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?)

  • $\begingroup$ 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

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.


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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