How would you integrate Cantor's function

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On the interval from 0 to 1?
Thanks
 
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integrating cantor's function

How would you integrate it from 0 to 1? For those of you who don't know what it is here is a link http://en.wikipedia.org/wiki/Cantor_function.
Is it possible to do this as some sort of geometric series? Any help would be appreaciated.
Thanks
 
It looks like itself upside down.
 
Well how can I integrate it? It's almost like an infinite piece wise function.
 
If its integral (over the range [0,1]) exists, let it be I, StatusX's hint was that 1-f(x) is the same function but going downhill not uphill, so the integral of 1-f(x) over [0,1] is also I. You can do the rest from here surely. The point is the function is very symmetric, so you can exploit that symmetry, just like you can integrate sin(x) from -t to t for any t without knowing the indefinite integral of sin(x).
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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