Is this really true?

Any insights on this monstrous integral?

\[\int\frac{x}{\sqrt{x^4+10x^2-96x-71}} \, dx = -\frac18 \ln((x^6+15x^4-80x^3+27x^2-528x+781)\sqrt{x^4+10x^2-96x-71}-(x^8+20x^6-128x^5+54x^4-1408x^3+3124x^2+10001))+constant\]

Note by Inquisitor Math
8 months, 1 week ago

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Hi, chris! I’m so happy to see you again! There is not much context for this thing, but this integral is said to be found by ‘Manuel Bronstein’ who tried to find an elementary function that has an elementary derivative.(tho this thing is not easy to find. how did he come up with this thing??)

Inquisitor Math - 8 months ago

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OK, so Manuel Bronstein was heavily involved in the development of computational symbolic integration. This - and the fact Alpha struggles with the integral - suggests it's a hard problem!!

The best I can come up with is working backwards: you might observe that integrals of the form xP(x)dx\int \frac{x}{\sqrt{P(x)}} dx

have a form something like alog(Q(x)P(x)R(x))+Ca \log \left(Q(x) \sqrt{P(x)}-R(x) \right)+C

where P,Q,RP,Q,R are monic polynomials, aa is a constant and CC is an arbitrary constant. Let's assume P(x)P(x) is not a perfect square.

Differentiating this, we get aQ(x)P(x)+P(x)Q(x)2P(x)R(x)Q(x)P(x)R(x)=xP(x)a \frac{Q'(x) \sqrt{P(x)} + \frac{P'(x) Q(x)}{2\sqrt{P(x)}} - R'(x)}{Q(x) \sqrt{P(x)}-R(x)} = \frac{x}{\sqrt{P(x)}}

Multiplying up, a(P(x)Q(x)+12P(x)Q(x)R(x)P(x))=xQ(x)P(x)xR(x)a \left(P(x) Q'(x) + \frac12 P'(x) Q(x) - R'(x) \sqrt{P(x)} \right) = xQ(x) \sqrt{P(x)}-xR(x)

We want this to be an identity that holds for all xx. This means we can equate coefficients, and also the terms that are multiples of P(x)\sqrt{P(x)} (remember we said PP wasn't a square).

So a(P(x)Q(x)+12P(x)Q(x))=xR(x)a \left(P(x) Q'(x) + \frac12 P'(x) Q(x)\right) = -xR(x)

and aR(x)=xQ(x)-a R'(x) = xQ(x)

(it's easy to verify these hold for the example). This certainly looks neater. We can see from here how the degrees of the polynomials are related: deg(P)+deg(Q)1=deg(R)+1,      deg(R)1=deg(Q)+1\deg(P) + \deg(Q)-1=\deg(R)+1,\;\;\;\deg(R)-1=\deg(Q)+1

so in fact we have to have deg(P)=4\deg(P)=4. OK, that was a surprise!

Also, we get a=1deg(R)a=\frac{-1}{\deg(R)} (the two equations both give the same result).

The sensible thing to do here seems to be to pick a polynomial QQ, use the second equation to get RR (up to a constant of integration), then equate coefficients in the first equation. The trickier part (and perhaps why the polynomials in the example have strange looking coefficients) would be getting all of the polynomials to have integer coefficients.

That's as far as I can get at the moment - I'd be interested in your thoughts. Do you think this was the approach? The fact that deg(P)\deg(P) has to be 44 is making me reconsider. It'd be worth trying to solve those two equations to see if there are other examples.

Chris Lewis - 7 months, 4 weeks ago

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I just tried this out. If we set Q(x)=x2+3x+4Q(x)=x^2+3x+4 (this was arbitrary, though I chose a multiple of 33 for the xx term to make RR nicer), we get a=14a=-\frac14, R(x)=x4+4x3+8x2132R(x)=x^4 + 4x^3 + 8x^2 - \frac{13}{2} and P(x)=x4+2x3+3x212x+8P(x)=x^4 + 2x^3 + 3x^2 - 12x + 8

and you can check that xx4+2x3+3x212x+8=14log((x2+3x+4)x4+2x3+3x212x+8(x4+4x3+8x2132))+C\int \frac{x}{\sqrt{x^4 + 2x^3 + 3x^2 - 12x + 8}}=-\frac14 \log \left( \left(x^2+3x+4 \right) \sqrt{x^4 + 2x^3 + 3x^2 - 12x + 8} - \left(x^4 + 4x^3 + 8x^2 - \frac{13}{2} \right) \right)+C

Again, Wolfram|Alpha couldn't do the integral, but can do the differentiation.

