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jsandifer
 Post subject: Integrate 1/[x(x^2-2x-8)]
PostPosted: Wed, 9 May 2012 14:14:28 UTC 
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How can I integrate 1/{xsqrt[(x^2-2x-8]} without looking it up in a table or using a calculator? Correction!


Last edited by jsandifer on Wed, 9 May 2012 14:20:28 UTC, edited 1 time in total.

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outermeasure
 Post subject: Re: Integrate 1/[x(x^2-2x-8)]
PostPosted: Wed, 9 May 2012 14:15:28 UTC 
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jsandifer wrote:
How can I integrate 1/[x(x^2-2x-8)] without looking it up in a table or using a calculator?


Partial fractions, remembering to factorise x^2-2x-8 first.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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jsandifer
 Post subject: Re: Integrate 1/[xsqrt(x^2-2x-8)]
PostPosted: Wed, 9 May 2012 15:41:45 UTC 
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Note the correction. I forgot to include the square root.
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outermeasure
 Post subject: Re: Integrate 1/[xsqrt(x^2-2x-8)]
PostPosted: Wed, 9 May 2012 16:40:21 UTC 
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jsandifer wrote:
Note the correction. I forgot to include the square root.


Well, it depends on what you know. The easiest is to look at the meromorphic differential \omega=\dfrac{\mathrm{d}x}{x\sqrt{x^2-2x-8}} as a differential on the conic C=(Y^2=X^2-2XZ-8Z^2)\subseteq\mathbb{P}^2, and thus there is a local coordinate t on C\cong\mathbb{P}^1 such that \omega=q(t)\,\mathrm{d}t with q(t) a meromorphic function of t.

If you carry that through, you get t=\sqrt{\dfrac{x-4}{x+2}}, or its reciprocal, is a substitution to try (in other words, you mapped the branch points of the square-root to 0 and infinity).

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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jsandifer
 Post subject: Re: Integrate 1/[x(x^2-2x-8)]
PostPosted: Sat, 12 May 2012 15:47:22 UTC 
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Didn't work. Any other ideas?
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outermeasure
 Post subject: Re: Integrate 1/[x(x^2-2x-8)]
PostPosted: Sat, 12 May 2012 15:49:57 UTC 
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jsandifer wrote:
Didn't work. Any other ideas?


What do you mean by "didn't work"? It clearly does work.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Shadow
 Post subject: Re: Integrate 1/[x(x^2-2x-8)]
PostPosted: Sat, 12 May 2012 18:09:12 UTC 
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outermeasure wrote:
jsandifer wrote:
Didn't work. Any other ideas?


What do you mean by "didn't work"? It clearly does work.


My thoughts exactly.

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jsandifer
 Post subject: Re: Integrate 1/[x(x^2-2x-8)]
PostPosted: Sat, 12 May 2012 22:46:33 UTC 
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More specifically, if I use t=sqrt((x-4)/(x+2)) I get for an answer: Arctan(t/sqrt2)/sqrt(2). Wolfram Mathematica gives me: -Arctan((x+8)/(2sqrt(2)sqrt(x^2-2x-8))/(2sqrt(2)) and the answer I was given is: (5/3)Arcsec(abs(x-1)/3). Have you guys worked this out?
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helmut
 Post subject: Re: Integrate 1/[x(x^2-2x-8)]
PostPosted: Sun, 13 May 2012 07:03:19 UTC 
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This confirms what I thought all along. The problem should have been $\int \frac{dx}{(x-1)\sqrt{x^2-2x-8}}$ instead; then you end up with something like $\sec^{-1}\left(\frac{x-1}{3}\right)$

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Shadow
 Post subject: Re: Integrate 1/[x(x^2-2x-8)]
PostPosted: Sun, 13 May 2012 07:57:12 UTC 
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helmut wrote:
This confirms what I thought all along. The problem should have been $\int \frac{dx}{(x-1)\sqrt{x^2-2x-8}}$ instead; then you end up with something like $\sec^{-1}\left(\frac{x-1}{3}\right)$


And what a difference that makes. Another point in favor of careful checking. :D

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jsandifer
 Post subject: Re: Integrate 1/[x(x^2-2x-8)]
PostPosted: Tue, 15 May 2012 02:01:10 UTC 
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Looks like everybody came out the same door, but I still wonder how Wolfram Mathematical got their answer.
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helmut
 Post subject: Re: Integrate 1/[x(x^2-2x-8)]
PostPosted: Tue, 15 May 2012 03:51:16 UTC 
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Mathematica does not like arcsec; all the arc functions are related anyway...

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