"For which alpha does the integral int_2^+inf ( e^(alpha x)/((x−1)^alpha ln x) converge? For which values of α does the integral int_2^+inf ( e^(alpha x)/((x−1)^alpha ln x) dx converge?

saucletbh 2022-09-20 Answered
For which α does the integral 2 + e α x ( x 1 ) α ln x d x converge?
For which values of α does the integral
2 + e α x ( x 1 ) α ln x d x
converge?
I'm lost here.
For simplicity, let us denote the above integral as I 1
I was able to prove (I noticed that it's kind of standard exercise though) that:
I 2 = 2 + 1 ( x 1 ) α ln β x
I2 converges for α > 1 and   β.
converges for α = 1 and β > 1. For β 1 it diverges.
diverges for α < 1 and   β.
So my strategy was to use I 2 to prove the converges/divergence of I 1 , by means of inequalities, but I get nothing from this.
For instance I did the following:
For α > 1 we have that
0 < 2 + d x x α ln x 2 + e α x ( x 1 ) α ln x d x
That is, for β = 1
0 < I 2 I 1
From 1. we know that I 2 converges, but this doesn't tell us anything about the convergence of I 1 .
The answer given by my textbook is that I 1 converges for α < 0, but I don't know how to arrive at that conclusion.
You can still ask an expert for help

Want to know more about Exponential growth and decay?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Cremolinoer
Answered 2022-09-21 Author has 11 answers
The main point is that exponentials are always going to dominate polynomials Specifically, for any ϵ > 0 there exists a constant C ϵ such that e x C ϵ ( x 1 ) ϵ for all x 2. This implies that for α > 0, e α x C ϵ α ( 1 x ) α ϵ , so choosing ϵ < 1 such that 0 < α ( 1 ϵ ) < 1, we find
e α x ( x 1 ) α log x C ϵ α 1 ( x 1 ) α ( 1 ϵ ) log x
and so by comparison with I 2 you know I 1 diverges. Similarly, if α < 0 then e α x / 2 C 2 α ( x 1 ) 2 α and hence
e α x ( x 1 ) α log x C 2 α / 2 1 ( x 1 ) α log x ,
so by comparison with I 2 you know I 1 converges.
Did you like this example?
Subscribe for all access
hotonglamoz
Answered 2022-09-22 Author has 1 answers
What you've shown about the integral w/o the exponential factor is actually somewhat harder conceptually than the integral at hand. If α < 0 then the exponential factor is an exponential decay which overpowers any of the other behavior and will make the integral converge. If α > 0 ,, it's an exponential growth that will likewise make it diverge. α = 0 you've already handled.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-02-21

Suppose that  f  is an exponential function with a percentage growth rate of  2% , and with f(0)=147. Find a formula for  f .
a) f(t)=0.02t+147
b) f(t)=1.02(1.47)t
c) f(t)=147(1.02)t
d) f(t)=147(2)t
e) f(t)=147(1.20)t

asked 2022-08-04
A population doubles in size every 15 years. assuming exponential growth, find the
a) annual growth rate
b) continuous growth rate
asked 2020-12-17

In 2017, the population of a district was 20,800. With a continuous annual growth rate of approximately 6%, what will the population be in 2032 according to the exponential growth function?

asked 2022-09-10
what happened to the other constant of integration?
The standard equation for exponential growth and decay starts and is derived like this:
d P d t = k P
d P P = k d t
d P P = k d t
ln | P | = k t + C
I don't understand the left hand side at this point, isn't 1 x d x = ln | x | + C? Where did the constant of integration from the left integral go?
asked 2022-01-31
Describe what the values of C and k represent in the exponential growth and decay model y=Cekt.
asked 2021-05-21
Determine whether the function represents exponential growth or exponential decay. y=5(0.95)x
asked 2022-09-05
What is the formula for exponential growth with a decay rate?
Exponential growth can be modeled as
b ( 1 + r ) N
For b your starting quantity, (1+r) your rate of growth, and N the number of periods. But for N , this formula can get out of control.
Is there a traditional way of controlling for this by factoring in some notion of a decay factor (so that for periods N past some threshold, you stop growing asymptotically)?