Proving $int_0^infty logleft (1-2fraccos 2thetax^2+frac1x^4 right)dx =2pi sin theta$Computing $int_0^pilnleft(1-2acos x+a^2right) , dx$Find the integral of log(1 - a/x^2 + 1/x^4) from 0 to infinityEvaluating $int_0^largefracpi4 logleft( cos xright) , mathrmdx $$iint_D cos left( fracx-yx+y right),dA$Evaluate: $int_0^pi ln left( sin theta right) dtheta$Computation of $int_0^pi fracsin^n theta(1+x^2-2x cdot cos theta)^fracn2 , dtheta$A Challenging Integral $int_0^fracpi2log left( x^2+log^2(cos x)right)dx$Log Sine: $int_0^pi theta^2 ln^2big(2sinfractheta2big)d theta.$Prove $cos(2theta) + cosleft(2 left(fracpi3 + thetaright)right) +cosleft(2 left(frac2pi3 + thetaright)right) = 0$Closed form of $int_0^pi/2 fracarctan^2 (sin^2 theta)sin^2 theta,dtheta$Interesting integral: $I=int_0^1 int_0^1 logleft( cos(pi x)^2 + cos(pi y)^2 right)dxdy$Substitutions for the trigonometric integral $int cos theta cos^5left(sin thetaright) dtheta$
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Proving $int_0^infty logleft (1-2fraccos 2thetax^2+frac1x^4 right)dx =2pi sin theta$
Computing $int_0^pilnleft(1-2acos x+a^2right) , dx$Find the integral of log(1 - a/x^2 + 1/x^4) from 0 to infinityEvaluating $int_0^largefracpi4 logleft( cos xright) , mathrmdx $$iint_D cos left( fracx-yx+y right),dA$Evaluate: $int_0^pi ln left( sin theta right) dtheta$Computation of $int_0^pi fracsin^n theta(1+x^2-2x cdot cos theta)^fracn2 , dtheta$A Challenging Integral $int_0^fracpi2log left( x^2+log^2(cos x)right)dx$Log Sine: $int_0^pi theta^2 ln^2big(2sinfractheta2big)d theta.$Prove $cos(2theta) + cosleft(2 left(fracpi3 + thetaright)right) +cosleft(2 left(frac2pi3 + thetaright)right) = 0$Closed form of $int_0^pi/2 fracarctan^2 (sin^2 theta)sin^2 theta,dtheta$Interesting integral: $I=int_0^1 int_0^1 logleft( cos(pi x)^2 + cos(pi y)^2 right)dxdy$Substitutions for the trigonometric integral $int cos theta cos^5left(sin thetaright) dtheta$
$begingroup$
Prove $$int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dx =2pi sin theta$$where $thetain[0,pi]$.
I've met another similar problem,
$$ int_0^2pi log(1-2rcos theta +r^2) dtheta=2pi log^+(r^2) $$
I am curious whether there is any relationship between them.
And I got stuck on the proposition in the title. I found that
$$1-2fraccos 2thetax^2+frac1x^4 =left(frac1x-e^ithetaright)left(frac1x+e^ithetaright)left(frac1x-e^-ithetaright)left(frac1x+e^-ithetaright)$$
But I couldn't move on.
Any hints? Thanks in advance.
calculus integration trigonometry definite-integrals logarithms
$endgroup$
add a comment |
$begingroup$
Prove $$int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dx =2pi sin theta$$where $thetain[0,pi]$.
I've met another similar problem,
$$ int_0^2pi log(1-2rcos theta +r^2) dtheta=2pi log^+(r^2) $$
I am curious whether there is any relationship between them.
And I got stuck on the proposition in the title. I found that
$$1-2fraccos 2thetax^2+frac1x^4 =left(frac1x-e^ithetaright)left(frac1x+e^ithetaright)left(frac1x-e^-ithetaright)left(frac1x+e^-ithetaright)$$
But I couldn't move on.
Any hints? Thanks in advance.
calculus integration trigonometry definite-integrals logarithms
$endgroup$
$begingroup$
Maybe you can complete a square and make a substitution afterwards.
$endgroup$
– mathreadler
Mar 11 at 10:30
1
$begingroup$
I am not sure about a relationship between them, but both of them can be solved by similar tehniques (Feynman's trick).
$endgroup$
– Zacky
Mar 11 at 11:11
add a comment |
$begingroup$
Prove $$int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dx =2pi sin theta$$where $thetain[0,pi]$.
I've met another similar problem,
$$ int_0^2pi log(1-2rcos theta +r^2) dtheta=2pi log^+(r^2) $$
I am curious whether there is any relationship between them.
