Integral from Unsolvable Equation

[TheFool]

I'm a bit stumped with a problem I have recently seen. Here it is:

There is a continuous and decreasing function $$y(x):[0,1] \to [0,1],\mbox{ }0<a<b$$ and $$x^a-x^b=y^a-y^b$$
Prove that $$\int_{0}^{1} \frac{\ln y}{x} dx=-\frac{\pi ^2}{3ab}$$

The trivial solution of y=x causes the integral to diverge. Frankly, I'm at a loss on how to approach this problem. Clearly you cannot solve for y in general as a and b can also take on non-integer values.

 

[DonAntonio]

I'm a bit stumped with a problem I have recently seen. Here it is:

There is a continuous and decreasing function $$y(x):[0,1] \to [0,1],\mbox{ }0<a<b$$ and $$x^a-x^b=y^a-y^b$$
Prove that $$\int_{0}^{1} \frac{\ln y}{x} dx=-\frac{\pi ^2}{3ab}$$

The trivial solution of y=x causes the integral to diverge. Frankly, I'm at a loss on how to approach this problem. Clearly you cannot solve for y in general as a and b can also take on non-integer values.

Besides not being very-very clear, you seem to have given an incorrect integral:
$$\int_0^1\frac{\log y}{x}\,dx=\log y\int_0^1\frac{1}{x}\,dx$$

and that improper integral diverges...perhaps you meant:

$$\,\,1)\,\,\,\forall\,\,0<a<b<1\,\,\,,\,\,x^a-x^b=y^a-y^b$$

$$\,\,2)\,\,\,\int_0^1 \log\frac{y}{x}\,dx\,\,?\,$$

Anyway, the above integral is an ugly beast...please clarify this.

DonAntonio

 

[TheFool]

That's how the problem was presented. Also, y is a function of x, not a constant. As such, it cannot be removed from the integral.

I have implicitly plotted $$x^a-x^b=y^a-y^b$$ for various values of a and b and I can see the two solutions for y. One is x, and the other fits the definition of y(x) given in my original post.

 

[TheFool]

In fact, it almost looks like a circle. Putting in $$y=\sqrt{1-x^{2}}$$ into the integral produces $$\int_{0}^{1} \frac{\ln y}{x} dx=-\frac{\pi ^2}{24}$$

 

[DonAntonio]

That's how the problem was presented. Also, y is a function of x, not a constant. As such, it cannot be removed from the integral.

I have implicitly plotted $$x^a-x^b=y^a-y^b$$ for various values of a and b and I can see the two solutions for y. One is x, and the other fits the definition of y(x) given in my original post.

Oh, I see now...and those values of $\,0<a<b\,$ are constant and fixed, uh? Sorry, I don't have the faintest idea, though

I tried to make a substitution $$bx+(1-x)a=u\Longrightarrow dx=\frac{du}{b-a}$$ to change the integral's limits so that $\,a,b\,$ get into play, but I can't see how to get them as powers of neither $\,x,y\,$...good luck!

DonAntonio

 

[DonAntonio]

In fact, it almost looks like a circle. Putting in $$y=\sqrt{1-x^{2}}$$ into the integral produces $$\int_{0}^{1} \frac{\ln y}{x} dx=-\frac{\pi ^2}{24}$$

What is what looks "almost" as a circle??

DonAntonio

 

[TheFool]

What is what looks "almost" as a circle??

DonAntonio

I said in a previous post:

...

I have implicitly plotted $$x^a-x^b=y^a-y^b$$ for various values of a and b and I can see the two solutions for y. One is x, and the other fits the definition of y(x) given in my original post.

I'm talking about the non-trivial solution for y. It's almost a circle.

 

[PAllen]

Just to clarify the problem, I worked the trivial case of a=1, b=2. Then you get either y=x, for which the improper integral diverges, or y=(1-x) for which the improper integral is π^2/6, consistent with the stated general formula. I don't have any suggestion for where to go from there ... just that the problem as stated seems plausible.

