Mixing
Finding $int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2} dx$
$begingroup$
Calculate $$int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2}dx$$
I have tried to put $displaystyle x=frac{1}{t}$ and $displaystyle dx=-frac{1}{t^2}dt$
$$ int^{infty}_{0}frac{t^2ln^2(t)}{(t^2-1)^2}dt$$
$$frac{1}{2}int^{infty}_{0}tln^2(t)frac{2t}{(t^2-1)^2}dt$$
$$ frac{1}{2}bigg[-tln^2(t)frac{1}{t^2-1}+int^{infty}_{0}frac{ln^2(t)}{t^2-1}+2int^{infty}_{0}frac{ln(t)}{t^2-1}dtbigg]$$
How can I solve it?
integration definite-integrals
edited Feb 2 at 13:29
Zacky
7,87511062
asked Feb 2 at 10:59
jackyjacky
1,349816
$endgroup$
add a comment |
$begingroup$
Calculate $$int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2}dx$$
I have tried to put $displaystyle x=frac{1}{t}$ and $displaystyle dx=-frac{1}{t^2}dt$
$$ int^{infty}_{0}frac{t^2ln^2(t)}{(t^2-1)^2}dt$$
$$frac{1}{2}int^{infty}_{0}tln^2(t)frac{2t}{(t^2-1)^2}dt$$
$$ frac{1}{2}bigg[-tln^2(t)frac{1}{t^2-1}+int^{infty}_{0}frac{ln^2(t)}{t^2-1}+2int^{infty}_{0}frac{ln(t)}{t^2-1}dtbigg]$$
How can I solve it?
integration definite-integrals
edited Feb 2 at 13:29
Zacky
7,87511062
asked Feb 2 at 10:59
jackyjacky
1,349816
$endgroup$
$begingroup$
Are you sure about the limits of the integral? Because the integrant has a singularity of the order $1$ at $x=1$.
$endgroup$
– Mundron Schmidt
Feb 2 at 11:15
$begingroup$
@MundronSchmidt W|A gives $pi^2/4$ as the result
$endgroup$
– TheSimpliFire
Feb 2 at 11:19
5
$begingroup$
Note that $$int_0^inftyfrac{ln^2x}{(1-x^2)^2},dx=int_{-infty}^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx$$ and begin{align}int_0^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx&=int_0^{infty}left(x^2e^{-3x}sum_{n=0}^{infty}left(left(1+nright)e^{-2nx}right)right),dx\&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}=frac{pi^2-7zeta(3)}8end{align}
$endgroup$
– TheSimpliFire
Feb 2 at 11:45
add a comment |
$begingroup$
Calculate $$int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2}dx$$
I have tried to put $displaystyle x=frac{1}{t}$ and $displaystyle dx=-frac{1}{t^2}dt$
$$ int^{infty}_{0}frac{t^2ln^2(t)}{(t^2-1)^2}dt$$
$$frac{1}{2}int^{infty}_{0}tln^2(t)frac{2t}{(t^2-1)^2}dt$$
$$ frac{1}{2}bigg[-tln^2(t)frac{1}{t^2-1}+int^{infty}_{0}frac{ln^2(t)}{t^2-1}+2int^{infty}_{0}frac{ln(t)}{t^2-1}dtbigg]$$
How can I solve it?
integration definite-integrals
edited Feb 2 at 13:29
Zacky
7,87511062
asked Feb 2 at 10:59
jackyjacky
1,349816
$endgroup$
Calculate $$int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2}dx$$
I have tried to put $displaystyle x=frac{1}{t}$ and $displaystyle dx=-frac{1}{t^2}dt$
$$ int^{infty}_{0}frac{t^2ln^2(t)}{(t^2-1)^2}dt$$
$$frac{1}{2}int^{infty}_{0}tln^2(t)frac{2t}{(t^2-1)^2}dt$$
$$ frac{1}{2}bigg[-tln^2(t)frac{1}{t^2-1}+int^{infty}_{0}frac{ln^2(t)}{t^2-1}+2int^{infty}_{0}frac{ln(t)}{t^2-1}dtbigg]$$
How can I solve it?
integration definite-integrals
integration definite-integrals
edited Feb 2 at 13:29
Zacky
7,87511062
asked Feb 2 at 10:59
jackyjacky
1,349816
edited Feb 2 at 13:29
Zacky
7,87511062
asked Feb 2 at 10:59
jackyjacky
1,349816
edited Feb 2 at 13:29
Zacky
7,87511062
edited Feb 2 at 13:29
Zacky
7,87511062
edited Feb 2 at 13:29
Zacky
7,87511062
7,87511062
asked Feb 2 at 10:59
jackyjacky
1,349816
asked Feb 2 at 10:59
jackyjacky
1,349816
asked Feb 2 at 10:59
jackyjacky
1,349816
1,349816
$begingroup$
Are you sure about the limits of the integral? Because the integrant has a singularity of the order $1$ at $x=1$.
