6

Evaluate

$$\displaystyle \int _{ 0 }^{ \pi /2 }{ \log(\cos(x))\log(\sin(x)) \ dx }$$

Consider : $$\displaystyle F(m,n)=\int _{ 0 }^{ \pi /2 }{ \sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx } $$

$$\sin^{2}x = t$$ :

$$ \displaystyle F(m,n) =\frac{1}{2} \int _{ 0 }^{ 1 }{ { t }^{ m-1 }{ (1-t) }^{ n-1 }dt }=\frac{\beta (m,n)}{2} $$

Where $\beta(m,n)$ is the beta function.

$$\displaystyle F(m,n) = \frac { \Gamma (m)\Gamma (n) }{2 \Gamma (m+n) } $$

Hence $$\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } = 2\int _{ 0 }^{ \pi /2 }{ \sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx }$$

Differentiating with respect to $m$:

$$\displaystyle \frac { \Gamma (n) }{ ({ \Gamma (m+n)) }^{ 2 } } (\Gamma '(m)\Gamma (m+n)-\Gamma (m)\Gamma '(m+n)) = 4\int _{ 0 }^{ \pi /2 }{ log(sin(x))\sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx } $$

$$\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (\psi (m)-\psi (m+n))=4\int _{ 0 }^{ \pi /2 }{ log(sin(x))\sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx }$$

$$\psi(x)$$ is the digamma function.

Differentiate with respect to $n$ :

$$\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (((\psi (m)-\psi (m+n))(\psi (n)-\psi (m+n))-\psi '(m+n))$$

$$\displaystyle =8\int _{ 0 }^{ \pi /2 }{ log(sin(x))log(cos(x))\sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x } dx } $$

$m=n=\dfrac{1}{2}$:

$$\displaystyle \frac { { \Gamma }^{ 2 }(1/2) }{ \Gamma (1) } ({ (\psi (1/2)-\psi (1)) }^{ 2 }-\psi '(1))$$

$$\displaystyle =8\int _{ 0 }^{ \pi /2 }{ log(cos(x))log(sin(x))dx } $$

$$\displaystyle \Gamma (1/2)=\sqrt { \pi } ,\Gamma (1)=1,\psi (1/2)=-\gamma -log(4),\psi (1)=-\gamma ,\psi '(1)=\frac { { \pi }^{ 2 } }{ 6 } $$

Any other method to do this?

User1234
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  • Have you tried expanding the fourier series for $\log (2\sin \theta)$ and $\log (2\cos \theta)$? – r9m Jun 03 '15 at 18:38
  • $\displaystyle -\log(\cos(\theta))=\sum_{k=1}^\infty(-1)^k\frac{\cos(2k \theta)}{k}+\log(2)$ and $\displaystyle -\log(\sin(\theta))=\sum_{k=1}^\infty \frac{\cos(2k \theta)}{k}+\log(2)$, if we multiply the two and use the orthogonality of $\cos 2n\theta$, we could get somewhere I guess .. – r9m Jun 03 '15 at 18:42
  • the result should be $-\frac{1}{48} \pi \left(\pi ^2-6 \log ^2(4)\right)$ – Dr. Sonnhard Graubner Jun 03 '15 at 18:43
  • @BetterWorld I'm not actually used to Feynman's trick all that much .. I'll have to think about it. Interesting question (+1) – r9m Jun 03 '15 at 18:51
  • @BetterWorld , i think one parametrisation could be $$I(\alpha ,\beta )=\int_0^{\pi/2}\ln (\alpha \cos x) \ln (\beta \sin x)$$ dx . Take partial differentiation wrt both $\alpha$ and $\beta$ and note that what is $I(\sec x, \beta )$ and $I(\alpha , \csc x)$ is – Someone Jun 03 '15 at 19:07
  • No never mind, i just realize that $sec x$ and etc can't be put into. it will make them function of x and hence invalidate result – Someone Jun 04 '15 at 09:49
  • It can be used, never the less. Just need to find a better value. – Someone Jun 04 '15 at 09:50
  • For e.g, another way is that $I'(1)=\int_0^{\pi/2}(\ln \sin x + \ln \cos x) $ , where only $\alpha$ is introduced and product rule .etc – Someone Jun 04 '15 at 09:51
  • Better value than $sec x $ to determine the constant after evaluation ^^ – Someone Jun 04 '15 at 09:52
  • I am getting $I'(\alpha)=\frac{\pi}{\alpha}\ln \frac{\alpha}{2}$ ,though rough calculations and not sure. , by (another way) @BetterWorld ,still need to find a way to integrate and evaluate this. ^^ – Someone Jun 04 '15 at 09:58
  • $I(\alpha)=\int_{0}^{\pi/2}\ln (\alpha \sin x) \ln (\alpha \cos x)$ after differentiation use, $\ln(ab)=\ln a + \ln b$ to separate function of x from $\alpha$ , but still no useful initial value i am able to find. @BetterWorld – Someone Jun 04 '15 at 13:24
  • Because then your differentiation is invalid, since $\alpha$ is a function of $x$ . Hence you'd have to apply chain rule, which'd make it harder. But try that if it helps. – Someone Jun 04 '15 at 13:45
  • Thought so :/ . – Someone Jun 04 '15 at 13:58

1 Answers1

5

We can use the same method used in this answer. $$ \begin{align} &\int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\left(\log(2)+\sum_{j=1}^\infty\frac{\cos(2jx)}{j}\right)\left(\log(2)+\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}\right)\,\mathrm{d}x\\ &=\frac\pi2\log(2)^2+2\log(2)\int_0^{\pi/2}\sum_{k=1}^\infty\frac{\cos(4kx)}{2k}\,\mathrm{d}x+\int_0^{\pi/2}\sum_{k=1}^\infty(-1)^k\frac{\cos^2(2kx)}{k^2}\,\mathrm{d}x\\ &=\frac\pi2\log(2)^2+\frac\pi4\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\\ &=\frac\pi2\log(2)^2-\frac{\pi^3}{48} \end{align} $$

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