I just had a quiz and I did not know how to do this question even though it looked so familiar.
Is the improper integral, from 0 to infinity,
convergent or divergent?
I tried using comparison theorem but it didn't work out.
And then I just tried solving the integral by substitution as in u=arctan(t) and du=1/1+t^2 dt but it didn't cancel out!
Please help! Thanks in advance!
Of course, it's a convergent integral. (Even an absolutely one.)
The integral converges as long as which in efect, it's convergent, since for all is Thus our integral is bounded by a convergent integral, hence the original integral converges.
I really don't care about computing its value, I'm just doing what the problem asks.
I can put and ask to show its convergence, and I know someone's gonna appear and batantly say which is far of the answer that I was looking for.
I know how to compute its value, but in this case, we're talking about convergence, besides, we don't know if the OP has covered computations on improper integrals.