$endgroup$
– Mundron Schmidt
Feb 2 at 11:15
$begingroup$
@MundronSchmidt W|A gives $pi^2/4$ as the result
$endgroup$
– TheSimpliFire
Feb 2 at 11:19
5
$begingroup$
Note that $$int_0^inftyfrac{ln^2x}{(1-x^2)^2},dx=int_{-infty}^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx$$ and begin{align}int_0^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx&=int_0^{infty}left(x^2e^{-3x}sum_{n=0}^{infty}left(left(1+nright)e^{-2nx}right)right),dx\&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}=frac{pi^2-7zeta(3)}8end{align}
$endgroup$
– TheSimpliFire
Feb 2 at 11:45
add a comment |
$begingroup$
Are you sure about the limits of the integral? Because the integrant has a singularity of the order $1$ at $x=1$.
$endgroup$
– Mundron Schmidt
Feb 2 at 11:15
$begingroup$
@MundronSchmidt W|A gives $pi^2/4$ as the result
$endgroup$
– TheSimpliFire
Feb 2 at 11:19
5
$begingroup$
Note that $$int_0^inftyfrac{ln^2x}{(1-x^2)^2},dx=int_{-infty}^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx$$ and begin{align}int_0^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx&=int_0^{infty}left(x^2e^{-3x}sum_{n=0}^{infty}left(left(1+nright)e^{-2nx}right)right),dx\&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}=frac{pi^2-7zeta(3)}8end{align}
$endgroup$
– TheSimpliFire
Feb 2 at 11:45
$begingroup$
Are you sure about the limits of the integral? Because the integrant has a singularity of the order $1$ at $x=1$.
$endgroup$
– Mundron Schmidt
Feb 2 at 11:15
$begingroup$
Are you sure about the limits of the integral? Because the integrant has a singularity of the order $1$ at $x=1$.
$endgroup$
– Mundron Schmidt
Feb 2 at 11:15
$begingroup$
@MundronSchmidt W|A gives $pi^2/4$ as the result
$endgroup$
– TheSimpliFire
Feb 2 at 11:19
$begingroup$
@MundronSchmidt W|A gives $pi^2/4$ as the result
$endgroup$
– TheSimpliFire
Feb 2 at 11:19
5
5
$begingroup$
Note that $$int_0^inftyfrac{ln^2x}{(1-x^2)^2},dx=int_{-infty}^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx$$ and begin{align}int_0^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx&=int_0^{infty}left(x^2e^{-3x}sum_{n=0}^{infty}left(left(1+nright)e^{-2nx}right)right),dx\&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}=frac{pi^2-7zeta(3)}8end{align}
$endgroup$
– TheSimpliFire
Feb 2 at 11:45
$begingroup$
Note that $$int_0^inftyfrac{ln^2x}{(1-x^2)^2},dx=int_{-infty}^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx$$ and begin{align}int_0^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx&=int_0^{infty}left(x^2e^{-3x}sum_{n=0}^{infty}left(left(1+nright)e^{-2nx}right)right),dx\&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}=frac{pi^2-7zeta(3)}8end{align}
$endgroup$
– TheSimpliFire
Feb 2 at 11:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You're definetly on the right track with that substitution of $x=frac1t$
Basically we have: $$I=int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2}dx=int_0^infty frac{x^2ln^2 x}{(1-x^2)^2}dx$$
Now what if we add them up?
$$2I=int_0^infty ln^2 x frac{1+x^2}{(1-x^2)^2}dx$$
If you don't know how to deal easily with the integral $$int frac{1+x^2}{(1-x^2)^2}dx=frac{x}{1-x^2}+C$$
I recommend you to take a look here.
Anyway we have, integrating by parts:
$$2I= underbrace{frac{x}{1-x^2}ln^2x bigg|_0^infty}_{=0} +2underbrace{int_0^infty frac{ln x}{x^2-1}dx}_{large =frac{pi^2}{4}}$$
$$Rightarrow 2I= 2cdot frac{pi^2}{4} Rightarrow I=frac{pi^2}{4}$$
For the last integral see here for example.
answered Feb 2 at 13:20
ZackyZacky
7,87511062
$endgroup$
add a comment |
$begingroup$
This is a variant of TheSimpliFire's approach given in a comment.
