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
6,1631936
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
52612
$endgroup$
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
6,1631936
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
52612
$endgroup$
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
6,1631936
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
52612
$endgroup$
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
6,1631936
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
52612
edited Dec 23 '18 at 7:20
Eevee Trainer
6,1631936
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
52612
edited Dec 23 '18 at 7:20
Eevee Trainer
6,1631936
edited Dec 23 '18 at 7:20
Eevee Trainer
6,1631936
edited Dec 23 '18 at 7:20
Eevee Trainer
6,1631936
6,1631936
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
52612
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
52612
asked Dec 23 '18 at 7:15
Fareed AFFareed AF
52612
52612
add a comment |
add a comment |
1 Answer
1
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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 $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
edited Dec 23 '18 at 7:40
DavidG
2,1321724
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
43813
$endgroup$
$begingroup$
@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
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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 $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
edited Dec 23 '18 at 7:40
DavidG
2,1321724
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
43813
$endgroup$
$begingroup$
@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 $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
edited Dec 23 '18 at 7:40
DavidG
2,1321724
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
43813
$endgroup$
$begingroup$
@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 $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
edited Dec 23 '18 at 7:40
DavidG
2,1321724
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
43813
$endgroup$
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 $z:=-x$.
$$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$
edited Dec 23 '18 at 7:40
DavidG
2,1321724
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
43813
edited Dec 23 '18 at 7:40
DavidG
2,1321724
edited Dec 23 '18 at 7:40
DavidG
2,1321724
edited Dec 23 '18 at 7:40
DavidG
2,1321724
2,1321724
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
43813
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
43813
answered Dec 23 '18 at 7:25
Awe Kumar JhaAwe Kumar Jha
43813
43813
$begingroup$
@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 |
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