 

[TheFool]

It seems unlikely that there will be a closed form solution for y since a and b can take on non-integer values as well. However, there may be a series representation of the logarithm of y. That can be integrated.

 

[PAllen]

Where's Ramanujan when you need him ...

 

[haruspex]

Change of variable, x = e^u+v, y = e^u-v. Equation becomes
e^(b-a)u = sinh(av)/sinh(bv)
Meanwhile, by symmetry about the u-axis, the integral becomes
I = ∫₀^∞(v-u).dv = ∫₀^∞((av - ln(sinh(av)) + bv - ln(sinh(bv)))/(b-a))dv
∫₀^∞(av - ln(sinh(av)))dv = ∫₀^∞(v - ln(sinh(v)))dv/a
I = ∫₀^∞(v - ln(sinh(v)))dv/ab
v - ln(sinh(v)) = -ln(1-e^-2v) = Ʃe^-2vn/n
I = Ʃ[-e^-2vn/2n²]_v=0^∞/ab
= π²/12ab
Can anyone see where I lost the factor of 4?

 

[Mute]

Change of variable, x = e^u+v, y = e^u-v. Equation becomes
e^(b-a)u = sinh(av)/sinh(bv)
Meanwhile, by symmetry about the u-axis, the integral becomes
I = ∫₀^∞(v-u).dv = ∫₀^∞((av - ln(sinh(av)) + bv - ln(sinh(bv)))/(b-a))dv
∫₀^∞(av - ln(sinh(av)))dv = ∫₀^∞(v - ln(sinh(v)))dv/a
I = ∫₀^∞(v - ln(sinh(v)))dv/ab
v - ln(sinh(v)) = -ln(1-e^-2v) = Ʃe^-2vn/n
I = Ʃ[-e^-2vn/2n²]_v=0^∞/ab
= π²/12ab
Can anyone see where I lost the factor of 4?

How did you get to the bolded line? It looks to me like you're missing a factor from the change of variables, $dx/x = dv\left[du(v)/dv + 1\right]$. Where did that go?

 

[haruspex]

How did you get to the bolded line? It looks to me like you're missing a factor from the change of variables, $dx/x = dv\left[du(v)/dv + 1\right]$. Where did that go?

It's easiest in two stages. First x = e^r, y = e^s. The integral becomes ∫_-∞⁰s.dr.
Next, r = u+v, s = u-v. The tricky part here is the integral. It's best to look at it graphically. In the r-s plane, the curve is in the -/- quadrant, asymptotic to both axes. It's symmetrical about the line r=s (the u-axis). We want the area between the curve and the r-s axes. Cut the area with the line r=s, and fix on the half of the area touching the r axis.
At a given v, the length of the area in the u direction is v-u. (I think I might have reversed the sign of u somewhere; it looks like this should be u+v, but v-u works.) So the (half) integral is ∫(v-u)dv. OTOH, the switch to u, v has changed the scale: du.dv = 2dr.ds. So I think ∫(v-u)dv is actually the whole integral. But if I have that wrong it might explain one or two factors of 2.

 

[TheFool]

Change of variable, x = e^u+v, y = e^u-v. ...

Is that substitution of y valid as it's an unknown function of x? What is u and v?

 

[haruspex]

Is that substitution of y valid as it's an unknown function of x? What is u and v?

It's just a change of variables. In graphical terms, a rotation through 45 degrees, but with a bit of scaling too. (To avoid the scaling it would be (u+v)/√2, etc.)

 

[PAllen]

Change of variable, x = e^u+v, y = e^u-v. Equation becomes
e^(b-a)u = sinh(av)/sinh(bv)
Meanwhile, by symmetry about the u-axis, the integral becomes
I = ∫₀^∞(v-u).dv = ∫₀^∞((av - ln(sinh(av)) + bv - ln(sinh(bv)))/(b-a))dv
∫₀^∞(av - ln(sinh(av)))dv = ∫₀^∞(v - ln(sinh(v)))dv/a
I = ∫₀^∞(v - ln(sinh(v)))dv/ab
v - ln(sinh(v)) = -ln(1-e^-2v) = Ʃe^-2vn/n
I = Ʃ[-e^-2vn/2n²]_v=0^∞/ab
= π²/12ab
Can anyone see where I lost the factor of 4?

