Using the Monotone Convergence Theorem,compute the integral $\int_0^1 \frac{1}{2-x^2} dm$ expressing the result as an infinite series. Can someone help me in doing this by explaining how to solve this. I have understood the theorem but how can i apply this? How to express this in form of an infinite series?
Answer: $\frac{1}{2-x^2} = \frac{1}{2} \frac{1}{1-\frac{x^2}{2}} = \frac{1}{2}\sum_{n\in N}(\frac{x^2}{2})^n$
Applying MCT, $\int_0^1\frac{1}{2-x^2} dm = \frac{1}{2}\int_0^1\sum_{n=0}^\infty (\frac{x^2}{2})^n dm\\ =\frac{1}{2}\sum_{[0,1]}\int(\frac{x^2}{2})^n dm\\ =\frac{1}{2}\sum\int_0^1(\frac{x^2}{2})^n dx \\ =\frac{1}{2}\sum\Big[\frac{x^{2n+1}}{2^n .2^{n+1}}\Big]_0^1\ =\sum\frac{1}{2^{n+1}}\Big[\frac{x^{2n+1}}{2^n .2^{n+1}}\Big]_0^1\\$
Hint: Write $$\frac{1}{2-x^2} = \frac{1}{2} \frac{1}{1-\frac{x^2}{2}} = \frac{1}{2} \sum_{n \in \mathbb{N}_0} \left( \frac{x^2}{2} \right)^n$$ and apply the monotone convergence theorem.