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I have some big O()'s in an integral. How can i compute or estimate such an integral?

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It would help us if you write the integral. – Davide Giraudo Mar 5 '12 at 12:57

The only thing one can do using the Landau symbols, given what you stated in the question, is to go to the definition.

$$ f(x) = O(g(x)) \iff \exists M,c ~\textrm{such that}~ \forall x > c\quad |f(x)| \leq M |g(x)| $$

Hence if $a < c < b$:

$$ \left|\int_a^b f(x)\mathrm{d}x\right| \leq \int_a^b |f(x)|\mathrm{d}x \leq \int_a^c |f(x)|\mathrm{d}x + \int_c^b |g(x)|\mathrm{d}x $$

An example of this is that if

$$ f(x) = O(x^\alpha) $$

Then, if $\alpha < -1$, the best you can say is that $\int_0^x f(y)\mathrm{d}y$ is $O(1)$. While if $\alpha > -1$, you can say that $\int_0^x f(y)\mathrm{d}y = O(x^{\alpha + 1})$ since

$$ \int_0^c |f(y)| \mathrm{d}y = O(1) $$

and (for $\alpha \neq -1$)

$$ \int_c^x |f(y)|\mathrm{d}y \leq \int_c^x y^\alpha \mathrm{d}y = \frac{1}{\alpha +1} \left( x^{\alpha + 1} - c^{\alpha+1}\right) = O(x^{\alpha+1}) + O(1)$$

One important remark is that while for integrals one can obtain estimates of the integral based on the integrand, the same cannot be done for derivatives. The derivative of a function $f(x) = O(g(x))$ need not satisfy $f'(x) = O(g'(x))$. Observe that if $f(x) = O(g(x))$ and $h(x) = \sin(x^n)f(x)$, then $h(x) = O(g(x))$ also. But note that $h'(x) = O(x^{n-1} g(x)) + O(f'(x))$, the first term of it can conceivably grow very much faster than $f'(x)$.

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