1. ## Comparison Theorem

Not sure if I am using this theorem correctly...

To evaluate x^2/(1+x^6) from negative infinity to positive infinity, I used x^2/x^6 as the comparison. After integrating, I took the limit from t to 0 as t -> - inf and the limit from 0 to s as s -> +inf of -1/(3x^3).

So, if I have done this much correctly, would the limit be convergent with them both equal to 0? Or divergent with them both equal to infinity?

2. Originally Posted by veronicak5678
Not sure if I am using this theorem correctly...

To evaluate x^2/(1+x^6) from negative infinity to positive infinity, I used x^2/x^6 as the comparison. After integrating, I took the limit from t to 0 as t -> - inf and the limit from 0 to s as s -> +inf of -1/(3x^3).

So, if I have done this much correctly, would the limit be convergent with them both equal to 0? Or divergent with them both equal to infinity?
Let $t=x^3$.

3. I'm sorry, I don't know what you mean. Do you mean the initial comparison should be to x^3?

4. Originally Posted by veronicak5678
I'm sorry, I don't know what you mean. Do you mean the initial comparison should be to x^3?
$\int_{-\infty}^{\infty} \frac{x^2}{x^6+1} dx = \frac{1}{3}\int_{-\infty}^{\infty} \frac{(x^3)'}{(x^3)^2+1} dx = \frac{1}{3}\int_{-\infty}^{\infty} \frac{dx}{x^2+1} = \frac{\pi}{3}$