# How calculate $\int_{0}^{\pi}|\sin(x)|dx$ from its indefinite integral

I'm having trouble understanding why it is that when I try to calculate $$\int_{0}^{\pi}|\sin(x)|dx$$ from it's indefinite integral I seem to get the wrong result of $$0$$.

Of course $$|\sin(x)| = \sin(x)$$ on $$[0, \pi]$$ so I know the result should be $$2$$. But considering the indefinite integral

$$\int |\sin(x)| = -\cos(x)sgn(\sin(x))$$

($$sgn$$ is the sign function) I use some fallacious reasoning to conclude that

$$\int_{0}^{\pi}|\sin(x)|dx = [-\cos(x)sgn(\sin(x))]_{0}^{\pi} = -(-1)(0) -(-1)(0)) = 0$$

I think the problem is how I use the sign function but as far as I know $$sgn(\sin(0)) = sgn(\sin(\pi)) = 0$$ and $$|\sin(x)|$$ is continuous on $$[0, \pi]$$. What am I doing wrong here?

• You can't apply the "sgn" function only to the endpoints like that. Since you recognize that sin(x) is positive for $0< x< \pi$ I don't see why you do not simply say that [tex]\int_0^\pi |sin(x)|dx= \nt_0^\pi sin(x) dx[/itex]. Jul 29, 2019 at 11:49
• @user247327 The point of the question was to understand why my reasoning was wrong. As I wrote I know how I can calculate the integral correctly the way you suggest but that's beside the point.
– user578018
Jul 29, 2019 at 11:52
• Let $F(x) = -\cos x \operatorname{sgn} \sin x$. Then $\int_0^\pi |\sin x| dx = F(\pi - 0) - F(+0)$. The integral from $0$ to $2 \pi$ would be $F(2 \pi - 0) - F(\pi + 0) + F(\pi - 0) - F(+0)$. Jul 29, 2019 at 19:19

An indefinite integral, aka antiderivative, is necessarily a continuous function. If you graph the function $$-\cos xsgn(\sin x)$$, you'll see that it is discontinuous at the multiples of $$\pi$$. However, you can make it continuous by adding on the step function $$s(x)=2\lfloor x/\pi\rfloor$$. More precisely, we have

$$\int|\sin x|\,dx=C+ \begin{cases} -\cos xsgn{(\sin x)}+2\lfloor x/\pi\rfloor\quad\text{for }x\not\in\pi\mathbb{Z}\\ 2\lfloor x/\pi\rfloor-1\quad\text{for }x\in\pi\mathbb{Z} \end{cases}$$

where $$C$$ is an arbitrary constant.

Note, it's easy to see that the derivative of this piecewise-defined function is equal to $$|\sin x|$$ in each interval $$k\pi\lt x\lt(k+1)\pi$$ with $$k\in\mathbb{Z}$$, since $$sgn(\sin x)$$ and $$2\lfloor x/\pi\rfloor$$ are constant in each such interval. It's a good exercise to verify that the derivative exists, and is equal to $$0$$, at the multiples of $$\pi$$.

For the given definite integral, we have

$$\int_0^\pi|\sin x|\,dx=(C+2\lfloor\pi/\pi\rfloor-1)-(C+2\lfloor0/\pi\rfloor-1)=(C+2-1)-(C+0-1)=2$$

which agrees, of course, with the simpler calculation

$$\int_0^\pi|\sin x|\,dx=2\int_0^{\pi/2}\sin x\,dx=-2\cos x\big|_0^{\pi/2}=0-(-2)=2$$

Again, the fallacy lay in thinking that the formula $$-\cos xsgn(\sin x)$$, which defines a discontinuous function, could serve as "the" indefinite integral for $$|\sin x|$$.