Integration math problem
- Thread starter bigman5
- Start date
- Joined
- Jan 15, 2006
- Sythe Gold
- 0
- Messages
- 6,989
That integral would be a headache to evaluate directly - even using a computer, it looks like a mess. Just looking at the graph of the function though, I suppose you could estimate it with a riemann sum from 0 to about 5.
To show it's convergent, though:
arctan(x) < Pi/2 for all x
So 0 to ∞ ∫(arctan(x))/(2+e^x)dx < 0 to ∞ ∫(Pi/2)/(2+e^x)dx
We know 0 to ∞ ∫(Pi/2)/(2+e^x)dx = -Pi/4 * e^(-x) evaluated from 0 to ∞
-Pi/4 * e^(-x) evaluated from 0 to ∞ = 0 - (-Pi/4) = Pi/4
So 0 to ∞ ∫(Pi/2)/(2+e^x)dx = Pi/4
Thus, ∫(arctan(x))/(2+e^x)dx < Pi/4, so it's convergent.
To show it's convergent, though:
arctan(x) < Pi/2 for all x
So 0 to ∞ ∫(arctan(x))/(2+e^x)dx < 0 to ∞ ∫(Pi/2)/(2+e^x)dx
We know 0 to ∞ ∫(Pi/2)/(2+e^x)dx = -Pi/4 * e^(-x) evaluated from 0 to ∞
-Pi/4 * e^(-x) evaluated from 0 to ∞ = 0 - (-Pi/4) = Pi/4
So 0 to ∞ ∫(Pi/2)/(2+e^x)dx = Pi/4
Thus, ∫(arctan(x))/(2+e^x)dx < Pi/4, so it's convergent.