I think there are several things wrong with this, as suggested by Mute. However, it looks promising. I suggest working it out with the trivial case I gave (a=1,b=2), to clarify the issues. If I apply my understanding of what you suggest above to this case, it leads to nonsense. Meanwhile, directly dealing with:

y=(1-x), you get polylog(2,x) for the antiderivative, and this is zero for zero and π^2/6 for 1, giving the right answer. Working out your suggestion for this simple case should track down issues you (I think) have with limits of integration, dropped terms, etc.

 

[haruspex]

If I apply my understanding of what you suggest above to this case, it leads to nonsense.

Then I haven't explained it well enough. How does it lead to nonsense?

 

[PAllen]

Then I haven't explained it well enough. How does it lead to nonsense?

For starters, the limits of integration you propose are wrong for this case after your substitution. I get that they should be -∞ to +∞, which could explain one factor of 2. But then, I may be misunderstanding your method.

What would be helpful (to make sure we are on the same page) is to outline your approach in detail using the simple case. We know that direct computation of this simple case produces the right answer. Applying your method to it should help track down the issues.

 

[TheFool]

Haruspex, could you use [tex] so we can better see what you're doing? It'll improve readability.

 

[haruspex]

For starters, the limits of integration you propose are wrong for this case after your substitution. I get that they should be -∞ to +∞, which could explain one factor of 2.

The first substitution (x = exp(r)) makes the range r = -∞ to 0. The area now sits in the third quadrant and the curve resembles 1/r. I cut that in half at its axis of symmetry, r = s, and consider only the half adjacent to the negative r-axis. With the second substitution, v = r - s, the range becomes v = -∞ to 0 (I think I had v reversed in my head, so probably had a wrong sign there). The area of interest is now bounded by the curve, the u-axis, and by the line u = v (the r-axis). Hence the height, for purposes of integration, is v-u (both being negative).

 

[haruspex]

(In response to a request to rewrite in tex, correcting a few typos, and apparently finding my scaling error as a bonus)
Change of variable, x = e^u+v, y = e^u-v. Equation becomes
$e^{(b-a)u} = sinh(av)/sinh(bv)$
Meanwhile, by symmetry about the u-axis, the integral becomes
$I = \int^{0}_{-∞}(v-u).dv = (4/(b-a))\int^{0}_{-∞}(-av + ln(sinh(av)) + bv - ln(sinh(bv)))dv$
(The region being integrated in the uv plane lies between the lines u = v < 0, u = -v > 0, and the curve. This is symmetric about the u-axis, so is twice area bounded by that axis, u = -v > 0 and the curve. For a given v < 0, the range of u for the area lies between the curve and -v. Hence the value to be integrated is the difference of the two.
The halving of the range provides a factor 2 in the expression for I. The other factor of 2 is because the substitution introduces a scaling: dxdy = 2dudv... got that backwards originally)
$\int^{0}_{-∞}(-av + ln(sinh(av)))dv = a^{-1}\int^{-∞}_{0}(v - ln(sinh(v)))dv$
Applying this to both the a and b terms, the factor b-a cancels:
$I = 4(ab)^{-1}\int^{-∞}_{0}(v - ln(sinh(v)))dv$
$v - ln(sinh(v)) = -ln(1-e^{-2v}) = \sum e^{-2vn}/n$
$I = 4(ab)^{-1}\sum[-e^{-2vn}/2n^2]_{v=0}^∞$
= π²/3ab

 

[PAllen]

The part I don't see is that after your r,s substitution, then u,v substitution, in my simple case, I get:

∫ (u-v)((du/dv)+1) dv with limits -∞ to +∞

I'm stuck on the same point Mute was: I don't buy how you get just (u-v)dv. I also, as noted, get different integration limits. I agree, after r,s substitution, we have:

∫sdr, with limits -∞ to 0. However, from here, I get the above instead of what you get.