By letting $x=e^t$ we get
$$begin{align*}
int_0^inftyfrac{ln^2(x)}{(1-x^2)^2},dx
&=int_{-infty}^inftyfrac{t^2e^{t}}{(1-e^{2t})^2},dt\
&=int_{0}^{+infty}frac{t^2e^{-t}}{(1-e^{-2t})^2},dt+int_{0}^inftyfrac{t^2e^{-3t}}{(1-e^{-2t})^2},dt\
&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+1)t},dt+sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\
&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+1)^3}+sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}\
&=left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}+sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)+left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}-sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)\
&=2sum_{n=0}^inftyfrac{1}{(2n+1)^2}=2left(sum_{n=0}^inftyfrac{1}{n^2}-sum_{n=0}^inftyfrac{1}{(2n)^2}right)\&=2left(1-frac{1}{4}right)sum_{n=0}^inftyfrac{1}{n^2}=frac{3}{2}cdot frac{pi^2}{6}=frac{pi^2}{4}.
end{align*}$$
answered Feb 9 at 11:31
Robert ZRobert Z
102k1072145
$endgroup$
$begingroup$
Could you provide a reference for those two fascinating sums (the ones with the $(2n+1)^3$ and $(2n+3)^3$ in the denominators)? Thanks
$endgroup$
– clathratus
Feb 9 at 16:15
1
$begingroup$
@clathratus Actually we don't need them. See my edit!
$endgroup$
– Robert Z
Feb 9 at 16:43
$begingroup$
Beautiful! Still though, any sources regarding those sums would be appreciated. They are intriguing indeed.
$endgroup$
– clathratus
Feb 9 at 20:35
1
$begingroup$
$sum_{n=0}^inftyfrac{1}{(2n+1)^3}=sum_{n=1}^inftyfrac{1}{n^3}-sum_{n=1}^inftyfrac{1}{(2n)^3}=frac{7zeta(3)}{8}.$
$endgroup$
– Robert Z
Feb 9 at 21:14
add a comment |
$begingroup$
Here is yet another slight variation on a theme.
Let
$$I = int_0^infty frac{ln^2}{(1 - x^2)^2} , dx$$
then
begin{align}
I &= int_0^1 frac{ln^2 x}{(1 - x^2)^2} , dx + int_1^infty frac{ln^2 x}{(1 - x^2)^2} , dx = int_0^1 frac{(1 + x^2) ln^2 x}{(1 - x^2)^2} , dx tag1,
end{align}
after a substitution of $x mapsto 1/x$ has been enforced in the second of the integrals.
As
$$frac{1}{1 - x^2} = sum_{n = 0}^infty x^{2n}, qquad |x| < 1,$$
differentiating with respect to $x$ gives
$$frac{1}{(1 - x^2)^2} = sum_{n = 1}^infty n x^{2n - 2}.$$
On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have
$$I = sum_{n = 1}^infty n int_0^1 (x^{2n - 2} + x^{2n}) ln^2 x , dx.$$
Integrating by parts twice, we are left with
$$I = sum_{n = 1}^infty left (frac{2n}{(2n - 1)^3} + frac{2n}{(2n + 1)^3} right ).$$
As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have
begin{align}
I &= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} + frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 0}^infty left [frac{1}{(2n + 1)^2} + frac{1}{(2n + 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= 2 + 2 sum_{n = 1}^infty frac{1}{(2n + 1)^2}\
&= 2 sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&= 2 left [sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty frac{1}{(2n)^2} right ]\
&= 2 left (1 - frac{1}{4} right ) sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} cdot frac{pi^2}{6}\
&= frac{pi^2}{4},
end{align}
as expected.
answered Feb 9 at 21:44
omegadotomegadot
6,2692829
$endgroup$
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%2f3097187%2ffinding-int-infty-0-frac-ln2x1-x22-dx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're definetly on the right track with that substitution of $x=frac1t$
Basically we have: $$I=int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2}dx=int_0^infty frac{x^2ln^2 x}{(1-x^2)^2}dx$$
Now what if we add them up?
$$2I=int_0^infty ln^2 x frac{1+x^2}{(1-x^2)^2}dx$$
If you don't know how to deal easily with the integral $$int frac{1+x^2}{(1-x^2)^2}dx=frac{x}{1-x^2}+C$$
I recommend you to take a look here.
