4
$\begingroup$

How to prove the (numerically correct) identity $$\int\limits_0^1 \frac{\sqrt{x-x^3} \log (x)}{x \left(x^2+1\right)} \, dx+\int\limits_0^{\frac{\pi }{2}} \frac{x \sqrt{\cos (x)}}{\sin (x)} \, dx =-\frac{\pi ^2}{4}$$ I tried elementary techniques but none of them works. Maybe a clever contour integration is what we need? Any help will be appreciated.

$\endgroup$
3
  • $\begingroup$ @MHZ: Ooopsss... labeled them backwards (thanks)... but they're still unequal. $\endgroup$ – David G. Stork Feb 26 '20 at 7:39
  • $\begingroup$ @MHZ, my apologies. I shouldn't have assumed. However, a good way to avoid misunderstandings in problems like this would be to make it clear in the post that numerically the formula checks out, and to what precision did you check :) very interesting problem $\endgroup$ – Yuriy S Feb 26 '20 at 7:41
  • $\begingroup$ @DavidG.Stork, I also verified the formula numerically. Your value for Pi^2/4 is incorrect $\endgroup$ – Yuriy S Feb 26 '20 at 7:51
4
$\begingroup$

In order to prove that $S=-\frac{\pi^2}{4}$ , where \begin{equation*} S=\int_{0}^{1}\dfrac{\sqrt{x-x^3}\log(x)}{x(x^2+1)}\, dx +\int_{0}^{\frac{\pi}{2}}\dfrac{x\sqrt{\cos(x)}}{\sin(x)}\, dx, \end{equation*} we will use Cauchy's integral theorem. The integrand will be \begin{equation*} f(z)= \dfrac{2i\log(z)\sqrt{\frac{1+z^2}{2}}}{\sqrt{z}(1-z^2)} \end{equation*} where \begin{equation*} \log(z) = \ln|z| +i\arg(z) \quad \mbox{ with } -\pi<\arg[z)<\pi \end{equation*} and \begin{equation*} \sqrt{z} =e^{\frac{1}{2}\log(z)}. \end{equation*} Let $C$ be the contour of the first quadrant of the unit circle. Then \begin{equation*} \int_{C}f(z)\, dz = 0.\tag{1} \end{equation*} Some additional justifications are probably needed.

Split $C=C_1+C_2+C_3$ where $C_1$ is the path along the real axis from $0$ to $1$, $C_2$ is the arc from $1$ to $i$ and $C_3$ is the path along the imaginary axis from $i$ to $0$.

The singularity in $1$ is removable.

Close to $0$ we have to slightly modify $C_2$ and $C_3$. At the point $ir$ we leave $C_3$ and follow a small arc $C_r $ with radius $r$ to $r$ on $C_1$. According to the ML inequality \begin{equation*} \int_{C_r}f(z)\, dz \to 0 \quad \mbox{ as } r\to 0. \end{equation*}

The point $i$ can be treated analogously.

We will now study (1). \begin{equation*} \int_{C_1}f(z)\, dz = 2i\int_{0}^{1}\dfrac{\ln(x)\sqrt{\frac{1+x^2}{2}}}{\sqrt{x}(1-x^2)}\, dx \end{equation*} with real part $= 0$.

The arc $C_2$ can be described as $z=e^{it}, \quad 0 <t< \frac{\pi}{2}$. Then \begin{gather*} \int_{C_2}f(z)\, dz = \int_{0}^{\frac{\pi}{2}}\dfrac{2i\log(e^{it})\sqrt{\frac{1+e^{i2t}}{2}}}{\sqrt{e^{it}}(1-e^{i2t})}ie^{it}\, dt =\\[2ex]\int_{0}^{\frac{\pi}{2}}\dfrac{2i^{2}(\ln|e^{it}|+it)\sqrt{\cos(t)}e^{it/2}}{e^{it/2}(e^{-it}-e^{it})}\, dt= \int_{0}^{\frac{\pi}{2}}\dfrac{t\sqrt{\cos(t)}}{\sin(t)}\, dt. \end{gather*}

