Separation of an integral Suppose I have the following integral, with $a$, $b$, $\alpha$, and $\beta$ constants and $f(x)$ monotone increasing, continuous, always positive, and bounded on $[\alpha,\beta]$.
$$\int_{\alpha}^{\beta} f(x)^{ab} \, dx$$
Is is possible to separate this integral into:
$$\int_{\alpha}^{\beta}f(x)^a \, dx\int_{\alpha}^{\beta}f(x)^b \, dx$$
If so, what are the conditions under which this can be done?
NOTE: This is not a homework problem.
EDIT: Fixed non-essential typo.
 A: Not in some cases.
For example, take 
$$\int x^6 dx$$
By your method, $f(x)=x$, $a=2$, $b=3$:
$$\int x^{2 \times 3}dx=\int x^6dx$$
$$\int x^6 dx=(?)\int (x^2)^3 dx=\left( \int x^2 dx \right) \left( \int x^3 dx \right)$$
But
$$\frac{x^7}{7}+C\neq \left(\frac{x^3}{3} \times \frac{x^4}{4}\right)+C$$
Make it a definite integral and the same thing applies.
You can see that the same thing happens in other cases. I haven't found one where your method works.

Actually, take your example:
$$\int f(x)^{ab}dx=\frac{f(x)^{ab+1}}{ab}+C$$
But
$$\left( \int f(x)^a dx\right) \left( \int f(x)^b dx \right)=\left(\frac{f(x)^{a+1}}{a+1} \right) \left(\frac{f(x)^{b+1}}{b+1} \right)=\frac{f(x)^{a+b+2}}{(a+1)(b+1)}$$
Which does not necessarily equal the first result.
A: Even the simple case $f(x)=x$ leads to the complicated equation, $${\beta^{ab+1}-\alpha^{ab+1}\over ab+1}={\beta^{a+1}-\alpha^{a+1}\over a+1}{\beta^{b+1}-\alpha^{b+1}\over b+1}$$ and I don't think you are going to find any nice conditions for when this does and doesn't hold. Simplify even further by taking the special case $\alpha=0$, and you will find that, given $a$ and $b$, the equation holds if and only if $$\beta=\left({ab+1\over(a+1)(b+1)}\right)^{1/(ab-a-b-1)}$$
Another special case, $f(x)=e^x$, leads to $$e^{ab\beta}-e^{ab\alpha}=(e^{a\beta}-e^{a\alpha})(e^{b\beta}-e^{b\alpha})$$ which looks a little neater but it still doesn't seem to be easy to say anything useful about it. 