Anyway we have, integrating by parts:
$$2I= underbrace{frac{x}{1-x^2}ln^2x bigg|_0^infty}_{=0} +2underbrace{int_0^infty frac{ln x}{x^2-1}dx}_{large =frac{pi^2}{4}}$$
$$Rightarrow 2I= 2cdot frac{pi^2}{4} Rightarrow I=frac{pi^2}{4}$$
For the last integral see here for example.
answered Feb 2 at 13:20
ZackyZacky
7,87511062
$endgroup$
add a comment |
$begingroup$
You're definetly on the right track with that substitution of $x=frac1t$
Basically we have: $$I=int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2}dx=int_0^infty frac{x^2ln^2 x}{(1-x^2)^2}dx$$
Now what if we add them up?
$$2I=int_0^infty ln^2 x frac{1+x^2}{(1-x^2)^2}dx$$
If you don't know how to deal easily with the integral $$int frac{1+x^2}{(1-x^2)^2}dx=frac{x}{1-x^2}+C$$
I recommend you to take a look here.
Anyway we have, integrating by parts:
$$2I= underbrace{frac{x}{1-x^2}ln^2x bigg|_0^infty}_{=0} +2underbrace{int_0^infty frac{ln x}{x^2-1}dx}_{large =frac{pi^2}{4}}$$
$$Rightarrow 2I= 2cdot frac{pi^2}{4} Rightarrow I=frac{pi^2}{4}$$
For the last integral see here for example.
answered Feb 2 at 13:20
ZackyZacky
7,87511062
$endgroup$
add a comment |
$begingroup$
You're definetly on the right track with that substitution of $x=frac1t$
Basically we have: $$I=int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2}dx=int_0^infty frac{x^2ln^2 x}{(1-x^2)^2}dx$$
Now what if we add them up?
$$2I=int_0^infty ln^2 x frac{1+x^2}{(1-x^2)^2}dx$$
If you don't know how to deal easily with the integral $$int frac{1+x^2}{(1-x^2)^2}dx=frac{x}{1-x^2}+C$$
I recommend you to take a look here.
Anyway we have, integrating by parts:
$$2I= underbrace{frac{x}{1-x^2}ln^2x bigg|_0^infty}_{=0} +2underbrace{int_0^infty frac{ln x}{x^2-1}dx}_{large =frac{pi^2}{4}}$$
$$Rightarrow 2I= 2cdot frac{pi^2}{4} Rightarrow I=frac{pi^2}{4}$$
For the last integral see here for example.
answered Feb 2 at 13:20
ZackyZacky
7,87511062
$endgroup$
You're definetly on the right track with that substitution of $x=frac1t$
Basically we have: $$I=int^{infty}_{0}frac{ln^2(x)}{(1-x^2)^2}dx=int_0^infty frac{x^2ln^2 x}{(1-x^2)^2}dx$$
Now what if we add them up?
$$2I=int_0^infty ln^2 x frac{1+x^2}{(1-x^2)^2}dx$$
If you don't know how to deal easily with the integral $$int frac{1+x^2}{(1-x^2)^2}dx=frac{x}{1-x^2}+C$$
I recommend you to take a look here.
Anyway we have, integrating by parts:
$$2I= underbrace{frac{x}{1-x^2}ln^2x bigg|_0^infty}_{=0} +2underbrace{int_0^infty frac{ln x}{x^2-1}dx}_{large =frac{pi^2}{4}}$$
$$Rightarrow 2I= 2cdot frac{pi^2}{4} Rightarrow I=frac{pi^2}{4}$$
For the last integral see here for example.
answered Feb 2 at 13:20
ZackyZacky
7,87511062
edited Feb 2 at 13:25
answered Feb 2 at 13:20
ZackyZacky
7,87511062
answered Feb 2 at 13:20
ZackyZacky
7,87511062
answered Feb 2 at 13:20
ZackyZacky
7,87511062
7,87511062
add a comment |
add a comment |
$begingroup$
This is a variant of TheSimpliFire's approach given in a comment.
By letting $x=e^t$ we get
$$begin{align*}
int_0^inftyfrac{ln^2(x)}{(1-x^2)^2},dx
&=int_{-infty}^inftyfrac{t^2e^{t}}{(1-e^{2t})^2},dt\
&=int_{0}^{+infty}frac{t^2e^{-t}}{(1-e^{-2t})^2},dt+int_{0}^inftyfrac{t^2e^{-3t}}{(1-e^{-2t})^2},dt\
&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+1)t},dt+sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\
&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+1)^3}+sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}\
&=left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}+sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)+left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}-sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)\
&=2sum_{n=0}^inftyfrac{1}{(2n+1)^2}=2left(sum_{n=0}^inftyfrac{1}{n^2}-sum_{n=0}^inftyfrac{1}{(2n)^2}right)\&=2left(1-frac{1}{4}right)sum_{n=0}^inftyfrac{1}{n^2}=frac{3}{2}cdot frac{pi^2}{6}=frac{pi^2}{4}.
end{align*}$$
answered Feb 9 at 11:31
Robert ZRobert Z
102k1072145
$endgroup$
$begingroup$
Could you provide a reference for those two fascinating sums (the ones with the $(2n+1)^3$ and $(2n+3)^3$ in the denominators)? Thanks
$endgroup$
– clathratus
Feb 9 at 16:15
1
$begingroup$
@clathratus Actually we don't need them. See my edit!