[Edit: also, we've both been sloppy with signs. The integral is negative. With the limits zero to one, my simple case is actually negative polylog(2,x), so the result is -π^2/6, consistent with post #1. The desired general result is -π^2/3ab]

 

[PAllen]

To get limits of integration going from r,s to u,v we need (implicitly or explicitly) v(r) applied to -∞ and 0. In my simple case, I can get this explicitly, and it leads to limits of -∞ and +∞.

 

[haruspex]

The part I don't see is that after your r,s substitution, then u,v substitution, in my simple case, I get:

∫ (u-v)((du/dv)+1) dv with limits -∞ to +∞

Draw yourself a diagram. In the rs view, the area required lies in the 3rd quadrant. It is bounded by the -ve r-axis, -ve s-axis, and a curve resembling r = 1/s. It is symmetric about r = s. Now cut it in half along r = s and consider only the portion adjacent to the r-axis. The v-coordinate is the distance from the origin in a "NW" direction (-ve v). For the purposes of integration, treat this as the independent variable.
At a given v, take an element width δv. This is a slice in SW/NE orientation. It cuts the area of interest at two points, u = v (i.e. s = 0) and at the curved boundary, u = u(v). The elemental area is therefore |(-v + u(v))δv|. For this halved region, the range is v = -∞ to 0. Note that if you wanted to integrate the whole region instead of cutting it in half, you'd have to involve the r = 0 boundary. This makes the integrand messier: ||v| - |u(v)||.

 

[PAllen]

Draw yourself a diagram. In the rs view, the area required lies in the 3rd quadrant. It is bounded by the -ve r-axis, -ve s-axis, and a curve resembling r = 1/s. It is symmetric about r = s. Now cut it in half along r = s and consider only the portion adjacent to the r-axis. The v-coordinate is the distance from the origin in a "NW" direction (-ve v). For the purposes of integration, treat this as the independent variable.
At a given v, take an element width δv. This is a slice in SW/NE orientation. It cuts the area of interest at two points, u = v (i.e. s = 0) and at the curved boundary, u = u(v). The elemental area is therefore |(-v + u(v))δv|. For this halved region, the range is v = -∞ to 0. Note that if you wanted to integrate the whole region instead of cutting it in half, you'd have to involve the r = 0 boundary. This makes the integrand messier: ||v| - |u(v)||.

diagram, shmiagram. What is the mathematical justification? To go from dr to dv, given r=u+v, u an unknown function of v, you need: (du/dv+1) dv. I don't see how any diagram changes this.

 

[Mute]

(In response to a request to rewrite in tex, correcting a few typos, and apparently finding my scaling error as a bonus)
Change of variable, x = e^u+v, y = e^u-v. Equation becomes
$e^{(b-a)u} = sinh(av)/sinh(bv)$
Meanwhile, by symmetry about the u-axis, the integral becomes
$I = \int^{0}_{-∞}(v-u).dv = (4/(b-a))\int^{0}_{-∞}(-av + ln(sinh(av)) + bv - ln(sinh(bv)))dv$
(The region being integrated in the uv plane lies between the lines u = v < 0, u = -v > 0, and the curve. This is symmetric about the u-axis, so is twice area bounded by that axis, u = -v > 0 and the curve. For a given v < 0, the range of u for the area lies between the curve and -v. Hence the value to be integrated is the difference of the two.
The halving of the range provides a factor 2 in the expression for I. The other factor of 2 is because the substitution introduces a scaling: dxdy = 2dudv... got that backwards originally)
$\int^{0}_{-∞}(-av + ln(sinh(av)))dv = a^{-1}\int^{-∞}_{0}(v - ln(sinh(v)))dv$
Applying this to both the a and b terms, the factor b-a cancels:
$I = 4(ab)^{-1}\int^{-∞}_{0}(v - ln(sinh(v)))dv$
$v - ln(sinh(v)) = -ln(1-e^{-2v}) = \sum e^{-2vn}/n$
$I = 4(ab)^{-1}\sum[-e^{-2vn}/2n^2]_{v=0}^∞$
= π²/3ab