Chris Lewis - 7 months, 4 weeks ago

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@Chris Lewis Hah, that one looks nasty too! Kudos!

Inquisitor Math - 7 months, 4 weeks ago

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At first glance, I thought this involved some crazy substitution, but this is a nice approach! I will ponder more about this and see where this goes! (By the way have you looked at the other link too? That one is a question posed by me.)

Inquisitor Math - 7 months, 4 weeks ago

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@Inquisitor Math There still might be a way involving substitution, but again, if Alpha can't do it, I don't think I will be able to! (By the way, have you seen the form that Alpha comes out with for the original one? Not nice.)

I wonder what the full set of solutions to this looks like. I mean, PP has to be a quartic, but I don't think it can be an arbitrary one.

Chris Lewis - 7 months, 4 weeks ago

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I did some simple calculations and here are the following results: let the degree of polynomial RR be nn. We choose some nice polynomial (degree 4) for PP such that the following has a nice polynomial solution. 2PQ+5PQ+(4P2n2x2)Q+(P6n2x)Q=02PQ^{\prime\prime\prime}+5P^{\prime}Q^{\prime\prime}+(4P ^{\prime\prime}-2n^2x^2)Q^{\prime}+(P^{\prime\prime\prime} -6n^2x)Q=0 (IS THIS USEFUL?)

After solving for QQ we finally get RR. (Slightly different from your approach, notice that I am starting from P, then to Q, then to R. However this doesn’t seem to yield any meaningful results since a. we still do not know what P should be b.we do not know the value of n.) In fact the examples above (which chose the right PP) fits the bill!

Inquisitor Math - 6 months, 2 weeks ago

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Just in case you are interested, for the example you tried out, if we set n=4,P(x)=x4+2x3+3x212x+8n=4, P(x)=x^4+2x^3+3x^2-12x+8 the eq is 2(x4+2x3+3x212x+8)Q+5(4x3+6x2+6x12)Q+(16x2+48x+24)Q+(1272x)Q=02 (x^4+2x^3+3x^2-12x+8)Q^{\prime\prime\prime}+5(4x^3+6x^2+6x-12)Q^{\prime\prime}+(16x^2+48x+24)Q^{\prime}+(12-72x)Q=0 and Q(x)=x2+3x+4Q(x)=x^2+3x+4 is indeed a solution!

Inquisitor Math - 6 months, 2 weeks ago

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Hi again! I tried your way of going Q to R to P (P to R) and I have some questions. A)How did you determine the constant of R (after setting Q as a degree 2 polynomial?) B) Did you ‘just’ try out a degree 2 polynomial to make things simple?

This is just my opinion but for degQ=2 I think this may be generalized. (Have you tried it?)

Inquisitor Math - 5 months, 1 week ago

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1a2xxQ12PQPQ=0\frac1{a^2}x \int xQ-\frac12P^{\prime}Q-PQ^ {\prime}=0

As a side note this is a result from wolfram. I’m no expert at these stuff... so I hope you can check it out and see if that can be of any help.

I’ve updated some questions on these as well. LINK

Inquisitor Math - 5 months, 1 week ago

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@Inquisitor Math Hi - I need to remind myself a bit of what I did on this! I'll have a look and get back to you over the weekend.

Chris Lewis - 5 months ago

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@Chris Lewis Thank you very much! I think the most helpful info here would be the link from wolfram.

Idk if I am asking too much of you... Recently I’m struggling on this problem... I would be greatful if you check it now if you can. LINK1

Also we had a discussion about this one too... LINK2

Inquisitor Math - 5 months ago

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Any motivation,backstory behind this? I have a strong feeling that this is not arbitrary.

Inquisitor Math - 8 months ago

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Hi again - you asked in a solution comment the other day for some help on this one but I'm not sure how much help I can be. Where did you find this question? It might be useful to start there.

Just some observations, though:

1) Yes, it's really true. Wolfram|Alpha couldn't cope with the integral itself, but differentiating the result does give the right integrand

2) The roots of the polynomial x4+10x296x71=0x^4+10x^2-96x-71=0 are fairly "nice" (the only surds needed are 2\sqrt2 and 3\sqrt3) which probably isn't a coincidence

3) That would suggest trying to find integrals of similar functions, but I haven't managed to get Alpha to do this

4) This might involve contour integration, though I'm not sure how it'd work for indefinite integrals

Chris Lewis - 8 months ago

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