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Compute the integral $\int_{-1}^1 f(x)\,\ dg(x)$, where $f(x)= \dfrac1{1+x^2}$ and $g(x) = \begin{cases} \phantom{-}1 &\text{if }x<0, \\ \phantom{-}0 &\text{if }x=0, \\ -1 &\text{if }x>0.\end{cases}$

My attempt: Looking at $g(x)$, you can see two jump discontinuities at $x=0$. This has been confusing me the most.

$$\int_{-1}^1 \dfrac1{1+x^2}\,\ dg(x)= f(0)(g(0+) - g(0)) = (1)(-1) = -1$$

But I feel like I should be incorporating $[g(0) - g(0-) = 0 - 1 = -1]$ into the equation somehow.

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it really helps to format questions in LaTex / MathJax (see FAQ). Regards –  Amzoti Mar 17 '13 at 20:30
    
@Drake, in the FAQ section there are directions to use LaTeX here. –  DonAntonio Mar 17 '13 at 20:38
    
Thank you for the directions on how to format questions. I will make sure to do that in the future. Thanks to Gigili for editing it for me! –  Drake Mar 17 '13 at 20:57
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Less important here than in Wikipdia articles, but the point actually arose in formatting this question: $\setminus$begin{cases} blahblah \end{cases} followed by a full-stop looks like this: $\begin{cases} a & \text{if whatever} \\ b & \text{if whatever} \\ c & \text{if whatever}\end{cases}$. ${}\qquad{}$ But it's better like this: $\begin{cases} a & \text{if whatever}, \\ b & \text{if whatever}, \\ c & \text{if whatever}.\end{cases}$ ${}\qquad{}$ –  Michael Hardy Mar 17 '13 at 21:13

1 Answer 1

up vote 0 down vote accepted

The total distance jumped at $x=0$ is $-2$, so the integral will be $-2f(0)$.

If you look at a partition of the interval from $-1$ to $1$, you'll have either an interval with $0$ in its interior or two intervals whose boundary is at $0$. In the first case, you'll be multiplying the value of $f$ at some point near $0$ by the size of the jump, $-2$, and as the mesh of the partition goes to $0$, the value of $f$ at the chosen point goes to $f(0)$. In the other case, you have two points near $0$, and the value of $f$ at each of them is multiplied by the jump, $-1$, and then they're added. And again, the values of $f$ at those points go to $f(0)$ as the mesh goes to $0$.

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