And I got stuck on the proposition in the title. I found that
$$1-2fraccos 2thetax^2+frac1x^4 =left(frac1x-e^ithetaright)left(frac1x+e^ithetaright)left(frac1x-e^-ithetaright)left(frac1x+e^-ithetaright)$$
But I couldn't move on.
Any hints? Thanks in advance.
calculus integration trigonometry definite-integrals logarithms
$endgroup$
Prove $$int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dx =2pi sin theta$$where $thetain[0,pi]$.
I've met another similar problem,
$$ int_0^2pi log(1-2rcos theta +r^2) dtheta=2pi log^+(r^2) $$
I am curious whether there is any relationship between them.
And I got stuck on the proposition in the title. I found that
$$1-2fraccos 2thetax^2+frac1x^4 =left(frac1x-e^ithetaright)left(frac1x+e^ithetaright)left(frac1x-e^-ithetaright)left(frac1x+e^-ithetaright)$$
But I couldn't move on.
Any hints? Thanks in advance.
calculus integration trigonometry definite-integrals logarithms
calculus integration trigonometry definite-integrals logarithms
edited Mar 11 at 12:18
Zero
asked Mar 11 at 10:20
ZeroZero
47210
47210
$begingroup$
Maybe you can complete a square and make a substitution afterwards.
$endgroup$
– mathreadler
Mar 11 at 10:30
1
$begingroup$
I am not sure about a relationship between them, but both of them can be solved by similar tehniques (Feynman's trick).
$endgroup$
– Zacky
Mar 11 at 11:11
add a comment |
$begingroup$
Maybe you can complete a square and make a substitution afterwards.
$endgroup$
– mathreadler
Mar 11 at 10:30
1
$begingroup$
I am not sure about a relationship between them, but both of them can be solved by similar tehniques (Feynman's trick).
$endgroup$
– Zacky
Mar 11 at 11:11
$begingroup$
Maybe you can complete a square and make a substitution afterwards.
$endgroup$
– mathreadler
Mar 11 at 10:30
$begingroup$
Maybe you can complete a square and make a substitution afterwards.
$endgroup$
– mathreadler
Mar 11 at 10:30
1
1
$begingroup$
I am not sure about a relationship between them, but both of them can be solved by similar tehniques (Feynman's trick).
$endgroup$
– Zacky
Mar 11 at 11:11
$begingroup$
I am not sure about a relationship between them, but both of them can be solved by similar tehniques (Feynman's trick).
$endgroup$
– Zacky
Mar 11 at 11:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We start off by some $xrightarrow frac1x$ substitutions while derivating under the integral sign: $$I(theta)=int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dxoversetxrightarrow frac1x=int_0^infty fracln(1- 2cos(2theta) x^2 +x^4)x^2dx$$
$$I'(theta)=4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dxoversetxrightarrow frac1x=4int_0^infty fracsin(2theta)x^2x^4-2cos(2theta)x^2+1dx$$
Now summing up the two integrals from above gives us:
$$Rightarrow 2I'(theta)=4int_0^infty fracsin(2theta)(1+x^2)x^4-2cos(2theta)x^2+1dx=4int_0^infty fracsin(2theta)left(frac1x^2+1right)x^2+frac1x^2-2cos(2theta)dx$$
$$Rightarrow I'(theta)=2int_0^infty fracsin(2theta)left(x-frac1xright)'left(x-frac1xright)^2 +2(1-cos(2theta))dxoversetlarge x- frac1x=t=2int_-infty^infty fracsin(2theta)t^2 +4sin^2 (theta)dt$$
$$=2 fracsin(2theta)2sin(theta)arctanleft(fract2sin(theta)right)bigg|_-infty^infty=2cos(theta) cdot pi$$
$$Rightarrow I(theta) = 2pi int cos(theta) dtheta =2pi sin theta +C$$
But $I(0)=0$ (see J.G. answer), thus:$$I(0)=0+CRightarrow C=0 Rightarrow boxedI(theta)=2pisin(theta)$$
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$begingroup$
Nice idea! But how to deal with $I'(0)$?
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– Zero
Mar 11 at 12:50
$begingroup$
Well, just plugg $theta =0$ here: $$4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dx$$
$endgroup$
– Zacky
Mar 11 at 13:02
$begingroup$
Sorry, I didn't understand. I thought if $theta=0$, it equals to $4int_0^infty frac0(x^2-1)^2dx$ , which is not integrable. Did I miss something?
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– Zero
Mar 12 at 0:27
$begingroup$
Wait, why doesn't it vanish after plugging $theta=0$?
