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Let $a$ and $b$ be real numbers such that $ 0<a<b$. The decreasing continuous function $y:[0,1] \to [0,1]$ is implicitly defined by the equation $y^a-y^b=x^a-x^b.$ Prove $$\int_0^1 \frac {\ln (y)} x \, dx=- \frac {\pi^2} {3ab}. $$

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  • $\begingroup$ @Sharkos: I think about that. $\endgroup$ – user64494 May 30 '13 at 16:31
  • $\begingroup$ I'm close to it but made an algebra error. The hint I would offer is that $$\int_0^1 \frac{\ln x}{1-x}\,dx= -\frac{\pi^2}6\,.$$ $\endgroup$ – Ted Shifrin May 30 '13 at 21:36
  • $\begingroup$ @Ted Shifrin : Thank you for the interest to the question and your interesting suggestion. I uderstand it suchwise: let us treat the integral under consideration as a line integral, continue the path, and apply the Green formula. Having tried that, I could not calculate the double integral. $\endgroup$ – user64494 Jun 3 '13 at 11:16
  • $\begingroup$ I tried that, as well. Still stuck, actually. Where did you find this? $\endgroup$ – Ted Shifrin Jun 3 '13 at 12:24
  • $\begingroup$ It is a real tough one. I tried most of the parametrization techniques but no progress. $\endgroup$ – AnilB Jun 4 '13 at 7:36
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OK, at long last, I have a solution. Thanks to @Occupy Gezi and my colleague Robert Varley for getting me on the right track. As @Occupy Gezi noted, some care is required to work with convergent integrals.

Consider the curve $x^a-x^b=y^a-y^b$ (with $y(0)=1$ and $y(1)=0$). We want to exploit the symmetry of the curve about the line $y=x$. Let $x=y=\tau$ be the point on the curve where $x=y$, and let's write $$\int_0^1 \ln y \frac{dx}x = \int_0^\tau \ln y \frac{dx}x + \int_\tau^1 \ln y \frac{dx}x\,.$$ We make a change of coordinates $x=yu$ to do the first integral: Since $\dfrac{dx}x = \dfrac{dy}y+\dfrac{du}u$, we get (noting that $u$ goes from $0$ to $1$ as $x$ goes from $0$ to $\tau$) \begin{align*} \int_0^\tau \ln y \frac{dx}x &= -\int_\tau^1 \ln y \frac{dy}y + \int_0^1 \ln y \frac{du}u \\ &= -\frac12(\ln y)^2\Big]_\tau^1 + \int_0^1 \ln y \frac{du}u = \frac12(\ln\tau)^2 + \int_0^1 \ln y \frac{du}u\,. \end{align*}

Next, note that as $(x,y)\to (1,0)$ along the curve, $(\ln x)(\ln y)\to 0$ because, using $(\ln x)\ln(1-x^{b-a}) = (\ln y)\ln(1-y^{b-a})$, we have $$(\ln x)(\ln y) \sim \frac{(\ln y)^2\ln(1-y^{b-a})}{\ln (1-x^{b-a})} \sim \frac{(\ln y)^2 y^{b-a}}{a\ln y} = \frac1a y^{b-a}\ln y\to 0 \text{ as } y\to 0.$$ We now can make the "inverse" change of coordinates $y=xv$ to do the second integral. This time we must do an integration by parts first. \begin{align*} \int_\tau^1 \ln y \frac{dx}x &= (\ln x)(\ln y)\Big]_{(x,y)=(\tau,\tau)}^{(x,y)=(1,0)} + \int_0^\tau \ln x \frac{dy}y \\ & = -(\ln\tau)^2 + \int_0^\tau \ln x \frac{dy}y \\ &= -(\ln\tau)^2 - \int_\tau^1 \ln x \frac{dx}x + \int_0^1 \ln x \frac{dz}z \\ &= -\frac12(\ln\tau)^2 + \int_0^1 \ln x \frac{dz}z\,. \end{align*} Thus, exploiting the inherent symmetry, we have $$\int_0^1 \ln y\frac{dx}x = \int_0^1 \ln y \frac{du}u + \int_0^1 \ln x \frac{dz}z = 2\int_0^1 \ln x \frac{dz}z\,.$$ Now observe that $$x^a-x^b=y^a-y^b \implies x^a(1-x^{b-a}) = x^az^a(1-x^{b-a}z^{b-a}) \implies x^{b-a} = \frac{1-z^a}{1-z^b}\,,$$ and so, doing easy substitutions, \begin{align*} \int_0^1 \ln x \frac{dz}z &= \frac1{b-a}\left(\int_0^1 \ln(1-z^a)\frac{dz}z - \int_0^1 \ln(1-z^b)\frac{dz}z\right) \\ &=\frac1{b-a}\left(\frac1a\int_0^1 \ln(1-w)\frac{dw}w - \frac1b\int_0^1 \ln(1-w)\frac{dw}w\right) \\ &= \frac1{ab}\int_0^1 \ln(1-w)\frac{dw}w\,. \end{align*} By expansion in power series, one recognizes that this dilogarithm integral gives us, at long last, $$\int_0^1 \ln y\frac{dx}x = \frac2{ab}\int_0^1 \ln(1-w)\frac{dw}w = \frac2{ab}\left(-\frac{\pi^2}6\right) = -\frac{\pi^2}{3ab}\,.$$ (Whew!!)

