#What function should I use for the limit comparison test when u is between 0 and 1?
amber flame
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wind gale
Hmm...
Well I see two problem cases here
one of which is already solved seemingly
do you have any result for integrating x^alpha between 0 and 1?
or rather, I think it's well know that the integral of x^alpha from 1 to infinity converges iff alpha < -1
I'll leave it for you to check, but the second one is the one you want
in which case you want -alpha = u-1
alpha < 1
iff -alpha > -1
iff u-1 > -1
iff u > 0, which is always the case with your hypothesis
amber flame
Should I not worry about e^-x?
wind gale
We don't care, it's bounded by 1
exp(-x) is very well behaved overall so it's usually not a concern
amber flame
I think I'm picking up what you're putting down
amber flame
but that means we aren't doing the limit comparison test, right?
We're basically just using the P tests
wind gale
It's just a plain dumb comparison test yes, not summoning limits anywhere
wind gale
We're basically just using the P tests
Due to language barriers, I don't really know what a p-test is
aren't they exclusive to series?
Here it's just that f(x) = x^(u-1) exp(-x) is upper bounded by g(x) = x^(u-1) on ]0,1]
and g is integrable on ]0, 1]
therefore f is too
amber flame
Due to language barriers, I don't really know what a p-test is
There's one for both normal integrals and series, It's basically what the image you posted showed
Also I see what You're saying now
x^(u-1) is always bigger than x^(u-1)*e^(-x) if u is between 0 and 1
Then using x^(u-1) to see if it converges
wind gale
Correct
wind gale
x^(u-1) is always bigger than x^(u-1)*e^(-x) if u is between 0 and 1
if x is between 0 and 1
since x is the integration variable
amber flame
I'd still have to check 1 to infinity to see if it diverges
Since x is going from 0 to infinity
Then do something similar for u at 1 and bigger than 1
wind gale
Then do something similar for u at 1 and bigger than 1
For u >= 0 it's very easy already
because of compared growths between polynomials and exponentials
this is a theorem
amber flame
Okay, I'm sorry, I'm lost again
1 to infinity being just trivial is weird
Because wouldn't that make most of the possible integrals diverge?
Like let's make u-1 into -.5
The integral x^(-.5) would converge if we only look at 0 to 1
But diverges from 1 to infinity
wind gale
Because wouldn't that make most of the possible integrals diverge?
Sorry I poofed for a bit
And yes you're correct
wind gale
But diverges from 1 to infinity
On its own yes
but it's trivial due to the exponential being there
to regulate the whole thing
because the integral of exp(-x) is known to converge quite well
on [1 : infty[
and on that interval, 1/sqrt(x) is bounded too
for short:
- on [0, 1], everything is good thanks to exp(-x) being bounded and x^(u-1) being integrable
- on [1, infinity[, everything is good thanks to exp(-x) controlling the growth of the integrand, so it's integrable
just tag me again if you have more questions btw, i have this on mute
amber flame
@wind gale sorry for needing you again, but what exactly does part b mean?
wind gale
Hello, no worries, I told you to tag me so all good
Basically, you know that the integral of exp(-x) converges yeah?
amber flame
Yes
wind gale
so you look at x^(u-1) exp(-x)
if u-1 <= 0, x^(u-1) exp(-x) is bounded by exp(-x)
So all good, yes?
amber flame
Okay I think so
amber flame
I basically did that already because of the part a's u being u≥1
wind gale
why: it's a well known fact that for any alpha > 0, x^alpha << exp(x)
So you apply that to alpha = u+1 = (u-1) +2
x^(u+1) / exp(x) << 1
so x^(u-1) exp(-x) << x^(-2)
which is very cool because you also know that the integral of 1/x² converges
in simpler terms: exp(-x) is overriding the growth of x^(u-1)