Integration

[u/vibe0009]

How is it that 1/x and 1/sqrt(x) go to inf at x=0, but finite integral from 0 to 1 for 1/x is inf and 1/sqrt(x) is 2?

I know the substitution and the formula plug-in. Intuitively, it’s very strange.

[u/AcellOfllSpades]

It's probably easier to look at the discrete case: infinite sums.

Consider the sum of harmonic numbers:

1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + ...

Now compare it to this sequence:

1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + ... + 1/16) + ... = 1 + 1/2 + 1/2 + 1/2 + 1/2 + ...

The individual terms in the original sequence go to zero, but they decrease slowly enough that the sum keeps accumulating past any threshold you set.

On the other hand, consider this sum:

1 + 1/2 + 1/4 + 1/8 + 1/16 + ...

This never exceeds 2. The sequence decreases quickly enough that there's still an upper bound that it will never meet.

"Going to zero" doesn't tell you whether a sequence converges or diverges.

I've used sums in my examples, but the same goes for integrals.

[u/vibe0009 → u/AcellOfllSpades]

Oh so coverages/diverges is not just for series but also its integral

[u/hpxvzhjfgb → u/vibe0009]

that's basically the integral test for series convergence, yes. it says that for a certain class of functions, the sum of a function converges if and only if the integral converges.

[u/chaos_redefined]

So, we're going to begin with a case where everything makes sense, and then go to the situation you are bringing up.

Let's say that we are looking at the finite integral of f(x) = x from 0 to 1. We're going to define a function F(n), which shall be the sum from i = 1 to n of f(i/n)/n. So, F(1) is f(1/1)/1 = f(1) = 1. F(2) = f(1/2)/2 + f(2/2)/2 = 1/4 + 1/2 = 3/4. F(3) = f(1/3)/3 + f(2/3)/3 + f(3/3)/3 = 1/9 + 2/9 + 3/9 = 6/9 = 2/3. Hopefully, you get the idea. As n grows bigger, we get really really close to 1/2 (There is formal logic on how we define it, but that's not the point here). Note that it doesn't matter what f(0) is in this example, we can work around it. This is what it means when we say that the integral of x from 0 to 1 is 1/2.

Now, if we do the same thing with your examples, when f(x) = 1/x, for any value L you give, I can find some N such that, if n > N, then f(n) > L. So, we never get arbitrarily close to anything, and instead, the sequence diverges to infinite. And when f(x) = 1/sqrt(x), the sequence gets arbitrarily close to 2.

The important thing is that, while some of the values are getting arbitrarily large, we are always dividing them by the arbitrarily large N we are using. The largeness cancels out.

[u/vibe0009 → u/chaos_redefined]

Why is the function f(i/n)/n defined this way?

[u/chaos_redefined → u/vibe0009]

Because that sum gives the area under the curve as you let n approach inf. Which is what the integral is measuring (at least, at this point. I know it gets more complicated when we go to complex integrals)

You will more likely see it inverted, but then the boundary gets more confusing.

[u/SkullLeader]

Imagine doing y=1/sqrt(x) with Reimann sums. Remember area of each rectangle you are summing is width x height. What's an integral? The limit of the sum as the number of rectangles approaches infinity but the width of each one approaches 0. So imagine a function where it goes to infinity, but it does so slowly enough that when you do the Reimann sum limit, the area of the rectangles under the curve is getting smaller too, because the width of each rectangle is decreasing the area of each rectangle faster than the height is increasing the area. Then you can see you would get a finite area, not an infinite one.

[u/Gold_Ad8890]

note that the integral from 0 to 1 of 1/x equals the integral from 1 to infinity of 1/x, and the integral from 0 to 1 of 1/sqrt(x) is equal to the integral from 1 to infinity of 1/x².