How can this integral be solved? Is is an indefinite integral?
$begingroup$
The integral I want to solve is $$int_1^2 frac {2 ln(x)}{x+1} dx$$
I tried to integrate it by parts in 2 ways and I tried to do the integral by parts twice, I thought of a change of variable and they all didn't work.
Can any one help by giving a solution please?
integration indefinite-integrals
edited Dec 23 '18 at 7:20
Eevee Trainer
10.6k31842
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
972112
$endgroup$
add a comment |
$begingroup$
The integral I want to solve is $$int_1^2 frac {2 ln(x)}{x+1} dx$$
I tried to integrate it by parts in 2 ways and I tried to do the integral by parts twice, I thought of a change of variable and they all didn't work.
Can any one help by giving a solution please?
integration indefinite-integrals
edited Dec 23 '18 at 7:20
Eevee Trainer
10.6k31842
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
972112
$endgroup$
add a comment |
$begingroup$
The integral I want to solve is $$int_1^2 frac {2 ln(x)}{x+1} dx$$
I tried to integrate it by parts in 2 ways and I tried to do the integral by parts twice, I thought of a change of variable and they all didn't work.
Can any one help by giving a solution please?
integration indefinite-integrals
edited Dec 23 '18 at 7:20
Eevee Trainer
10.6k31842
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
972112
$endgroup$
The integral I want to solve is $$int_1^2 frac {2 ln(x)}{x+1} dx$$
I tried to integrate it by parts in 2 ways and I tried to do the integral by parts twice, I thought of a change of variable and they all didn't work.
Can any one help by giving a solution please?
integration indefinite-integrals
integration indefinite-integrals
edited Dec 23 '18 at 7:20
Eevee Trainer
10.6k31842
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
972112
edited Dec 23 '18 at 7:20
Eevee Trainer
10.6k31842
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
972112
edited Dec 23 '18 at 7:20
Eevee Trainer
10.6k31842
edited Dec 23 '18 at 7:20
Eevee Trainer
10.6k31842
edited Dec 23 '18 at 7:20
Eevee Trainer
10.6k31842
10.6k31842
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
972112
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
972112
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
972112
972112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your integral evaluates to,
$$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
Use the identity,
$$int (f.g') dx = f.g - int (f'.g) dx$$
And the definition of dilogarithm to get your result.
Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
638113
$endgroup$
$begingroup$
@Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
$endgroup$
– Awe Kumar Jha
Dec 23 '18 at 9:32
$begingroup$
Thank you :-), I appreciate your help
$endgroup$
– Fareed AF
Dec 23 '18 at 11:05
add a comment |
Your Answer
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%2f3050127%2fhow-can-this-integral-be-solved-is-is-an-indefinite-integral%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
$begingroup$
Your integral evaluates to,
$$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
Use the identity,
$$int (f.g') dx = f.g - int (f'.g) dx$$
And the definition of dilogarithm to get your result.
Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
638113
$endgroup$
$begingroup$
@Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
$endgroup$
– Awe Kumar Jha
Dec 23 '18 at 9:32
$begingroup$
Thank you :-), I appreciate your help
$endgroup$
– Fareed AF
Dec 23 '18 at 11:05
add a comment |
$begingroup$
Your integral evaluates to,
$$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
Use the identity,
$$int (f.g') dx = f.g - int (f'.g) dx$$
And the definition of dilogarithm to get your result.
Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
638113
$endgroup$
$begingroup$
@Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
$endgroup$
– Awe Kumar Jha
Dec 23 '18 at 9:32
$begingroup$
Thank you :-), I appreciate your help
$endgroup$
– Fareed AF
Dec 23 '18 at 11:05
add a comment |
$begingroup$
Your integral evaluates to,
$$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
Use the identity,
$$int (f.g') dx = f.g - int (f'.g) dx$$
And the definition of dilogarithm to get your result.
Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
638113
$endgroup$
Your integral evaluates to,
$$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
Use the identity,
$$int (f.g') dx = f.g - int (f'.g) dx$$
And the definition of dilogarithm to get your result.
Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
638113
edited Dec 23 '18 at 7:40
user150203
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
638113
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
638113
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
638113
638113
$begingroup$
@Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
$endgroup$
– Awe Kumar Jha
Dec 23 '18 at 9:32
$begingroup$
Thank you :-), I appreciate your help
$endgroup$
– Fareed AF
Dec 23 '18 at 11:05
add a comment |
$begingroup$
@Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
$endgroup$
– Awe Kumar Jha
Dec 23 '18 at 9:32
$begingroup$
Thank you :-), I appreciate your help
$endgroup$
– Fareed AF
Dec 23 '18 at 11:05
$begingroup$
@Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
$endgroup$
– Awe Kumar Jha
Dec 23 '18 at 9:32
$begingroup$
@Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
$endgroup$
– Awe Kumar Jha
Dec 23 '18 at 9:32
$begingroup$
Thank you :-), I appreciate your help
$endgroup$
– Fareed AF
Dec 23 '18 at 11:05
$begingroup$
Thank you :-), I appreciate your help
$endgroup$
– Fareed AF
Dec 23 '18 at 11:05
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%2f3050127%2fhow-can-this-integral-be-solved-is-is-an-indefinite-integral%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
