3
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

If you want to evaluate the integral

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

then you can follow the steps

1)

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

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

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

$\endgroup$
1
$\begingroup$

\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}

Thus

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

$\endgroup$

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.