The integral is one-dimensional. y is a function of x; similarly, either u must be a function of v or vice versa. I do not see how quoting a Jacobian result for $dxdy \rightarrow |\mathcal J| duduv$ (bolded in the quote) applies; furthermore, on the basis of your change of variables the Jacobian should be -2exp(2u), not just -2. If you did not mean a Jacobian by that expression, then you need to clarify what you did mean, because I am not following your argument.

It would be helpful if you drew the picture you have in mind when you make these symmetry arguments.

It would also be rather instructive if you could actually show mathematically, without using graphical arguments, that

$$\int_{-\infty}^\infty dv (v-u)\left[ \frac{du(v)}{dv}+1\right] =_? \int_{-\infty}^0 dv (v-u)$$
(there may be a sign error there, and I haven't double-checked the limits PAllen gets for v).

At the moment, I still find your symmetry argument confusing, and it looks to me like you are mixing results, or at least notation, from one and two dimensional integrals.

Also,

$$v - \ln(\sinh(v)) = -\ln(1-e^{-2x}) + \ln 2,$$
but in your derivation you leave out the ln2, which is integrated over and would give a divergent result. The integral of v - ln(sinh(v)) does not actually converge. On $(-\infty,0]$, the area is simply unbounded, as can be seen by plotting. These are the original limits you wrote down which you then switch to $[0,\infty)$, so perhaps you still have a typo, but even on this interval, the function levels off to ln2, so the area still diverges unless that ln2 is subtracted off, but that doesn't occur in the derivation offered.

 

[TheFool]

No wonder I'm confused by his substitutions.

 

[haruspex]

diagram, shmiagram. What is the mathematical justification? To go from dr to dv, given r=u+v, u an unknown function of v, you need: (du/dv+1) dv. I don't see how any diagram changes this.

∫_As.dr is an area. Any way you validly measure that area is justified. It doesn't necessarily automatically pop out as a result of the substitution - it may require further algebraic manipulation to arrive at the same result. If you wish to obtain it that way then go ahead; it must be possible to turn my geometric argument into algebra.

 

[PAllen]

∫_As.dr is an area. Any way you validly measure that area is justified. It doesn't necessarily automatically pop out as a result of the substitution - it may require further algebraic manipulation to arrive at the same result. If you wish to obtain it that way then go ahead; it must be possible to turn my geometric argument into algebra.

But that is for you to do. The other thread participants find your geometric argument some combination of incomprehensible and unconvincing. For me, I don't understand how any geometric argument can make your integral valid. I want to see analysis.

 

[haruspex]

The integral is one-dimensional. y is a function of x; similarly, either u must be a function of v or vice versa.

The original equation specifies a relationship between x and y. Either can be viewed as a function of the other. After the rs substitution, the integral becomes ∫s.dr, -∞<r<0. This can be interpreted as an area A bounded by the negative r-axis, the negative s-axis, and the curve of s as a function of r. This can equally be thought of as a two dimensional integral, ∫∫_Adrds.

I do not see how quoting a Jacobian result for $dxdy \rightarrow |\mathcal J| duduv$ (bolded in the quote) applies; furthermore, on the basis of your change of variables the Jacobian should be -2exp(2u), not just -2.

Sorry, typo: I meant drds = 2dudv.

It would be helpful if you drew the picture you have in mind when you make these symmetry arguments.

If you insist, but I really think I have described it in enough detail for you to draw your own.

$$v - \ln(\sinh(v)) = -\ln(1-e^{-2x}) + \ln 2,$$

Good catch - another typo in transcription. I should have written e^(b−a)u=(2sinh(av))/(2sinh(bv)) then worked with twice sinh throughout. This gets rid of the problem.

