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Let $\theta\in[0,1]$ be a constant and $f\in C^1[0,1] $.

\begin{align} & S_n=\sum_k^{n-1}f\left(\frac{k+\theta} n\right)-n\int_0^1 f(x)\,dx \\[10pt] ={} & n\sum_k^{n-1} \int_{k/n}^{(k+1)/n} \left( f \left(\frac{k+\theta} n\right)-f(x)\right) \, dx \\[10pt] = {} & n\sum_k^{n-1} \int_{k/n}^{(k+1)/n} \int_x^{(k+\theta)/n} f'(s) \, ds \, dx \\[10pt] = {} & n\sum_k^{n-1} \left( \int_{k/n}^{(k+\theta)/n} \int_x^{(k+\theta)/n}-\int_{k+\theta/n}^{(k+1)/n} \int_{(k+\theta)/n}^x \right)f'(s)\,ds\,dx \end{align}

I do not understand the last step

\begin{align} & n\sum_k^{n-1} \int_{k/n}^{(k+1)/n} \int_x^{(k+\theta)/n} f'(s) \, ds \, dx \\[10pt] = {} & n\sum_k^{n-1} \left( \int_{k/n}^{(k+\theta)/n}\int_x^{(k+\theta)/n}-\int_{k+\theta/n}^{(k+1)/n} \int_{(k+\theta)/n}^x \right) f'(s) \, ds \, dx \end{align}

It goes against my intuition of breaking integrals.

Question:

How did the author derive

\begin{align} n\sum_k^{n-1} \left(\int_{k/n}^{(k+\theta)/n}\int_x^{(k+\theta)/n}-\int_{k+\theta/n}^{(k+1)/n} \int_{(k+\theta)/n}^x \right) f'(s) \, ds \, dx \text{?} \end{align}

Thanks in advance!

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Note that

$$\int_{k/n}^{(k+1)/n}g(x)\,dx = \int_{k/n}^{(k+\theta)/n}g(x)\,dx +\int_{(k+\theta)/n}^{(k+1)/n}g(x)\,dx ,$$

and, hence,

$$\int_{k/n}^{(k+1)/n}\int_{x}^{(k+\theta)/n}f'(s)\,ds\,dx = \int_{k/n}^{(k+\theta)/n}\int_{x}^{(k+\theta)/n}f'(s)\,ds\,dx +\int_{(k+\theta)/n}^{(k+1)/n}\int_{x}^{(k+\theta)/n}f'(s)\,ds\,dx $$

Now reverse the limits in the last integral and change "+" to "-".

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