Convergence of Integral: Splitting and Comparing Functions

In summary, the integral is finite for any interval (k,1), where 0 < k < 1, because continuous, bounded functions are always integrable.
  • #1
Dell
590
0
i am given the following, and asked if it converges or diverges

[tex]\int[/tex]dx/ln(x) (from 0-1)

what i did was look for a similar function
i took 1/x which is bigger than 1/lnx,
but 1/x diverges from 0-1

can i split my function up and compare
[tex]\int[/tex]dx/ln(x) (from 0-0.5) + [tex]\int[/tex]dx/ln(x) (from 0.5-1)
now i know that 1/x converges from 0.5-1 so 1/lnx does as well

now i need to find a fuction to compare 1/lnx with from 0-0.5
1/lnx < 1/(x+1)
1/(x+1) converges
therefore 1/lnx does as well

therefore 1/lnx is made up of two converging parts from 0-1 so the whole also converges


is this correct?
 
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  • #2
The integral is finite for any interval (k,1), where 0 < k < 1, because continuous, bounded functions are always integrable. You can't compare 1/ln(x) to 1/x or 1/(x+1) for that matter because the technique you're working with only works for positive functions. We're trying to check that the integral isn't NEGATIVE infinity, so finding a function that's larger than it on an interval doesn't work. You need to find a function that decreases to -infinity as x goes to 0 SLOWER than 1/ln(x) on an arbitrary small interval around 0. Considering that ln(x) goes to -infinity as x goes to 0 at the same rate that e^x goes to infinity as x goes to infinity, what could you use?
 
  • #3
i don't know, 1/e^x ??
i understand what you are saying but don't understand how to put it into practice
 
  • #4
i tried 1/(e^1/x) but i found that this converges and is smaller than 1/lnx so it doesn't help, please give some idea of what to do
 
  • #5
use the integral test, by using integration by parts.
 
  • #6
how do i integrate in parts, i tried and i keep on going in cirlces!

i made t=lnx, dt=dx/x

then got

integral of ((e^t)/t)dt, made 1/t=u e^tdt=dv, and even tried the other way round, -nothing!
 

What is the definition of convergence of an integral?

The convergence of an integral refers to the behavior of the value of an integral as the limits of integration approach a certain point or as the number of terms in the integrand approach infinity. It determines whether the integral will have a finite or infinite value.

What are the different types of convergence of an integral?

There are three types of convergence of an integral: absolute convergence, conditional convergence, and divergence. Absolute convergence refers to the convergence of the integral regardless of the order of integration. Conditional convergence refers to the convergence of the integral only when the order of integration is specific. Divergence refers to the integral having an infinite value.

How is the convergence of an integral evaluated?

The convergence of an integral is evaluated using various tests such as the Comparison Test, Ratio Test, Root Test, and the Integral Test. These tests compare the given integral to a known convergent or divergent integral and determine the behavior of the integral.

What is the importance of convergence of an integral in mathematics?

The concept of convergence of an integral is crucial in mathematics as it helps in determining the existence of the integral and its value. It also allows for the evaluation of complex integrals and plays a significant role in applications such as physics, engineering, and economics.

What are some common applications of convergence of an integral?

The convergence of an integral is applied in various fields such as determining the area under a curve, calculating the volume of a solid, finding the center of mass, and solving differential equations. It is also used in statistics, probability, and signal processing.

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