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I've corrected typing error in the integral. I apologize for my mistake.

Reedited question:

Can anybody solve integral: $$\int_0^\frac{\pi}{2}x\ln(\sin x)~dx$$ I'm just trying to guess some simple formula for $\zeta(3)$. My "strategy" is simple: Find some conjectures and check them numerically.

Value of similar integral is well known: $$\int_0^\pi x\ln(\sin x)~dx=-\dfrac{\pi^2}{2}\ln(2) $$ Is there any idea for antiderivative of $x\ln(\sin x)$? It's too difficult to solve it for me. Any ideas?

Quanto
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Marek
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  • What have you done so far? – AlexR Dec 02 '13 at 12:58
  • This integral seems to have a serious issue with the point $;x=1;$ , by which it seems to diverge...have you already checked divergence/convergence matters here before wanting to know the integral's value? – DonAntonio Dec 02 '13 at 13:04
  • Do you really have the upper bound $\frac{\pi}{2}$? For arguments $x>1$ you have $\sin \pi x < 0$ and $\ln$ is complex! Anyway, for your integral Maple gives a complicated answer with polylogs and evaluated the answer is $\approx -0.72741684669932 - 5.344168447932846 i$ – gammatester Dec 02 '13 at 13:13
  • Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you [edit] your question to include some motivation, and an explanation of your own attempts. – Daniel R Dec 02 '13 at 13:15
  • Thank you. My mistake! (bad substitution), correct integral should be: $\int_{0}^{\frac{\pi}{2}} x\ln(\sin x)\ dx$. – Marek Dec 02 '13 at 13:57
  • @Marek You should [edit] your question and correct the error. While you're at it, please take a look at my previous comment and try to improve your question. – Daniel R Dec 02 '13 at 14:42
  • The answer should be ${\frac {7}{16}},\zeta \left(3\right)-\frac{1}{8},{\pi }^{2}\ln \left( 2\right)$ – Mhenni Benghorbal Dec 02 '13 at 16:03
  • You can use these techniques. – Mhenni Benghorbal Dec 02 '13 at 16:14
  • Conjecture(probably not correct): $\int_{0}^{\frac{\pi}2}x\ln(\sin x) dx=-\frac{{\pi}^2}{30}$. – Marek Dec 02 '13 at 16:21
  • @Marek: The answer I gave you is correct. – Mhenni Benghorbal Dec 02 '13 at 16:27

4 Answers4

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$$I=\int^{\frac{\pi}{2}}_0 x \log|\sin(x )| \, dx =\int^{\frac{\pi}{2}}_0 x \log|2\sin(x )| -\frac{\pi^2}{8}\log(2)$$

We relate the integral to the Clausen function \begin{align} \int^{\frac{\pi}{2}}_0 x \log|2\sin(x )| dx &=\frac{1}{2} \int^{\frac{\pi}{2}}_0 \mathrm{Cl}_2(2\theta)\, d\theta\\ &=\frac{1}{2} \int^{\frac{\pi}{2}}_0 \sum_{n=1}^{\infty}\frac{\sin(2n\theta)}{n^2}\, d\theta \\ &=-\frac{1}{4}\sum_{n=1}\frac{(-1)^n}{n^3}+\frac{1}{4}\sum_{n=1}\frac{1}{n^3}\\ &=\frac{7}{16}\zeta(3) \end{align}

Collecting that together we have

$$I=\frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\log(2)$$

I considered a more general case in this thread

Zaid Alyafeai
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Note

\begin{align}\int_0^\frac{\pi}{2}x\ln(2\sin x)~dx =\frac12 \int_0^\frac{\pi}{2}x\ln(2\sin 2x)~dx + \frac12 \int_0^\frac{\pi}{2}x\ln\frac{\sin x}{\cos x}~dx\tag1\\ \end{align} where \begin{align} &\int_0^\frac{\pi}{2} \overset{2x\to x} {x\ln(2\sin 2x)}~dx =\frac14 \int_0^{\pi} \overset{x\to\pi-x} {x\ln(2\sin x)}~{dx} =\frac\pi8 \int_0^{\pi}\ln(2\sin x)~dx =0 \end{align} and \begin{align} &\int_0^\frac{\pi}{2}x\ln\frac{\sin x}{\cos x} ~dx \overset{t=\tan x}= \int_0^\infty \frac{\ln t\tan^{-1}t}{1+t^2} dt \\ =& \int_0^\infty \frac{\ln t}{1+t^2} \int_0^1 \frac t{1+ t^2 y^2}dy ~dt \overset{t^2\to t}=\frac14\int_0^1 \int_0^\infty \frac{\ln t}{(1+t)(1+ t y^2)} \overset{t\to 1/(y^2t)}{dt ~dy}\\ =& \frac14\int_0^1 \int_0^\infty \frac{-\ln t -\ln y^2}{(1+t)(1+ t y^2)}dt ~dy = -\frac14\int_0^1 \int_0^\infty \frac{\ln y}{(1+t)(1+ t y^2)}dt ~dy\\ =& \frac12\int_0^1 \frac{\ln^2 y}{1-y^2}dy = \frac12\int_0^1 \frac{\ln^2 y}{1-y}dy - \frac12\int_0^1 \frac{y \ln^2 y}{1-y^2}\overset{y^2\to y}{dy}\\ =& \frac7{16}\int_0^1 \frac{\ln^2 y}{1-y}dy = \frac7{16}\cdot 2\zeta(3)=\frac 78\zeta(3) \end{align} Substitute above results into (1) to get $\int_0^\frac{\pi}{2}x\ln(2\sin x)~dx= \frac7{16}\zeta(3)$, which leads to $$ \int_0^\frac{\pi}{2}x\ln(\sin x)~dx= \frac7{16}\zeta(3)-\frac{\pi^2}8\ln2$$

Quanto
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The integral from $0$ to $\pi/2$ of $\ln(\sin x)$ is one of Euler's integrals - look here for some information. This paper explains why $$\int_{0}^{\pi/2}x\ln(\sin x)\ dx=\frac7{16}\zeta(3)-\frac{\pi^2}8\ln2. $$

Did
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stevie
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    Very interesting article.Thank uou very much. I'm just trying to guess some simple formula for the $\zeta(3)$. Only intuitively, and I'll check it numerically. – Marek Dec 02 '13 at 14:53
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    @Marek "Some simple formula for $\zeta(3)$"? This would make the headlines... http://en.wikipedia.org/wiki/Particular_values_of_Riemann_zeta_function – Did Dec 02 '13 at 16:16
  • To be more specific, http://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant. Finding a "simple formula" (whatever that means) for this constant would be interesting, for example it is not know if it is transcendental according to this article. – Sid Dec 02 '13 at 16:22
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Hint: Let $t=\sin x$, then integrate by parts, and you will be left with evaluating

$\displaystyle\int_0^1\frac{\arcsin^2t}tdt$, at which point you will make use of the formula $\displaystyle\sum_{n=1}^\infty\frac{(2t)^{2n}}{n^2\displaystyle{2n\choose n}}$

$=2\arcsin^2t,$ in conjunction with Apery's alternative series for $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n^3\displaystyle{2n\choose n}}=$

$=-\dfrac25\zeta(3)$, by switching the order of summation $\&$ integration. $\big($In case you're

wondering where the formula for $\arcsin^2t$ comes form, start with the expression

$\arcsin t=\displaystyle\int_0^t\frac{du}{\sqrt{1-u^2}}$, expand the integrand using the binomial series, then

switch the order of summation and integration, and square it using the Cauchy

product$\big)$.

Lucian
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