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  • $\begingroup$ @ Ted Shifrin : I have seen your answer today. $\endgroup$ – user64494 Jul 12 '13 at 19:02
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$$F(x,y)=y^a-y^b-x^a+x^b=0$$ by line parameterization $x=m\,y$ $$F(m\,y,y)=m^ay^a-m^by^b-y^a+y^b=0$$ $$\Rightarrow (m^a-1)y^a-(m^b-1)y^b=0$$ $$\Rightarrow \frac{m^a-1}{m^b-1}=y^{b-a}\Rightarrow y=\bigg(\frac{m^a-1}{m^b-1}\bigg)^{\frac 1{b-a}}$$ and $$x=m\,y=m\bigg(\frac{m^a-1}{m^b-1}\bigg)^{\frac 1{b-a}}$$ You can see that the integral limits for $m$ are $0$ and $+\infty$.

We can transform the integral such that $$x=m\,y\Rightarrow dx=dm\,y+m\,dy$$ $$\int \frac {\ln (y)} x \, dx=\int \frac {\ln (y)} {m\,y} \, (dm\,y+m\,dy)=\int_0^{\infty} \frac{\ln\bigg(\frac{m^a-1}{m^b-1}\bigg)^{\frac 1{b-a}}}{m}dm+\int_0^1 \frac{\ln(y)}{y}dy $$

If you try to calculate the integral for any value of $0\lt a\lt b$ you can see that the integrals are not convergent. Therefore I guess that it is a wrong statement?

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  • $\begingroup$ @ Anil Baseski : I find it to be a big step. In this way I obtain the convergent integral $$ -\int _{0}^{\infty }\!\ln \left( \left( {\frac {{m}^{a}-1}{{m}^{b}-1 }} \right) ^{1/ \left( b-1 \right) } \right) \frac { \left( b{m}^{b+a}+2\, a{m}^{b}-{m}^{b}b-{m}^{a}b-{m}^{2\,b}a-a+b \right)} {\left( a-b \right) \left( {m}^{b}-1 \right) ^{2}{m} \left( {m}^{a}-1 \right) }{dm} $$ which I can't calculate. However, I vote up your answer. $\endgroup$ – user64494 Jun 4 '13 at 18:20
  • $\begingroup$ @user64494 I guess there is a short-cut which we are missing... $\endgroup$ – AnilB Jun 4 '13 at 18:26
  • $\begingroup$ The line parametrization won't work, $y$ is a decreasing function. $\endgroup$ – Shuhao Cao Jun 17 '13 at 21:29
  • $\begingroup$ @ Shuhao Cao: Could you explain it in details? $\endgroup$ – user64494 Jun 18 '13 at 3:59
  • $\begingroup$ @ShuhaoCao You can make parametrization to any kind of function. Even a circle can be parametrized this way. mathnow.wordpress.com/2009/11/06/… $\endgroup$ – AnilB Jun 18 '13 at 7:49

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