- #1

- 959

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After reading http://en.wikipedia.org/wiki/Weierstrass_function it occurred to me that I could do the same thing to an integral:

[tex]

\int \sum_{i=0}^\infty \frac{sin(\frac{x}{3^i})}{2^i} dx

[/tex]

[tex]

= \sum_{i=0}^\infty \int \frac{sin(\frac{x}{3^i})}{2^i} dx

[/tex]

[tex]

= \sum_{i=0}^\infty -cos(\frac{x}{3^i})(\frac{3}{2})^i

[/tex]

which obviously diverges.

If we turn this into a definite integral and take the limit as n goes to infinity, it becomes a conditionally convergent sum. So we have the odd property that the indefinite integral doesn't exist, but the definite one does.

[tex]

\lim_{n \rightarrow \infty} \int_a^b \sum_{i=0}^n \frac{sin(\frac{x}{3^i})}{2^i} dx

[/tex]

[tex]

= \lim_{n \rightarrow \infty} \sum_{i=0}^n (cos(\frac{a}{3^i}) - cos(\frac{b}{3^i}))(\frac{3}{2})^i dx

[/tex]

[tex]

= \lim_{n \rightarrow \infty} \sum_{i=0}^n -2sin(\frac{a+b}{2*3^i})sin(\frac{a - b}{2*3^i})(\frac{3}{2})^i dx

[/tex]

which should converge because sin(x) becomes linear as x approaches 0.

What exactly is the explanation for this? I thought all continuous functions could be integrated (they definitely told us that!).

[tex]

\int \sum_{i=0}^\infty \frac{sin(\frac{x}{3^i})}{2^i} dx

[/tex]

[tex]

= \sum_{i=0}^\infty \int \frac{sin(\frac{x}{3^i})}{2^i} dx

[/tex]

[tex]

= \sum_{i=0}^\infty -cos(\frac{x}{3^i})(\frac{3}{2})^i

[/tex]

which obviously diverges.

If we turn this into a definite integral and take the limit as n goes to infinity, it becomes a conditionally convergent sum. So we have the odd property that the indefinite integral doesn't exist, but the definite one does.

[tex]

\lim_{n \rightarrow \infty} \int_a^b \sum_{i=0}^n \frac{sin(\frac{x}{3^i})}{2^i} dx

[/tex]

[tex]

= \lim_{n \rightarrow \infty} \sum_{i=0}^n (cos(\frac{a}{3^i}) - cos(\frac{b}{3^i}))(\frac{3}{2})^i dx

[/tex]

[tex]

= \lim_{n \rightarrow \infty} \sum_{i=0}^n -2sin(\frac{a+b}{2*3^i})sin(\frac{a - b}{2*3^i})(\frac{3}{2})^i dx

[/tex]

which should converge because sin(x) becomes linear as x approaches 0.

What exactly is the explanation for this? I thought all continuous functions could be integrated (they definitely told us that!).

Last edited: