1. ## Integral convergance

I am suppose to evaluate the integral of 1/(x^2 + 1) from 1 to infinity.

So I evaluated it from 1 to b and then took the antiderivative and got the inverse tan of b - inverse tan of 1. I think this integral converges at -pi/2 but the answer shows it converges at pi/4. Can somone tell me what I am doing wrong? I know that as be gets very large the inverse tan of it goes to zero but I can not figure out how to get pi/4!!!!! Thanks to all of you out there who help!

2. Originally Posted by Frostking
I am suppose to evaluate the integral of 1/(x^2 + 1) from 1 to infinity.

So I evaluated it from 1 to b and then took the antiderivative and got the inverse tan of b - inverse tan of 1. I think this integral converges at -pi/2 but the answer shows it converges at pi/4. Can somone tell me what I am doing wrong? I know that as be gets very large the inverse tan of it goes to zero but I can not figure out how to get pi/4!!!!! Thanks to all of you out there who help!
You are correct. The antiderivative is arctangent.

However, by the FTC, we have $\displaystyle {\color{blue}\lim_{b\to\infty}\tan^{-1}\left(b\right)}-\tan^{-1}\left(1\right)$

The limit (in blue) yields the value of $\displaystyle \frac{\pi}{2}$

Thus, we are left with $\displaystyle \frac{\pi}{2}-\tan^{-1}\left(1\right)=\frac{\pi}{2}-\frac{\pi}{4}=\color{red}\boxed{\frac{\pi}{4}}$

Does this make sense?

--Chris

3. ## Integral convergence at pi/4

So, can I think of it as arctan equal to zero degrees which would be at pi/2 and then subtracting the pi/4 to get pi/4??? or is this too simplistic a view. My trig class was back about 30 years ago so I am rusty to say the least. Thanks very much for your time and willingness to help!