $i2pi k$ term because $k in mathbb{Z}$ and $exp(z) = exp(z+2pi i)$. Switching $x$ with $-x$ replaces the argument of $exp$ with its complex conjugate, and $e^overline{z} = overline{e^z}$, so the imaginary part of the integral over $(-infty, 0]$ cancels the imaginary part of the integral over $[0, infty)$, and we are done.
edited Nov 29 '18 at 21:26
fpmoo
368113
answered Nov 29 '18 at 21:02
Connor HarrisConnor Harris
4,350723
add a comment |
Introduce a new variable $x = y - k$:
begin{align*}int_mathbb{R} exp(ky-y^2/2 + i2pi y),dy &= int_mathbb{R} exp left( -frac{1}{2} (y-k)^2 + frac{k^2}{2} + i 2pi yright),dy \
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x + i 2pi kright),dx\
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x right),dx
end{align*}
We can drop the $i2pi k$ term because $k in mathbb{Z}$ and $exp(z) = exp(z+2pi i)$. Switching $x$ with $-x$ replaces the argument of $exp$ with its complex conjugate, and $e^overline{z} = overline{e^z}$, so the imaginary part of the integral over $(-infty, 0]$ cancels the imaginary part of the integral over $[0, infty)$, and we are done.
edited Nov 29 '18 at 21:26
fpmoo
368113
answered Nov 29 '18 at 21:02
Connor HarrisConnor Harris
4,350723
Introduce a new variable $x = y - k$:
begin{align*}int_mathbb{R} exp(ky-y^2/2 + i2pi y),dy &= int_mathbb{R} exp left( -frac{1}{2} (y-k)^2 + frac{k^2}{2} + i 2pi yright),dy \
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x + i 2pi kright),dx\
&= int_mathbb{R} exp left( -frac{x^2}{2} + frac{k^2}{2} + i 2pi x right),dx
end{align*}
We can drop the $i2pi k$ term because $k in mathbb{Z}$ and $exp(z) = exp(z+2pi i)$. Switching $x$ with $-x$ replaces the argument of $exp$ with its complex conjugate, and $e^overline{z} = overline{e^z}$, so the imaginary part of the integral over $(-infty, 0]$ cancels the imaginary part of the integral over $[0, infty)$, and we are done.
edited Nov 29 '18 at 21:26
fpmoo
368113
answered Nov 29 '18 at 21:02
Connor HarrisConnor Harris
4,350723
edited Nov 29 '18 at 21:26
fpmoo
368113
edited Nov 29 '18 at 21:26
fpmoo
368113
edited Nov 29 '18 at 21:26
fpmoo
368113
368113
answered Nov 29 '18 at 21:02
Connor HarrisConnor Harris
4,350723
answered Nov 29 '18 at 21:02
Connor HarrisConnor Harris
4,350723
answered Nov 29 '18 at 21:02
Connor HarrisConnor Harris
4,350723
4,350723
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.
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%2f3019157%2fimaginary-part-of-integral-of-expkx-x2-2-2-pi-i-x-vanishes-over-the-re%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
