Prove $int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$
Prove: $$A=int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$$
(I think) that a suggestion was made to calculate and later use it: $B=intfrac{{e^{az}}}{z}$, over the path gamma $gamma=e^{it}, tin[-pi,pi]$, i think this integral is $2ipi$ for all a, because of Cauchy theorem (correct me please if I am wrong) , then I tried to make it look similar to the second integral:
$$B=int_{-pi}^pi e^{acos(x)}dx+int_{-pi}^pi e^{isin(x)}dx,$$ but do not know how to proceed. Any help appreciated
complex-analysis functional-analysis complex-numbers complex-integration
edited Nov 28 '18 at 13:32
Asaf Karagila♦
302k32425756
asked Nov 28 '18 at 13:28
ryszard eggink
308110
add a comment |
Prove: $$A=int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$$
(I think) that a suggestion was made to calculate and later use it: $B=intfrac{{e^{az}}}{z}$, over the path gamma $gamma=e^{it}, tin[-pi,pi]$, i think this integral is $2ipi$ for all a, because of Cauchy theorem (correct me please if I am wrong) , then I tried to make it look similar to the second integral:
$$B=int_{-pi}^pi e^{acos(x)}dx+int_{-pi}^pi e^{isin(x)}dx,$$ but do not know how to proceed. Any help appreciated
complex-analysis functional-analysis complex-numbers complex-integration
edited Nov 28 '18 at 13:32
Asaf Karagila♦
302k32425756
asked Nov 28 '18 at 13:28
ryszard eggink
308110
add a comment |
Prove: $$A=int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$$
(I think) that a suggestion was made to calculate and later use it: $B=intfrac{{e^{az}}}{z}$, over the path gamma $gamma=e^{it}, tin[-pi,pi]$, i think this integral is $2ipi$ for all a, because of Cauchy theorem (correct me please if I am wrong) , then I tried to make it look similar to the second integral:
$$B=int_{-pi}^pi e^{acos(x)}dx+int_{-pi}^pi e^{isin(x)}dx,$$ but do not know how to proceed. Any help appreciated
complex-analysis functional-analysis complex-numbers complex-integration
edited Nov 28 '18 at 13:32
Asaf Karagila♦
302k32425756
asked Nov 28 '18 at 13:28
ryszard eggink
308110
Prove: $$A=int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$$
(I think) that a suggestion was made to calculate and later use it: $B=intfrac{{e^{az}}}{z}$, over the path gamma $gamma=e^{it}, tin[-pi,pi]$, i think this integral is $2ipi$ for all a, because of Cauchy theorem (correct me please if I am wrong) , then I tried to make it look similar to the second integral:
$$B=int_{-pi}^pi e^{acos(x)}dx+int_{-pi}^pi e^{isin(x)}dx,$$ but do not know how to proceed. Any help appreciated
complex-analysis functional-analysis complex-numbers complex-integration
complex-analysis functional-analysis complex-numbers complex-integration
edited Nov 28 '18 at 13:32
Asaf Karagila♦
302k32425756
asked Nov 28 '18 at 13:28
ryszard eggink
308110
edited Nov 28 '18 at 13:32
Asaf Karagila♦
302k32425756
asked Nov 28 '18 at 13:28
ryszard eggink
308110
edited Nov 28 '18 at 13:32
Asaf Karagila♦
302k32425756
edited Nov 28 '18 at 13:32
Asaf Karagila♦
302k32425756
edited Nov 28 '18 at 13:32
Asaf Karagila♦
302k32425756
302k32425756
asked Nov 28 '18 at 13:28
ryszard eggink
308110
asked Nov 28 '18 at 13:28
ryszard eggink
308110
asked Nov 28 '18 at 13:28
ryszard eggink
308110
308110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
From
$$
2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
$$
we get
$$
int_0^{2pi}e^{ae^{it}},dt=2,pi.