$endgroup$
– Robert Z
Feb 9 at 16:43
$begingroup$
Beautiful! Still though, any sources regarding those sums would be appreciated. They are intriguing indeed.
$endgroup$
– clathratus
Feb 9 at 20:35
1
$begingroup$
$sum_{n=0}^inftyfrac{1}{(2n+1)^3}=sum_{n=1}^inftyfrac{1}{n^3}-sum_{n=1}^inftyfrac{1}{(2n)^3}=frac{7zeta(3)}{8}.$
$endgroup$
– Robert Z
Feb 9 at 21:14
add a comment |
$begingroup$
This is a variant of TheSimpliFire's approach given in a comment.
By letting $x=e^t$ we get
$$begin{align*}
int_0^inftyfrac{ln^2(x)}{(1-x^2)^2},dx
&=int_{-infty}^inftyfrac{t^2e^{t}}{(1-e^{2t})^2},dt\
&=int_{0}^{+infty}frac{t^2e^{-t}}{(1-e^{-2t})^2},dt+int_{0}^inftyfrac{t^2e^{-3t}}{(1-e^{-2t})^2},dt\
&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+1)t},dt+sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\
&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+1)^3}+sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}\
&=left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}+sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)+left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}-sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)\
&=2sum_{n=0}^inftyfrac{1}{(2n+1)^2}=2left(sum_{n=0}^inftyfrac{1}{n^2}-sum_{n=0}^inftyfrac{1}{(2n)^2}right)\&=2left(1-frac{1}{4}right)sum_{n=0}^inftyfrac{1}{n^2}=frac{3}{2}cdot frac{pi^2}{6}=frac{pi^2}{4}.
end{align*}$$
answered Feb 9 at 11:31
Robert ZRobert Z
102k1072145
$endgroup$
$begingroup$
Could you provide a reference for those two fascinating sums (the ones with the $(2n+1)^3$ and $(2n+3)^3$ in the denominators)? Thanks
$endgroup$
– clathratus
Feb 9 at 16:15
1
$begingroup$
@clathratus Actually we don't need them. See my edit!
$endgroup$
– Robert Z
Feb 9 at 16:43
$begingroup$
Beautiful! Still though, any sources regarding those sums would be appreciated. They are intriguing indeed.
$endgroup$
– clathratus
Feb 9 at 20:35
1
$begingroup$
$sum_{n=0}^inftyfrac{1}{(2n+1)^3}=sum_{n=1}^inftyfrac{1}{n^3}-sum_{n=1}^inftyfrac{1}{(2n)^3}=frac{7zeta(3)}{8}.$
$endgroup$
– Robert Z
Feb 9 at 21:14
add a comment |
$begingroup$
This is a variant of TheSimpliFire's approach given in a comment.
By letting $x=e^t$ we get
$$begin{align*}
int_0^inftyfrac{ln^2(x)}{(1-x^2)^2},dx
&=int_{-infty}^inftyfrac{t^2e^{t}}{(1-e^{2t})^2},dt\
&=int_{0}^{+infty}frac{t^2e^{-t}}{(1-e^{-2t})^2},dt+int_{0}^inftyfrac{t^2e^{-3t}}{(1-e^{-2t})^2},dt\
&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+1)t},dt+sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\
&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+1)^3}+sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}\
&=left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}+sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)+left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}-sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)\
&=2sum_{n=0}^inftyfrac{1}{(2n+1)^2}=2left(sum_{n=0}^inftyfrac{1}{n^2}-sum_{n=0}^inftyfrac{1}{(2n)^2}right)\&=2left(1-frac{1}{4}right)sum_{n=0}^inftyfrac{1}{n^2}=frac{3}{2}cdot frac{pi^2}{6}=frac{pi^2}{4}.
end{align*}$$
answered Feb 9 at 11:31
Robert ZRobert Z
102k1072145
$endgroup$
This is a variant of TheSimpliFire's approach given in a comment.