$endgroup$
– Zacky
2 days ago
$begingroup$
My thought: If you put $I'(theta)=4int_0^infty fracsin (2theta)x^4-2cos(2theta)x^2+1dx$, then $I'(0)=4int_0^infty frac0(x^2-1)^2$ which doesn't equal to $2pi$
$endgroup$
– Zero
12 hours ago
|
show 2 more comments
$begingroup$
For $theta in [0;pi]$,
beginalign
J(theta)&=int_0^infty lnleft(1-frac2cos(2theta)x^2+frac1x^4right) ,dx
endalign
Perform the change of variable $y=dfrac1x$,
beginalign
J(theta)&=int_0^infty fraclnleft(1-2cos(2theta)x^2+x^4right)x^2 ,dx
endalign
For $ageq -1$, define the function $F$ by,
beginalignF(a)&=int_0^infty fraclnleft(1+2ax^2+x^4right)x^2 ,dx\
&=left[-fraclnleft(1+2ax^2+x^4right)xright]_0^infty+int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
&=int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
endalign
Perform the change of variable $y=dfrac1x$,
beginalignF(a)&=int_0^infty frac4left( frac1x^2+aright) x^2left(1+frac2ax^2+frac1x^4right) ,dx\
&=int_0^infty frac4left( 1+ax^2right) x^4+2ax^2+1 ,dx\
endalign
Therefore,
beginalignF(a)&=int_0^infty frac2(a+1)left( 1+x^2right) x^4+2ax^2+1 ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)x^2+frac1x^2+2a ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)left(x-frac1xright)^2+2(a+1) ,dx\
endalign
Perform the change of variable $y=x-dfrac1x$,
beginalignF(a)&= 2(a+1)int_-infty^+inftyfrac1x^2+2(a+1),dx\
&=4(a+1)int_0^+inftyfrac1x^2+2(a+1),dx\
&=left[2sqrt2(a+1)arctanleft( fracxsqrt2(a+1) right)right]_0^infty\
&=boxedpisqrt2(1+a)
endalign
Observe that, $J(theta)=Fbig(-cos(2theta)big)$.
beginalign 2(1-cos(2theta))&=2(1-cos^2(theta)+sin^2 (theta))\
&=2times 2sin^2 (theta)\
&=4times sin^2 (theta)\
endalign
Since, for $theta in [0;pi],sin(theta)geq 0$ then $sqrt2(1-cos(2theta))=2sin(theta)$
Therefore,
beginalignboxedJ(theta)=2pi sin(theta)endalign
$endgroup$
add a comment |
$begingroup$
To complement Zacky's answer I'll add a proof that $I(0)=0$. Applying $xmapstofrac1x$ for $xge 1$ gives $$I(x)=2int_0^inftyln|1-x^-2|dx=2int_0^1left[(1+frac1x^2)ln(1-x^2)-2ln xright]\=-2int_0^1left[(1+x^2)sum_nge 0fracx^2nn+1+2ln xright]dx=-2int_0^1left[sum_nge 0left(fracx^2nn+1+fracx^2n+2n+1right)+2ln xright]dx\=-2left[sum_nge 0left(fracx^2n+1(n+1)(2n+1)+fracx^2n+3(n+1)(2n+3)right)+2xln x-2xright]_0^1\=-2left[sum_nge 0left(frac4(2n+1)(2n+3)right)-2right],$$which vanishes by partial fractions.
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
We start off by some $xrightarrow frac1x$ substitutions while derivating under the integral sign: $$I(theta)=int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dxoversetxrightarrow frac1x=int_0^infty fracln(1- 2cos(2theta) x^2 +x^4)x^2dx$$
$$I'(theta)=4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dxoversetxrightarrow frac1x=4int_0^infty fracsin(2theta)x^2x^4-2cos(2theta)x^2+1dx$$
Now summing up the two integrals from above gives us:
$$Rightarrow 2I'(theta)=4int_0^infty fracsin(2theta)(1+x^2)x^4-2cos(2theta)x^2+1dx=4int_0^infty fracsin(2theta)left(frac1x^2+1right)x^2+frac1x^2-2cos(2theta)dx$$
$$Rightarrow I'(theta)=2int_0^infty fracsin(2theta)left(x-frac1xright)'left(x-frac1xright)^2 +2(1-cos(2theta))dxoversetlarge x- frac1x=t=2int_-infty^infty fracsin(2theta)t^2 +4sin^2 (theta)dt$$
$$=2 fracsin(2theta)2sin(theta)arctanleft(fract2sin(theta)right)bigg|_-infty^infty=2cos(theta) cdot pi$$
$$Rightarrow I(theta) = 2pi int cos(theta) dtheta =2pi sin theta +C$$
But $I(0)=0$ (see J.G. answer), thus:$$I(0)=0+CRightarrow C=0 Rightarrow boxedI(theta)=2pisin(theta)$$
$endgroup$
$begingroup$
Nice idea! But how to deal with $I'(0)$?
