I was asked to find the integral $\iint_A \log(\sin(x-y))dxdy$ where $A$ is the triangle $y=0, x=\pi, y=x$ in the first quadrant.

I was given a hint: evaluate $\int_{0}^{\pi}\log(\sin(t))dt$ using symmetry.

What I did:

I inferred from the hint that the variable change $t=x-y$ is the way to go, so $t=x-y$, and since we integrate by $y$ first, then $x$ is a "constant" and $dt=-dy$, and since $y$ transitions from $0$ to $x$, then $t$ transitions from $x$ to $0$, so we can rewrite the integral:

$$\int_{0}^{\pi} \int_{0}^{x}\log(\sin(x-y))dydx=\int_{0}^{\pi}\int_{x}^{0}-\log(\sin( t))dtdx=\int_{0}^{\pi}\int_{0}^{x}\log(\sin( t))dtdx$$

And here I am stuck. Firstly, I don't know how $\int_{0}^{\pi}\log(\sin(t))dt$ is related to the question, since in the question the limits are $0$ and $x$. not $0$ and $\pi$. They are not the same thing (even though $x$ transitions from $0$ to $\pi$).

But even if I did, how would I evaluate $\int_{0}^{\pi}\log(\sin(t))dt$??


At the point

$$\int_0^\pi \int_0^x \log (\sin t)\,dt\,dx,$$

changing the order of integration is a tempting thing to do:

$$\begin{align} \int_0^\pi \int_0^x \log (\sin t)\,dt\,dx &= \int_0^\pi \int_t^\pi \log (\sin t)\,dx\,dt\\ &= \int_0^\pi (\pi - t)\log (\sin t)\,dt\tag{a}\\ &= \int_0^\pi u\log (\sin (\pi-u))\,du\\ &= \int_0^\pi u\log (\sin u)\,du\tag{b}. \end{align}$$

Now add $(a)$ and $(b)$, and use the symmetry.

  • $\begingroup$ Could you explain why $\int_{0}^{\pi} \int_{0}^{x} \log(\sin t)dtdx = \int_{0}^{\pi} \int_{t}^{\pi}\log(\sin t)dxdt$? Changing order of integration means you integrate by $x$ first rather than $t$. i dont see why the integral equality you wrote follows from that. $\endgroup$ – Oria Gruber Jun 17 '14 at 18:55
  • 1
    $\begingroup$ I find it easiest to denote the region of integration by a chain of inequalities in such cases. Here, we have $B = \{(x,t) : 0 \leqslant t \leqslant x \leqslant \pi\}$. So when we keep $x$ fixed for the inner integral, $t$ varies from $0$ to $x$. Changing the order of integration, we keep $t$ fixed for the inner integral, and thus $x$ varies from $t$ to $\pi$. $\endgroup$ – Daniel Fischer Jun 17 '14 at 18:59
  • $\begingroup$ Did you forget to include a $-$ sign? you changed variable from $t$ to $u$, so $u=\pi -t$, so $du=-dt$ but you didn't change the sign or the domain of integration. simple mistake of am i missing something? $\endgroup$ – Oria Gruber Jun 17 '14 at 19:06
  • $\begingroup$ The substitution $u = \pi - t$ also flips the integral bounds, so the minus sign from the derivative cancels with the swapping of the integral bounds needed to get $\int_0^\pi$ again. $$\int_0^\pi f(t)\,dt = \int_\pi^0 f(\pi-u)\,d(\pi-u) = -\int_\pi^0 f(\pi-u)\,du = \int_0^\pi f(\pi-u)\,du.$$ $\endgroup$ – Daniel Fischer Jun 17 '14 at 19:09
  • $\begingroup$ Yes yes I see... $\endgroup$ – Oria Gruber Jun 17 '14 at 19:11

If you want to evaluate the integral

$$ \int_{0}^{\pi}\log(\sin(t))dt $$

then you can follow the steps


$$\int_{0}^{\pi}\log(\sin(t))dt = \int_{0}^{\pi/2}\log(\sin(t))dt+ \int_{\pi/2}^{\pi}\log(\sin(t))dt = I_1+I_2 = 2I_1. $$

You can prove that $I_1=I_2$ by using the change of variables $t=\pi-u$.

2) use the change of variables $u=\sin(t)$ to get

$$ I_1 = \int_{0}^{1}\frac{\log(u)}{\sqrt{1-u^2}}dt. $$

3) To evaluate $ I_1 $ consider the integral

$$ F = \int_{0}^{1} \frac{u^{\alpha}}{\sqrt{1-u^2}} du = \frac{\sqrt {\pi }}{2}\,{\frac { \Gamma \left( \frac{\alpha}{2}+\frac{1}{2} \right) }{ \Gamma \left( \frac{\alpha}{2} + 1 \right) }}, $$

which can be evaluated using the $\beta$ function (see related technique). Finally

$$ I_1 = \lim_{ \alpha \to 0} F_\alpha . $$


\begin{align} \int_{0}^{\pi}\log{(\sin x)}dx &= 2\int_{0}^{\pi/2}\log{(\sin x)}dx \\ &= \int_{0}^{\pi/2}\log{(\sin x)}dx+\int_{0}^{\pi/2}\log{(\cos x)}dx\\ &= \int_{0}^{\pi/2}\log{(\sin x\cos x)}dx=\int_{0}^{\pi/2}[\log{(\sin 2x)}-\log(2)]dx \\ &=\frac1{2}\int_{0}^{\pi}\log{(\sin x)}dx- \pi\log(2) /2 \end{align}


$$\int_{0}^{\pi}\log{(\sin x)}dx=-\pi\log(2)$$


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