 

[Mute]

diagram, shmiagram. What is the mathematical justification? To go from dr to dv, given r=u+v, u an unknown function of v, you need: (du/dv+1) dv. I don't see how any diagram changes this.

Come on now, don't be patronizing. Just because we're not quite following haruspex's diagrammatical arguments doesn't mean they're wrong. That said, haruspex, I do feel that if you are going to be making a diagrammatical argument it helps if you supply the picture you have in mind, rather than trying to describe the picture you have in your mind in detail. A picture of what you are thinking of will come across much easier than a description of the picture you are thinking of. (The typos are also hurting the presentation of your argument).

I'm very short on time these couple of days which is why I haven't tried to fully work through it myself. I took a short break and tried working through haruspex's argument, and I can almost see it - the slicing in the u-v plane is the last thing I need to work out, but I really need to go finish some other things now.

I also tried working through it quickly mathematically, but I think I have a sign error somewhere. Perhaps overlooked the absolute value signs haruspex mentions. Basically, using the fact that u(v) is an even function of v and so u'(v) is an odd function of v, one can perhaps show that the u'(v)+1 factor cancels out.

 

[PAllen]

Come on now, don't be patronizing. Just because we're not quite following haruspex's diagrammatical arguments doesn't mean they're wrong. That said, haruspex, I do feel that if you are going to be making a diagrammatical argument it helps if you supply the picture you have in mind, rather than trying to describe the picture you have in your mind in detail. A picture of what you are thinking of will come across much easier than a description of the picture you are thinking of. (The typos are also hurting the presentation of your argument).

I'm very short on time these couple of days which is why I haven't tried to fully work through it myself. I took a short break and tried working through haruspex's argument, and I can almost see it - the slicing in the u-v plane is the last thing I need to work out, but I really need to go finish some other things now.

I also tried working through it quickly mathematically, but I think I have a sign error somewhere. Perhaps overlooked the absolute value signs haruspex mentions. Basically, using the fact that u(v) is an even function of v and so u'(v) is an odd function of v, one can perhaps show that the u'(v)+1 factor cancels out.

Until the last post, haruspex didn't even mention that he was treating a single integral as a double integral, and insisted 'it was just substitution'. The words used in his last post, for the first time, would have at least given a clue what was in mind, rather than mis-using established terminology. Substitution, integration by parts, double integral are not normally interchanged. Especially confusing is that the r,s substitution was just that - an ordinary substitution for single integrals.

To me, there was not only refusal to provide explanatory detail, there was misuse of terminology.

Seeing that a couple of us thought dr = (u'+1)dv was required, some real attempt to explain would be helpful, rather than ignoring this request.

 

[haruspex]

Seeing that a couple of us thought dr = (u'+1)dv was required, some real attempt to explain would be helpful, rather than ignoring this request.

Would it have been less confusing if I'd said 'change of co-ordinates'?
How do you substitute for dxdy when changing to polar? You don't write dx = cos(θ)dr - r sin(θ)dθ etc. You use the Jacobian. For rs to uv we get drds = 2dudv.

 

[haruspex]

Diagram attached.

 

[PAllen]

Would it have been less confusing if I'd said 'change of co-ordinates'?
How do you substitute for dxdy when changing to polar? You don't write dx = cos(θ)dr - r sin(θ)dθ etc. You use the Jacobian. For rs to uv we get drds = 2dudv.

Well for starters, I don't see how, for most of your posts, or starting from the original problem statement, anyone would be thinking about dxdy, drds, or dudv. You, yourself wrote:

∫sdr

and ended up with simple integral on dv. No mention of double integration until the very end of this discussion.So, yes, a great deal of your thought process was not presented, and words and expressions you did use pointed in a different direction.

My integral in post #22 must be correct (at least for the simple case of a=1,b=2), and it remains an interesting task to understand its equivalence to yours, on the hypothesis that yours is correct.