$endgroup$
– Zero
Mar 11 at 12:50
$begingroup$
Well, just plugg $theta =0$ here: $$4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dx$$
$endgroup$
– Zacky
Mar 11 at 13:02
$begingroup$
Sorry, I didn't understand. I thought if $theta=0$, it equals to $4int_0^infty frac0(x^2-1)^2dx$ , which is not integrable. Did I miss something?
$endgroup$
– Zero
Mar 12 at 0:27
$begingroup$
Wait, why doesn't it vanish after plugging $theta=0$?
$endgroup$
– Zacky
2 days ago
$begingroup$
My thought: If you put $I'(theta)=4int_0^infty fracsin (2theta)x^4-2cos(2theta)x^2+1dx$, then $I'(0)=4int_0^infty frac0(x^2-1)^2$ which doesn't equal to $2pi$
$endgroup$
– Zero
12 hours ago
|
show 2 more comments
$begingroup$
We start off by some $xrightarrow frac1x$ substitutions while derivating under the integral sign: $$I(theta)=int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dxoversetxrightarrow frac1x=int_0^infty fracln(1- 2cos(2theta) x^2 +x^4)x^2dx$$
$$I'(theta)=4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dxoversetxrightarrow frac1x=4int_0^infty fracsin(2theta)x^2x^4-2cos(2theta)x^2+1dx$$
Now summing up the two integrals from above gives us:
$$Rightarrow 2I'(theta)=4int_0^infty fracsin(2theta)(1+x^2)x^4-2cos(2theta)x^2+1dx=4int_0^infty fracsin(2theta)left(frac1x^2+1right)x^2+frac1x^2-2cos(2theta)dx$$
$$Rightarrow I'(theta)=2int_0^infty fracsin(2theta)left(x-frac1xright)'left(x-frac1xright)^2 +2(1-cos(2theta))dxoversetlarge x- frac1x=t=2int_-infty^infty fracsin(2theta)t^2 +4sin^2 (theta)dt$$
$$=2 fracsin(2theta)2sin(theta)arctanleft(fract2sin(theta)right)bigg|_-infty^infty=2cos(theta) cdot pi$$
$$Rightarrow I(theta) = 2pi int cos(theta) dtheta =2pi sin theta +C$$
But $I(0)=0$ (see J.G. answer), thus:$$I(0)=0+CRightarrow C=0 Rightarrow boxedI(theta)=2pisin(theta)$$
$endgroup$
$begingroup$
Nice idea! But how to deal with $I'(0)$?
$endgroup$
– Zero
Mar 11 at 12:50
$begingroup$
Well, just plugg $theta =0$ here: $$4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dx$$
$endgroup$
– Zacky
Mar 11 at 13:02
$begingroup$
Sorry, I didn't understand. I thought if $theta=0$, it equals to $4int_0^infty frac0(x^2-1)^2dx$ , which is not integrable. Did I miss something?
$endgroup$
– Zero
Mar 12 at 0:27
$begingroup$
Wait, why doesn't it vanish after plugging $theta=0$?
$endgroup$
– Zacky
2 days ago
$begingroup$
My thought: If you put $I'(theta)=4int_0^infty fracsin (2theta)x^4-2cos(2theta)x^2+1dx$, then $I'(0)=4int_0^infty frac0(x^2-1)^2$ which doesn't equal to $2pi$
$endgroup$
– Zero
12 hours ago
|
show 2 more comments
$begingroup$
We start off by some $xrightarrow frac1x$ substitutions while derivating under the integral sign: $$I(theta)=int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dxoversetxrightarrow frac1x=int_0^infty fracln(1- 2cos(2theta) x^2 +x^4)x^2dx$$
$$I'(theta)=4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dxoversetxrightarrow frac1x=4int_0^infty fracsin(2theta)x^2x^4-2cos(2theta)x^2+1dx$$
Now summing up the two integrals from above gives us:
$$Rightarrow 2I'(theta)=4int_0^infty fracsin(2theta)(1+x^2)x^4-2cos(2theta)x^2+1dx=4int_0^infty fracsin(2theta)left(frac1x^2+1right)x^2+frac1x^2-2cos(2theta)dx$$
$$Rightarrow I'(theta)=2int_0^infty fracsin(2theta)left(x-frac1xright)'left(x-frac1xright)^2 +2(1-cos(2theta))dxoversetlarge x- frac1x=t=2int_-infty^infty fracsin(2theta)t^2 +4sin^2 (theta)dt$$
$$=2 fracsin(2theta)2sin(theta)arctanleft(fract2sin(theta)right)bigg|_-infty^infty=2cos(theta) cdot pi$$
$$Rightarrow I(theta) = 2pi int cos(theta) dtheta =2pi sin theta +C$$
But $I(0)=0$ (see J.G. answer), thus:$$I(0)=0+CRightarrow C=0 Rightarrow boxedI(theta)=2pisin(theta)$$
