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Hmm. I think I posted the wrong question before. Consider $I=\int_0^\pi \log(\sin(t)) \ dt$. Then $I\neq 0$ but the substitution $u=\sin(t)$ makes the end points equal so I end up having $I=0$. What am I doing wrong here? I don't quite understand the explanation at (Why should the substitution be injective when integrating by substitution?)

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  • $\begingroup$ Because the integrand is negative except for one point. $\endgroup$ – zhw. May 12 '15 at 2:58
  • $\begingroup$ $I = -\pi \log2$ $\endgroup$ – user239828 May 12 '15 at 2:58
  • $\begingroup$ @zhw. Yeah, realized a second after and deleted the comment, my bad. $\endgroup$ – YoTengoUnLCD May 12 '15 at 2:58
  • $\begingroup$ I am told to look at this post (math.stackexchange.com/questions/351355/…) but also this doesn't make sense. I just want to know which theorem is not being met when we do such a substitution that we get the wrong answer. $\endgroup$ – user239828 May 12 '15 at 2:59
  • $\begingroup$ I am not interested in the evaluation of the integral. It is easily evaluated using Complex Analysis. The point is why is this substitution making sense only when I break the integral piecewise and not when I use the substitution as is on the entire $[0,\pi]$ $\endgroup$ – user239828 May 12 '15 at 3:01
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Substitution: We start with $\int_a^b f(t)\, dt.$ If $g:[c,d]\to [a,b]$ is $C^1$ and $g(c) = a, g(d) =b ,$ then

$$\int_a^b f(t)\, dt = \int_c^d f(g(u))g'(u)\, du.$$

That's the theorem in its basic form. Now, when you tried "$u = \sin t$" above, you weren't following this recipe. That's where the problem is.

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The basic question on substitution was already aptly addressed by zhw. Another way of seeing this is that upon making the substitution $u=\sin x$ you are tacitly making the substitution $x = \arcsin u.$

But, the arcsine function has range $(-\pi/2, \pi/2)$, while $x$ spans $(0,\pi)$. That is the reason for needing to split the integral into the sum of integrals over domains for which the appropriate inverse function can be identified.

The correct substitution is $x\to \arcsin x$ for $0<x<\pi/2$ and $x\to \pi -\arcsin x$ for $\pi/2<x<\pi$. Then,

$$\begin{align} I&=\int_0^{\pi} \log (\sin x) dx\\ &=\int_0^{\pi/2} \log (\sin x) dx+\int_{\pi/2}^{\pi} \log (\sin x) dx\\ &=2\int_0^1 \frac{\log u}{\sqrt{1-u^2}}du \end{align}$$

That said, we will proceed to evaluate the integral using the method of substitution.


Here, we will use a set of three simple substitutions that can be used in succession to evaluate this integral.

First, we split the integral into two pieces as

$$\begin{align} I&=\int_0^{\pi} \log (\sin x) dx\tag 1\\ &=\int_0^{\pi/2} \log (\sin x) dx+\int_{\pi/2}^{\pi} \log (\sin x) dx\tag 2 \end{align}$$

Substitution 1: We let $x\to \pi-x$ in the second integral of $(2)$ and find that

$$I=2 \int_0^{\pi/2} \log (\sin x) dx\tag 3$$

Substitution : Next, we make the substitution $x \to 2x$ into $(1)$ and find that

$$\begin{align} I&=2 \int_0^{\pi/2}\log(\sin(2x)) dx\\ &=2\int_0^{\pi/2}\log 2 dx+2\int_0^{\pi/2}\log(\sin x) dx+2\int_0^{\pi/2}\log(\cos x) dx\\ &=\pi\log 2+I+2\int_0^{\pi/2}\cos x dx \tag 4 \end{align}$$

where we used $(2)$ to arrive at $(4)$.

Substitution 3: Finally, we make the third substitution in the last integral of $(4)$. There, letting $x \to \pi/2-x$ reveals that

$$\begin{align} I&=\pi\log 2+I+2\int_0^{\pi/2}\log (\cos x) dx \\ &=\pi\log 2+I+2\int_0^{\pi/2}\log(\sin x) dx\\ &=\pi\log 2+2I \end{align}$$

whereupon solving for $I$ yields $I=-\pi\log 2$.

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