a definite integral identity? is it true? $\int_{a}^{b} f(x) dx $

Question

i came across this identity , i don't know if it is true or not

$$\int_{a}^{b} f(x) \ \mathrm{d}x = (b-a) \sum_{n=1}^{\infty} \sum_{k=1}^{2^n - 1} \dfrac{(-1)^{k+1}}{2^{n}} f \left( a+ \left(\frac{b-a}{2^n}\right) k \right)$$

i tried to use the Riemann Sum but i couldn't proceed

Anyone has an idea about it?

Answer 1

Note that: $$ \int_a^bf(x)dx=\lim_{N \to \infty}S_N \\ S_N=\sum_{i=1}^{2^N-1} \frac{b-a}{2^N}f(a+\frac{b-a}{2^N}i) = \sum_{n=1}^N \sum_{k=1}^{2^n-1}\frac{b-a}{2^n}(-1)^{k+1}f(a+\frac{b-a}{2^n}k) $$

Prove the last identity by induction:

For $N$=1, trivial.

For $N\to N+1$: $$ \sum_{n=1}^{N+1} \sum_{k=1}^{2^n-1}\frac{b-a}{2^n}(-1)^{k+1} f(a+\frac{b-a}{2^n}k)\\ =\sum_{n=1}^{N} \sum_{k=1}^{2^n-1}\frac{b-a}{2^n}(-1)^{k+1} f(a+\frac{b-a}{2^n}k) +\sum_{k=1}^{2^{N+1}-1}\frac{b-a}{2^{N+1}}(-1)^{k+1} f(a+\frac{b-a}{2^{N+1}}k)\\ =\sum_{i'=1}^{2^N-1} \frac{b-a}{2^N}f(a+\frac{b-a}{2^N}i') +\sum_{k=1}^{2^{N+1}-1}\frac{b-a}{2^{N+1}}(-1)^{k+1} f(a+\frac{b-a}{2^{N+1}}k)\\ =\sum_{i'=1}^{2^N-1} \frac{b-a}{2^N}f(a+\frac{b-a}{2^N}i') -\sum_{i'=1}^{2^{N}-1}\frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i')) +\sum_{i'=1}^{2^{N}}\frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'-1))\\ =\sum_{i'=1}^{2^N-1} \frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i')) +\sum_{i'=1}^{2^{N}}\frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'-1))\\ =\sum_{i=1}^{2^{N+1}-1} \frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}i) $$