Now we proceed to \begin{gather*} \int_{C_3}f(z)\, dz = -\int_{0}^{1}\dfrac{2i\log(iy)\sqrt{\frac{1-y^2}{2}}}{\sqrt{(iy)}(1+y^2)}i\, dy =\\[2ex] \int_{0}^{1}\dfrac{2\left(\ln(y)+i\frac{\pi}{2}\right)\sqrt{\frac{1-y^2}{2}}}{\frac{1+i}{\sqrt{2}}\sqrt{y}(1+y^2)}\, dy = \int_{0}^{1}\dfrac{(1-i)\left(\ln(y)+i\frac{\pi}{2}\right)\sqrt{1-y^2}}{\sqrt{y}(1+y^2)}\, dy=\\[2ex] \int_{0}^{1}\dfrac{\left(\ln(y)+\frac{\pi}{2}\right)\sqrt{1-y^2}}{\sqrt{y}(1+y^2)}\, dy+i\cdot\int_{0}^{1}\dfrac{\left(\frac{\pi}{2}-\ln(y)\right)\sqrt{1-y^2}}{\sqrt{y}(1+y^2)}\, dy. \end{gather*} Now we extract the real part of (1). We get \begin{equation*} \int_{0}^{\frac{\pi}{2}}\dfrac{t\sqrt{\cos(t)}}{\sin(t)}\, dt+\int_{0}^{1}\dfrac{\left(\ln(y)+\frac{\pi}{2}\right)\sqrt{1-y^2}}{\sqrt{y}(1+y^2)}\, dy =0. \end{equation*} Thus \begin{equation*} S=-\dfrac{\pi}{2}\int_{0}^{1}\dfrac{\sqrt{1-y^2}}{\sqrt{y}(1+y^2)}\, dy =[y=s^2] =-\pi\int_{0}^{1}\dfrac{\sqrt{1-s^4}}{1+s^4}\, ds \end{equation*} However, \begin{equation*} \dfrac{\sqrt{1-s^4}}{1+s^4} =\dfrac{1}{2}\dfrac{\sqrt{\frac{1}{s^2}-s^2}}{\left(\frac{1}{s^2}-s^2\right)^2+4}\cdot 2\dfrac{1+s^4}{s^3} = \dfrac{1}{2}\dfrac{\sqrt{u}}{u^2+4}\cdot \dfrac{du}{ds}\cdot(-1). \end{equation*} where \begin{equation*} u=\dfrac{1}{s^2}-s^2. \end{equation*} Consequently, if we make the substitution $u=\dfrac{1}{s^2}-s^2$ then \begin{equation*} S=-\pi\int_{0}^{\infty}\dfrac{\sqrt{u}}{2(u^2+4)}\, du = [u=2\sqrt{v}]=-\dfrac{\pi}{4\sqrt{2}}\int_{0}^{\infty}\dfrac{v^{\frac{3}{4}-1}}{(1+v)^{\frac{3}{4}+\frac{1}{4}}}\, dv. \end{equation*} Here we recognize the beta $\mathrm{B}$ function. See

https://en.wikipedia.org/wiki/Beta_function

If we combine this with Euler's reflection formula we get \begin{gather*} S= -\dfrac{\pi}{4\sqrt{2}}B\left(\frac{3}{4},\frac{1}{4}\right)=-\dfrac{\pi}{4\sqrt{2}}\dfrac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{4})}{\Gamma(1)} =\\[2ex] -\dfrac{\pi}{4\sqrt{2}}\Gamma\left(\frac{3}{4}\right)\Gamma\left(1-\frac{3}{4}\right)= -\dfrac{\pi}{4\sqrt{2}}\dfrac{\pi}{\sin(\frac{3\pi}{4})} = -\dfrac{\pi^2}{4}. \end{gather*}

$\endgroup$
2
  • $\begingroup$ Thank you. That's exactly the construction I was trying to find. $\endgroup$ – Iridescent Mar 3 '20 at 4:34
  • $\begingroup$ You are welcome! $\endgroup$ – JanG Mar 4 '20 at 9:35
0
$\begingroup$

The second integral is $$ 2\int_0^1\frac{\tan^{-1} x}{\sqrt{x}}\frac{\sqrt{x-x^3}}{x\sqrt{x^2+1}}dx $$ and the sum misses $-\pi^2/4$ by about 11 percent.

$\endgroup$
1
  • $\begingroup$ Using GP PARI: A=intnum(x=0,1,sqrt(x-x^3)*log(x)/(x*(1+x^2)))+intnum(x=0,Pi/2,x*sqrt(cos(x))/sin(x)) A+Pi^2/4 %3 = 8.144391135782186966999445735647815307695292574859547044951886690071901355890594051292095765871792356488798500013860358058370141584753778515405059170442551967940712225713928333018474192315745430818859308416707157537505527727904813412573396656292555061 E-258+2.91(...)E-753*I $\endgroup$ – FDP Feb 27 '20 at 23:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.