$endgroup$
We start off by some $xrightarrow frac1x$ substitutions while derivating under the integral sign: $$I(theta)=int_0^infty log left(1-2fraccos 2thetax^2+frac1x^4 right)dxoversetxrightarrow frac1x=int_0^infty fracln(1- 2cos(2theta) x^2 +x^4)x^2dx$$
$$I'(theta)=4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dxoversetxrightarrow frac1x=4int_0^infty fracsin(2theta)x^2x^4-2cos(2theta)x^2+1dx$$
Now summing up the two integrals from above gives us:
$$Rightarrow 2I'(theta)=4int_0^infty fracsin(2theta)(1+x^2)x^4-2cos(2theta)x^2+1dx=4int_0^infty fracsin(2theta)left(frac1x^2+1right)x^2+frac1x^2-2cos(2theta)dx$$
$$Rightarrow I'(theta)=2int_0^infty fracsin(2theta)left(x-frac1xright)'left(x-frac1xright)^2 +2(1-cos(2theta))dxoversetlarge x- frac1x=t=2int_-infty^infty fracsin(2theta)t^2 +4sin^2 (theta)dt$$
$$=2 fracsin(2theta)2sin(theta)arctanleft(fract2sin(theta)right)bigg|_-infty^infty=2cos(theta) cdot pi$$
$$Rightarrow I(theta) = 2pi int cos(theta) dtheta =2pi sin theta +C$$
But $I(0)=0$ (see J.G. answer), thus:$$I(0)=0+CRightarrow C=0 Rightarrow boxedI(theta)=2pisin(theta)$$
edited Mar 11 at 11:23
answered Mar 11 at 11:01
ZackyZacky
7,74511061
7,74511061
$begingroup$
Nice idea! But how to deal with $I'(0)$?
$endgroup$
– Zero
Mar 11 at 12:50
$begingroup$
Well, just plugg $theta =0$ here: $$4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dx$$
$endgroup$
– Zacky
Mar 11 at 13:02
$begingroup$
Sorry, I didn't understand. I thought if $theta=0$, it equals to $4int_0^infty frac0(x^2-1)^2dx$ , which is not integrable. Did I miss something?
$endgroup$
– Zero
Mar 12 at 0:27
$begingroup$
Wait, why doesn't it vanish after plugging $theta=0$?
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– Zacky
2 days ago
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My thought: If you put $I'(theta)=4int_0^infty fracsin (2theta)x^4-2cos(2theta)x^2+1dx$, then $I'(0)=4int_0^infty frac0(x^2-1)^2$ which doesn't equal to $2pi$
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– Zero
12 hours ago
|
show 2 more comments
$begingroup$
Nice idea! But how to deal with $I'(0)$?
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– Zero
Mar 11 at 12:50
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Well, just plugg $theta =0$ here: $$4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dx$$
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– Zacky
Mar 11 at 13:02
$begingroup$
Sorry, I didn't understand. I thought if $theta=0$, it equals to $4int_0^infty frac0(x^2-1)^2dx$ , which is not integrable. Did I miss something?
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– Zero
Mar 12 at 0:27
$begingroup$
Wait, why doesn't it vanish after plugging $theta=0$?
$endgroup$
– Zacky
2 days ago
$begingroup$
My thought: If you put $I'(theta)=4int_0^infty fracsin (2theta)x^4-2cos(2theta)x^2+1dx$, then $I'(0)=4int_0^infty frac0(x^2-1)^2$ which doesn't equal to $2pi$
$endgroup$
– Zero
12 hours ago
$begingroup$
Nice idea! But how to deal with $I'(0)$?
$endgroup$
– Zero
Mar 11 at 12:50
$begingroup$
Nice idea! But how to deal with $I'(0)$?
$endgroup$
– Zero
Mar 11 at 12:50
$begingroup$
Well, just plugg $theta =0$ here: $$4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dx$$
$endgroup$
– Zacky
Mar 11 at 13:02
$begingroup$
Well, just plugg $theta =0$ here: $$4int_0^infty fracsin(2theta)x^4-2cos(2theta)x^2+1dx$$
$endgroup$
– Zacky
Mar 11 at 13:02
$begingroup$
Sorry, I didn't understand. I thought if $theta=0$, it equals to $4int_0^infty frac0(x^2-1)^2dx$ , which is not integrable. Did I miss something?