By letting $x=e^t$ we get
$$begin{align*}
int_0^inftyfrac{ln^2(x)}{(1-x^2)^2},dx
&=int_{-infty}^inftyfrac{t^2e^{t}}{(1-e^{2t})^2},dt\
&=int_{0}^{+infty}frac{t^2e^{-t}}{(1-e^{-2t})^2},dt+int_{0}^inftyfrac{t^2e^{-3t}}{(1-e^{-2t})^2},dt\
&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+1)t},dt+sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\
&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+1)^3}+sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}\
&=left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}+sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)+left(sum_{n=0}^inftyfrac{1}{(2n+1)^2}-sum_{n=0}^inftyfrac{1}{(2n+1)^3}right)\
&=2sum_{n=0}^inftyfrac{1}{(2n+1)^2}=2left(sum_{n=0}^inftyfrac{1}{n^2}-sum_{n=0}^inftyfrac{1}{(2n)^2}right)\&=2left(1-frac{1}{4}right)sum_{n=0}^inftyfrac{1}{n^2}=frac{3}{2}cdot frac{pi^2}{6}=frac{pi^2}{4}.
end{align*}$$
answered Feb 9 at 11:31
Robert ZRobert Z
102k1072145
edited Feb 9 at 16:56
answered Feb 9 at 11:31
Robert ZRobert Z
102k1072145
answered Feb 9 at 11:31
Robert ZRobert Z
102k1072145
answered Feb 9 at 11:31
Robert ZRobert Z
102k1072145
102k1072145
$begingroup$
Could you provide a reference for those two fascinating sums (the ones with the $(2n+1)^3$ and $(2n+3)^3$ in the denominators)? Thanks
$endgroup$
– clathratus
Feb 9 at 16:15
1
$begingroup$
@clathratus Actually we don't need them. See my edit!
$endgroup$
– Robert Z
Feb 9 at 16:43
$begingroup$
Beautiful! Still though, any sources regarding those sums would be appreciated. They are intriguing indeed.
$endgroup$
– clathratus
Feb 9 at 20:35
1
$begingroup$
$sum_{n=0}^inftyfrac{1}{(2n+1)^3}=sum_{n=1}^inftyfrac{1}{n^3}-sum_{n=1}^inftyfrac{1}{(2n)^3}=frac{7zeta(3)}{8}.$
$endgroup$
– Robert Z
Feb 9 at 21:14
add a comment |
$begingroup$
Could you provide a reference for those two fascinating sums (the ones with the $(2n+1)^3$ and $(2n+3)^3$ in the denominators)? Thanks
$endgroup$
– clathratus
Feb 9 at 16:15
1
$begingroup$
@clathratus Actually we don't need them. See my edit!
$endgroup$
– Robert Z
Feb 9 at 16:43
$begingroup$
Beautiful! Still though, any sources regarding those sums would be appreciated. They are intriguing indeed.
$endgroup$
– clathratus
Feb 9 at 20:35
1
$begingroup$
$sum_{n=0}^inftyfrac{1}{(2n+1)^3}=sum_{n=1}^inftyfrac{1}{n^3}-sum_{n=1}^inftyfrac{1}{(2n)^3}=frac{7zeta(3)}{8}.$
$endgroup$
– Robert Z
Feb 9 at 21:14
$begingroup$
Could you provide a reference for those two fascinating sums (the ones with the $(2n+1)^3$ and $(2n+3)^3$ in the denominators)? Thanks
$endgroup$
– clathratus
Feb 9 at 16:15
$begingroup$
Could you provide a reference for those two fascinating sums (the ones with the $(2n+1)^3$ and $(2n+3)^3$ in the denominators)? Thanks
$endgroup$
– clathratus
Feb 9 at 16:15
1
1
$begingroup$
@clathratus Actually we don't need them. See my edit!
$endgroup$
– Robert Z
Feb 9 at 16:43
$begingroup$
@clathratus Actually we don't need them. See my edit!
$endgroup$
– Robert Z
Feb 9 at 16:43
$begingroup$
Beautiful! Still though, any sources regarding those sums would be appreciated. They are intriguing indeed.
$endgroup$
– clathratus
Feb 9 at 20:35
$begingroup$
Beautiful! Still though, any sources regarding those sums would be appreciated. They are intriguing indeed.
$endgroup$
– clathratus
Feb 9 at 20:35
1
1
$begingroup$
$sum_{n=0}^inftyfrac{1}{(2n+1)^3}=sum_{n=1}^inftyfrac{1}{n^3}-sum_{n=1}^inftyfrac{1}{(2n)^3}=frac{7zeta(3)}{8}.$
$endgroup$
– Robert Z
Feb 9 at 21:14
$begingroup$
$sum_{n=0}^inftyfrac{1}{(2n+1)^3}=sum_{n=1}^inftyfrac{1}{n^3}-sum_{n=1}^inftyfrac{1}{(2n)^3}=frac{7zeta(3)}{8}.$
$endgroup$
– Robert Z
Feb 9 at 21:14
add a comment |
$begingroup$
Here is yet another slight variation on a theme.