$$
Now
$$
e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
$$
Since $2pi$ is real, we get
$$
int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
$$
answered Nov 28 '18 at 13:41
Julián Aguirre
67.6k24094
I have been there :/, this is precisely what i wrote in the last line
– ryszard eggink
Nov 28 '18 at 13:42
It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
– Julián Aguirre
Nov 28 '18 at 13:57
Ouch right on! thank you very much :)
– ryszard eggink
Nov 28 '18 at 14:12
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%2f3017151%2fprove-int-0-piea-cosx-cosa-sinxdx-pi%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
From
$$
2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
$$
we get
$$
int_0^{2pi}e^{ae^{it}},dt=2,pi.
$$
Now
$$
e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
$$
Since $2pi$ is real, we get
$$
int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
$$
answered Nov 28 '18 at 13:41
Julián Aguirre
67.6k24094
I have been there :/, this is precisely what i wrote in the last line
– ryszard eggink
Nov 28 '18 at 13:42
It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
– Julián Aguirre
Nov 28 '18 at 13:57
Ouch right on! thank you very much :)
– ryszard eggink
Nov 28 '18 at 14:12
add a comment |
From
$$
2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
$$
we get
$$
int_0^{2pi}e^{ae^{it}},dt=2,pi.
$$
Now
$$
e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
$$
Since $2pi$ is real, we get
$$
int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
$$
answered Nov 28 '18 at 13:41
Julián Aguirre
67.6k24094
I have been there :/, this is precisely what i wrote in the last line
– ryszard eggink
Nov 28 '18 at 13:42
It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
– Julián Aguirre
Nov 28 '18 at 13:57
Ouch right on! thank you very much :)
– ryszard eggink
Nov 28 '18 at 14:12
add a comment |
From
$$
2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
$$
we get
$$
int_0^{2pi}e^{ae^{it}},dt=2,pi.
$$
Now
$$
e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
$$
Since $2pi$ is real, we get
$$
int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
$$
answered Nov 28 '18 at 13:41
Julián Aguirre
67.6k24094
From
$$
2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
$$
we get
$$
int_0^{2pi}e^{ae^{it}},dt=2,pi.
$$
Now
$$
e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
$$
Since $2pi$ is real, we get
$$
int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
$$
answered Nov 28 '18 at 13:41
Julián Aguirre
67.6k24094
edited Nov 28 '18 at 13:56
answered Nov 28 '18 at 13:41
Julián Aguirre
67.6k24094
answered Nov 28 '18 at 13:41
Julián Aguirre
67.6k24094
answered Nov 28 '18 at 13:41
Julián Aguirre
67.6k24094
67.6k24094
I have been there :/, this is precisely what i wrote in the last line
– ryszard eggink
Nov 28 '18 at 13:42
It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
– Julián Aguirre
Nov 28 '18 at 13:57
Ouch right on! thank you very much :)
– ryszard eggink
Nov 28 '18 at 14:12
add a comment |
I have been there :/, this is precisely what i wrote in the last line
– ryszard eggink
Nov 28 '18 at 13:42
It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
– Julián Aguirre
Nov 28 '18 at 13:57
Ouch right on! thank you very much :)
– ryszard eggink
Nov 28 '18 at 14:12
I have been there :/, this is precisely what i wrote in the last line
– ryszard eggink
Nov 28 '18 at 13:42
I have been there :/, this is precisely what i wrote in the last line
– ryszard eggink
Nov 28 '18 at 13:42
It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
– Julián Aguirre
Nov 28 '18 at 13:57
It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
– Julián Aguirre
Nov 28 '18 at 13:57
Ouch right on! thank you very much :)
– ryszard eggink
Nov 28 '18 at 14:12
Ouch right on! thank you very much :)
– ryszard eggink
Nov 28 '18 at 14:12
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%2f3017151%2fprove-int-0-piea-cosx-cosa-sinxdx-pi%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