$endgroup$
– Zero
Mar 12 at 0:27
$begingroup$
Sorry, I didn't understand. I thought if $theta=0$, it equals to $4int_0^infty frac0(x^2-1)^2dx$ , which is not integrable. Did I miss something?
$endgroup$
– Zero
Mar 12 at 0:27
$begingroup$
Wait, why doesn't it vanish after plugging $theta=0$?
$endgroup$
– Zacky
2 days ago
$begingroup$
Wait, why doesn't it vanish after plugging $theta=0$?
$endgroup$
– Zacky
2 days ago
$begingroup$
My thought: If you put $I'(theta)=4int_0^infty fracsin (2theta)x^4-2cos(2theta)x^2+1dx$, then $I'(0)=4int_0^infty frac0(x^2-1)^2$ which doesn't equal to $2pi$
$endgroup$
– Zero
12 hours ago
$begingroup$
My thought: If you put $I'(theta)=4int_0^infty fracsin (2theta)x^4-2cos(2theta)x^2+1dx$, then $I'(0)=4int_0^infty frac0(x^2-1)^2$ which doesn't equal to $2pi$
$endgroup$
– Zero
12 hours ago
|
show 2 more comments
$begingroup$
For $theta in [0;pi]$,
beginalign
J(theta)&=int_0^infty lnleft(1-frac2cos(2theta)x^2+frac1x^4right) ,dx
endalign
Perform the change of variable $y=dfrac1x$,
beginalign
J(theta)&=int_0^infty fraclnleft(1-2cos(2theta)x^2+x^4right)x^2 ,dx
endalign
For $ageq -1$, define the function $F$ by,
beginalignF(a)&=int_0^infty fraclnleft(1+2ax^2+x^4right)x^2 ,dx\
&=left[-fraclnleft(1+2ax^2+x^4right)xright]_0^infty+int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
&=int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
endalign
Perform the change of variable $y=dfrac1x$,
beginalignF(a)&=int_0^infty frac4left( frac1x^2+aright) x^2left(1+frac2ax^2+frac1x^4right) ,dx\
&=int_0^infty frac4left( 1+ax^2right) x^4+2ax^2+1 ,dx\
endalign
Therefore,
beginalignF(a)&=int_0^infty frac2(a+1)left( 1+x^2right) x^4+2ax^2+1 ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)x^2+frac1x^2+2a ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)left(x-frac1xright)^2+2(a+1) ,dx\
endalign
Perform the change of variable $y=x-dfrac1x$,
beginalignF(a)&= 2(a+1)int_-infty^+inftyfrac1x^2+2(a+1),dx\
&=4(a+1)int_0^+inftyfrac1x^2+2(a+1),dx\
&=left[2sqrt2(a+1)arctanleft( fracxsqrt2(a+1) right)right]_0^infty\
&=boxedpisqrt2(1+a)
endalign
Observe that, $J(theta)=Fbig(-cos(2theta)big)$.
beginalign 2(1-cos(2theta))&=2(1-cos^2(theta)+sin^2 (theta))\
&=2times 2sin^2 (theta)\
&=4times sin^2 (theta)\
endalign
Since, for $theta in [0;pi],sin(theta)geq 0$ then $sqrt2(1-cos(2theta))=2sin(theta)$
Therefore,
beginalignboxedJ(theta)=2pi sin(theta)endalign
$endgroup$
add a comment |
$begingroup$
For $theta in [0;pi]$,
beginalign
J(theta)&=int_0^infty lnleft(1-frac2cos(2theta)x^2+frac1x^4right) ,dx
endalign
Perform the change of variable $y=dfrac1x$,
beginalign
J(theta)&=int_0^infty fraclnleft(1-2cos(2theta)x^2+x^4right)x^2 ,dx
endalign
For $ageq -1$, define the function $F$ by,
beginalignF(a)&=int_0^infty fraclnleft(1+2ax^2+x^4right)x^2 ,dx\
&=left[-fraclnleft(1+2ax^2+x^4right)xright]_0^infty+int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
&=int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
endalign
Perform the change of variable $y=dfrac1x$,
beginalignF(a)&=int_0^infty frac4left( frac1x^2+aright) x^2left(1+frac2ax^2+frac1x^4right) ,dx\
&=int_0^infty frac4left( 1+ax^2right) x^4+2ax^2+1 ,dx\
endalign
Therefore,
beginalignF(a)&=int_0^infty frac2(a+1)left( 1+x^2right) x^4+2ax^2+1 ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)x^2+frac1x^2+2a ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)left(x-frac1xright)^2+2(a+1) ,dx\
endalign
Perform the change of variable $y=x-dfrac1x$,
beginalignF(a)&= 2(a+1)int_-infty^+inftyfrac1x^2+2(a+1),dx\
&=4(a+1)int_0^+inftyfrac1x^2+2(a+1),dx\
&=left[2sqrt2(a+1)arctanleft( fracxsqrt2(a+1) right)right]_0^infty\
&=boxedpisqrt2(1+a)
endalign
Observe that, $J(theta)=Fbig(-cos(2theta)big)$.