Let
$$I = int_0^infty frac{ln^2}{(1 - x^2)^2} , dx$$
then
begin{align}
I &= int_0^1 frac{ln^2 x}{(1 - x^2)^2} , dx + int_1^infty frac{ln^2 x}{(1 - x^2)^2} , dx = int_0^1 frac{(1 + x^2) ln^2 x}{(1 - x^2)^2} , dx tag1,
end{align}
after a substitution of $x mapsto 1/x$ has been enforced in the second of the integrals.
As
$$frac{1}{1 - x^2} = sum_{n = 0}^infty x^{2n}, qquad |x| < 1,$$
differentiating with respect to $x$ gives
$$frac{1}{(1 - x^2)^2} = sum_{n = 1}^infty n x^{2n - 2}.$$
On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have
$$I = sum_{n = 1}^infty n int_0^1 (x^{2n - 2} + x^{2n}) ln^2 x , dx.$$
Integrating by parts twice, we are left with
$$I = sum_{n = 1}^infty left (frac{2n}{(2n - 1)^3} + frac{2n}{(2n + 1)^3} right ).$$
As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have
begin{align}
I &= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} + frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 0}^infty left [frac{1}{(2n + 1)^2} + frac{1}{(2n + 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= 2 + 2 sum_{n = 1}^infty frac{1}{(2n + 1)^2}\
&= 2 sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&= 2 left [sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty frac{1}{(2n)^2} right ]\
&= 2 left (1 - frac{1}{4} right ) sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} cdot frac{pi^2}{6}\
&= frac{pi^2}{4},
end{align}
as expected.
answered Feb 9 at 21:44
omegadotomegadot
6,2692829
$endgroup$
add a comment |
$begingroup$
Here is yet another slight variation on a theme.
Let
$$I = int_0^infty frac{ln^2}{(1 - x^2)^2} , dx$$
then
begin{align}
I &= int_0^1 frac{ln^2 x}{(1 - x^2)^2} , dx + int_1^infty frac{ln^2 x}{(1 - x^2)^2} , dx = int_0^1 frac{(1 + x^2) ln^2 x}{(1 - x^2)^2} , dx tag1,
end{align}
after a substitution of $x mapsto 1/x$ has been enforced in the second of the integrals.
As
$$frac{1}{1 - x^2} = sum_{n = 0}^infty x^{2n}, qquad |x| < 1,$$
differentiating with respect to $x$ gives
$$frac{1}{(1 - x^2)^2} = sum_{n = 1}^infty n x^{2n - 2}.$$
On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have
$$I = sum_{n = 1}^infty n int_0^1 (x^{2n - 2} + x^{2n}) ln^2 x , dx.$$
Integrating by parts twice, we are left with
$$I = sum_{n = 1}^infty left (frac{2n}{(2n - 1)^3} + frac{2n}{(2n + 1)^3} right ).$$
As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have
begin{align}
I &= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} + frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 0}^infty left [frac{1}{(2n + 1)^2} + frac{1}{(2n + 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= 2 + 2 sum_{n = 1}^infty frac{1}{(2n + 1)^2}\
&= 2 sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&= 2 left [sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty frac{1}{(2n)^2} right ]\
&= 2 left (1 - frac{1}{4} right ) sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} cdot frac{pi^2}{6}\
&= frac{pi^2}{4},
end{align}
as expected.
answered Feb 9 at 21:44
omegadotomegadot
6,2692829
$endgroup$
add a comment |
$begingroup$
Here is yet another slight variation on a theme.
Let
$$I = int_0^infty frac{ln^2}{(1 - x^2)^2} , dx$$
then
begin{align}
I &= int_0^1 frac{ln^2 x}{(1 - x^2)^2} , dx + int_1^infty frac{ln^2 x}{(1 - x^2)^2} , dx = int_0^1 frac{(1 + x^2) ln^2 x}{(1 - x^2)^2} , dx tag1,
end{align}
after a substitution of $x mapsto 1/x$ has been enforced in the second of the integrals.
