For the integral of a function from infinity to negative infinity: split into two parts going from inifity to 0 and from 0 to minus infinity

negative-infinity
minus-infinity

#1

for the integral of a function from infinity to negative infinity: split into two parts going from inifity to 0 and from 0 to minus infinity. What if one part is convergent and the other is divergent?

Answer:

First, take the indefinite integral of [x*e^(-x^2)dx], and you will find that it is [e^(-x^2)]/2. Here’s proof:

Now to find the definite integral from x = negative infinity to x = positive infinity, we need to evaluate the indefinite integral at positive infinity and subtract the indefinite integral as evaluated at negative infinity.

So evaluated at positive infinity, we get -(e^(-x^2))/2 = -(1/2)*e^(- infinity^2) = -(1/2)*0 = 0.

Evaluated at negative infinity, we get THE SAME THING. Because we squared x, its sign didn’t matter. So the indefinite integral, evaluated at negative infinity is also equal to 0.

0-0 = 0. The integral from negative infinity to positive infinity is 0. That was a lot of work for 0…

I’m not sure what you did that made the function appear to diverge, but it shouldn’t have diverged anywhere. As x approaches negative or positive infinity, y approaches zero. Here’s a link to a picture of the function itself: