The question is evaluating the integral from negative infinity to positive infinity, of (30-x^4) dx. I think that the answer is negative infinity because I got a very large negative number, and that would be divergent, anyone think the same?
The question is evaluating the integral from negative infinity to positive infinity, of (30-x^4) dx. I think that the answer is negative infinity because I got a very large negative number, and that would be divergent, anyone think the same?
You got a very large negative number from where? As an approximation? Don't trust such methods if you want proof.
The integral is divergent. To demonstrate it, split the interval into two and show that either integral diverges by letting one of the limits of integration tend to infinity or -infinity. That is, show that , which would imply that diverges, so the integral over the whole real line also diverges.
Oh, okay, so integrating the integral gives you 30b-(b^5/5). So how do you know that it goes to negative infinity? This is probably a dumb question, but I don't really understand how you can tell as b goes to infinity, the integral goes to negative infinity? Please help me understand how to figure that out! Thanks!!
Nice!
However I have a doubt. Am I right in saying that my method is a sufficient condition for the integral to diverge?
Never heard of itUnless you want the Cauchy principal value, you need to treat the endpoints separately by splitting the integral or using a double limit.
So I will have to learn what "Cauchy principal value" is .... I will do that.
Thank you
Yes, it is sufficient, but not necessary. If the improper integral exists, then the Cauchy principal value exists and is equal to it, so by the contrapositive, the divergence of the principal value implies that the improper integral diverges.
Thank you, I should have been clearer about that. You can indeed choose any real value at which to split the interval. I chose 0 only because it was convenient to do so.