beginalign 2(1-cos(2theta))&=2(1-cos^2(theta)+sin^2 (theta))\
&=2times 2sin^2 (theta)\
&=4times sin^2 (theta)\
endalign
Since, for $theta in [0;pi],sin(theta)geq 0$ then $sqrt2(1-cos(2theta))=2sin(theta)$
Therefore,
beginalignboxedJ(theta)=2pi sin(theta)endalign
$endgroup$
add a comment |
$begingroup$
For $theta in [0;pi]$,
beginalign
J(theta)&=int_0^infty lnleft(1-frac2cos(2theta)x^2+frac1x^4right) ,dx
endalign
Perform the change of variable $y=dfrac1x$,
beginalign
J(theta)&=int_0^infty fraclnleft(1-2cos(2theta)x^2+x^4right)x^2 ,dx
endalign
For $ageq -1$, define the function $F$ by,
beginalignF(a)&=int_0^infty fraclnleft(1+2ax^2+x^4right)x^2 ,dx\
&=left[-fraclnleft(1+2ax^2+x^4right)xright]_0^infty+int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
&=int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
endalign
Perform the change of variable $y=dfrac1x$,
beginalignF(a)&=int_0^infty frac4left( frac1x^2+aright) x^2left(1+frac2ax^2+frac1x^4right) ,dx\
&=int_0^infty frac4left( 1+ax^2right) x^4+2ax^2+1 ,dx\
endalign
Therefore,
beginalignF(a)&=int_0^infty frac2(a+1)left( 1+x^2right) x^4+2ax^2+1 ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)x^2+frac1x^2+2a ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)left(x-frac1xright)^2+2(a+1) ,dx\
endalign
Perform the change of variable $y=x-dfrac1x$,
beginalignF(a)&= 2(a+1)int_-infty^+inftyfrac1x^2+2(a+1),dx\
&=4(a+1)int_0^+inftyfrac1x^2+2(a+1),dx\
&=left[2sqrt2(a+1)arctanleft( fracxsqrt2(a+1) right)right]_0^infty\
&=boxedpisqrt2(1+a)
endalign
Observe that, $J(theta)=Fbig(-cos(2theta)big)$.
beginalign 2(1-cos(2theta))&=2(1-cos^2(theta)+sin^2 (theta))\
&=2times 2sin^2 (theta)\
&=4times sin^2 (theta)\
endalign
Since, for $theta in [0;pi],sin(theta)geq 0$ then $sqrt2(1-cos(2theta))=2sin(theta)$
Therefore,
beginalignboxedJ(theta)=2pi sin(theta)endalign
$endgroup$
For $theta in [0;pi]$,
beginalign
J(theta)&=int_0^infty lnleft(1-frac2cos(2theta)x^2+frac1x^4right) ,dx
endalign
Perform the change of variable $y=dfrac1x$,
beginalign
J(theta)&=int_0^infty fraclnleft(1-2cos(2theta)x^2+x^4right)x^2 ,dx
endalign
For $ageq -1$, define the function $F$ by,
beginalignF(a)&=int_0^infty fraclnleft(1+2ax^2+x^4right)x^2 ,dx\
&=left[-fraclnleft(1+2ax^2+x^4right)xright]_0^infty+int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
&=int_0^infty frac4left(x^2+aright)1+2ax^2+x^4,dx\
endalign
Perform the change of variable $y=dfrac1x$,
beginalignF(a)&=int_0^infty frac4left( frac1x^2+aright) x^2left(1+frac2ax^2+frac1x^4right) ,dx\
&=int_0^infty frac4left( 1+ax^2right) x^4+2ax^2+1 ,dx\
endalign
Therefore,
beginalignF(a)&=int_0^infty frac2(a+1)left( 1+x^2right) x^4+2ax^2+1 ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)x^2+frac1x^2+2a ,dx\
&=2(a+1)int_0^infty fracleft(1+frac1x^2right)left(x-frac1xright)^2+2(a+1) ,dx\
endalign
Perform the change of variable $y=x-dfrac1x$,
beginalignF(a)&= 2(a+1)int_-infty^+inftyfrac1x^2+2(a+1),dx\
&=4(a+1)int_0^+inftyfrac1x^2+2(a+1),dx\
&=left[2sqrt2(a+1)arctanleft( fracxsqrt2(a+1) right)right]_0^infty\
&=boxedpisqrt2(1+a)
endalign
Observe that, $J(theta)=Fbig(-cos(2theta)big)$.