As
$$frac{1}{1 - x^2} = sum_{n = 0}^infty x^{2n}, qquad |x| < 1,$$
differentiating with respect to $x$ gives
$$frac{1}{(1 - x^2)^2} = sum_{n = 1}^infty n x^{2n - 2}.$$
On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have
$$I = sum_{n = 1}^infty n int_0^1 (x^{2n - 2} + x^{2n}) ln^2 x , dx.$$
Integrating by parts twice, we are left with
$$I = sum_{n = 1}^infty left (frac{2n}{(2n - 1)^3} + frac{2n}{(2n + 1)^3} right ).$$
As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have
begin{align}
I &= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} + frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 0}^infty left [frac{1}{(2n + 1)^2} + frac{1}{(2n + 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= 2 + 2 sum_{n = 1}^infty frac{1}{(2n + 1)^2}\
&= 2 sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&= 2 left [sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty frac{1}{(2n)^2} right ]\
&= 2 left (1 - frac{1}{4} right ) sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} cdot frac{pi^2}{6}\
&= frac{pi^2}{4},
end{align}
as expected.
answered Feb 9 at 21:44
omegadotomegadot
6,2692829
$endgroup$
Here is yet another slight variation on a theme.
Let
$$I = int_0^infty frac{ln^2}{(1 - x^2)^2} , dx$$
then
begin{align}
I &= int_0^1 frac{ln^2 x}{(1 - x^2)^2} , dx + int_1^infty frac{ln^2 x}{(1 - x^2)^2} , dx = int_0^1 frac{(1 + x^2) ln^2 x}{(1 - x^2)^2} , dx tag1,
end{align}
after a substitution of $x mapsto 1/x$ has been enforced in the second of the integrals.
As
$$frac{1}{1 - x^2} = sum_{n = 0}^infty x^{2n}, qquad |x| < 1,$$
differentiating with respect to $x$ gives
$$frac{1}{(1 - x^2)^2} = sum_{n = 1}^infty n x^{2n - 2}.$$
On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have
$$I = sum_{n = 1}^infty n int_0^1 (x^{2n - 2} + x^{2n}) ln^2 x , dx.$$
Integrating by parts twice, we are left with
$$I = sum_{n = 1}^infty left (frac{2n}{(2n - 1)^3} + frac{2n}{(2n + 1)^3} right ).$$
As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have
begin{align}
I &= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} + frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 1}^infty left [frac{1}{(2n - 1)^2} + frac{1}{(2n - 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= sum_{n = 0}^infty left [frac{1}{(2n + 1)^2} + frac{1}{(2n + 1)^3} right ] + sum_{n = 1}^infty left [frac{1}{(2n + 1)^2} - frac{1}{(2n + 1)^3} right ]\
&= 2 + 2 sum_{n = 1}^infty frac{1}{(2n + 1)^2}\
&= 2 sum_{n = 0}^infty frac{1}{(2n + 1)^2}\
&= 2 left [sum_{n = 1}^infty frac{1}{n^2} - sum_{n = 1}^infty frac{1}{(2n)^2} right ]\
&= 2 left (1 - frac{1}{4} right ) sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} sum_{n = 1}^infty frac{1}{n^2}\
&= frac{3}{2} cdot frac{pi^2}{6}\
&= frac{pi^2}{4},
end{align}
as expected.
answered Feb 9 at 21:44
omegadotomegadot
6,2692829
answered Feb 9 at 21:44
omegadotomegadot
6,2692829
answered Feb 9 at 21:44
omegadotomegadot
6,2692829
answered Feb 9 at 21:44
omegadotomegadot
6,2692829
6,2692829
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%2f3097187%2ffinding-int-infty-0-frac-ln2x1-x22-dx%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
$begingroup$
Are you sure about the limits of the integral? Because the integrant has a singularity of the order $1$ at $x=1$.
$endgroup$
– Mundron Schmidt
Feb 2 at 11:15
$begingroup$
@MundronSchmidt W|A gives $pi^2/4$ as the result
$endgroup$
– TheSimpliFire
Feb 2 at 11:19
5
$begingroup$
Note that $$int_0^inftyfrac{ln^2x}{(1-x^2)^2},dx=int_{-infty}^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx$$ and begin{align}int_0^inftyfrac{x^2e^{-3x}}{(1-e^{2x})^2},dx&=int_0^{infty}left(x^2e^{-3x}sum_{n=0}^{infty}left(left(1+nright)e^{-2nx}right)right),dx\&=sum_{n=0}^infty (1+n)int_0^infty t^2e^{-(2n+3)t},dt\&=sum_{n=0}^inftyfrac{2(1+n)}{(2n+3)^3}=frac{pi^2-7zeta(3)}8end{align}
$endgroup$
– TheSimpliFire
Feb 2 at 11:45