beginalign 2(1-cos(2theta))&=2(1-cos^2(theta)+sin^2 (theta))\
&=2times 2sin^2 (theta)\
&=4times sin^2 (theta)\
endalign
Since, for $theta in [0;pi],sin(theta)geq 0$ then $sqrt2(1-cos(2theta))=2sin(theta)$
Therefore,
beginalignboxedJ(theta)=2pi sin(theta)endalign
answered Mar 11 at 19:49
FDPFDP
6,15211829
6,15211829
add a comment |
add a comment |
$begingroup$
To complement Zacky's answer I'll add a proof that $I(0)=0$. Applying $xmapstofrac1x$ for $xge 1$ gives $$I(x)=2int_0^inftyln|1-x^-2|dx=2int_0^1left[(1+frac1x^2)ln(1-x^2)-2ln xright]\=-2int_0^1left[(1+x^2)sum_nge 0fracx^2nn+1+2ln xright]dx=-2int_0^1left[sum_nge 0left(fracx^2nn+1+fracx^2n+2n+1right)+2ln xright]dx\=-2left[sum_nge 0left(fracx^2n+1(n+1)(2n+1)+fracx^2n+3(n+1)(2n+3)right)+2xln x-2xright]_0^1\=-2left[sum_nge 0left(frac4(2n+1)(2n+3)right)-2right],$$which vanishes by partial fractions.
$endgroup$
add a comment |
$begingroup$
To complement Zacky's answer I'll add a proof that $I(0)=0$. Applying $xmapstofrac1x$ for $xge 1$ gives $$I(x)=2int_0^inftyln|1-x^-2|dx=2int_0^1left[(1+frac1x^2)ln(1-x^2)-2ln xright]\=-2int_0^1left[(1+x^2)sum_nge 0fracx^2nn+1+2ln xright]dx=-2int_0^1left[sum_nge 0left(fracx^2nn+1+fracx^2n+2n+1right)+2ln xright]dx\=-2left[sum_nge 0left(fracx^2n+1(n+1)(2n+1)+fracx^2n+3(n+1)(2n+3)right)+2xln x-2xright]_0^1\=-2left[sum_nge 0left(frac4(2n+1)(2n+3)right)-2right],$$which vanishes by partial fractions.
$endgroup$
add a comment |
$begingroup$
To complement Zacky's answer I'll add a proof that $I(0)=0$. Applying $xmapstofrac1x$ for $xge 1$ gives $$I(x)=2int_0^inftyln|1-x^-2|dx=2int_0^1left[(1+frac1x^2)ln(1-x^2)-2ln xright]\=-2int_0^1left[(1+x^2)sum_nge 0fracx^2nn+1+2ln xright]dx=-2int_0^1left[sum_nge 0left(fracx^2nn+1+fracx^2n+2n+1right)+2ln xright]dx\=-2left[sum_nge 0left(fracx^2n+1(n+1)(2n+1)+fracx^2n+3(n+1)(2n+3)right)+2xln x-2xright]_0^1\=-2left[sum_nge 0left(frac4(2n+1)(2n+3)right)-2right],$$which vanishes by partial fractions.
$endgroup$
To complement Zacky's answer I'll add a proof that $I(0)=0$. Applying $xmapstofrac1x$ for $xge 1$ gives $$I(x)=2int_0^inftyln|1-x^-2|dx=2int_0^1left[(1+frac1x^2)ln(1-x^2)-2ln xright]\=-2int_0^1left[(1+x^2)sum_nge 0fracx^2nn+1+2ln xright]dx=-2int_0^1left[sum_nge 0left(fracx^2nn+1+fracx^2n+2n+1right)+2ln xright]dx\=-2left[sum_nge 0left(fracx^2n+1(n+1)(2n+1)+fracx^2n+3(n+1)(2n+3)right)+2xln x-2xright]_0^1\=-2left[sum_nge 0left(frac4(2n+1)(2n+3)right)-2right],$$which vanishes by partial fractions.
answered Mar 11 at 11:08
J.G.J.G.
29.9k22947
29.9k22947
add a comment |
add a comment |
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$begingroup$
Maybe you can complete a square and make a substitution afterwards.
$endgroup$
– mathreadler
Mar 11 at 10:30
1
$begingroup$
I am not sure about a relationship between them, but both of them can be solved by similar tehniques (Feynman's trick).
$endgroup$
– Zacky
Mar 11 at 11:11