Solving a Logarithmic Integral: Is it Right?

  • Context: Graduate 
  • Thread starter Thread starter Alem2000
  • Start date Start date
  • Tags Tags
    Integral Logarithmic
Click For Summary
SUMMARY

The discussion centers on evaluating the convergence of the integral \(\int_{2}^{\infty}\frac{1}{n\ln(n)}dn\) using integration by parts. The initial approach mistakenly cancels logarithms, leading to an incorrect conclusion of convergence to \(\frac{1}{2}\). The correct method involves recognizing that the integral approaches 0 as \(n\) approaches infinity, confirming convergence. The key takeaway is the necessity of applying limits when assessing the convergence or divergence of integrals.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with logarithmic functions and their properties.
  • Knowledge of limits and their application in evaluating integrals.
  • Basic concepts of convergence and divergence in calculus.
NEXT STEPS
  • Study the application of limits in integral calculus.
  • Learn advanced integration techniques, including substitution and partial fractions.
  • Explore convergence tests for improper integrals.
  • Practice problems involving logarithmic integrals and their convergence properties.
USEFUL FOR

Students studying calculus, particularly those focusing on integral calculus and convergence analysis, as well as educators looking for examples of common mistakes in integration techniques.

Alem2000
Messages
117
Reaction score
0
I don't know what's wrong with me...I have been studing for 6 hrs straight...and i noticed I have been making stupid mistakes...so can someone tell me what's up with my work...this is what I did. [tex]\sum_{n=2}^{\infty}\frac{1}{nln(n)}[/tex] must show if it converges or diverges soooo this is what i did

[tex]\int_{2}^{\infty}\frac{1}{nln(n)}dn[/tex] integration by parts and

[tex]u=\frac{1}{ln(n)}[/tex] ,
[tex]du=\frac{1}{n(ln(n))^2}[/tex] ,
[tex]v=ln(n)[/tex] ,
[tex]dv=\frac{1}{x}[/tex]

[tex]\int_{2}^{\infty}\frac{1}{nln(n)}dn[/tex]=[tex]\frac{ln(n)}{ln(n)}-\int_{2}^{+\infty}\frac{ln(n)}{nln(n)^2}dn[/tex]

my logarithms cancel out and become [tex]1[/tex] and my integral of logorithms the top cancels and becomes one and i add it to the other side divide by two and i get [tex]\frac{1}{2}[/tex] now which say it converges did I do that right?
 
Physics news on Phys.org
darn it i realized what I have been doing wrong!

bad integration i should have done [tex]u=ln(n)[/tex]
[tex]du=\frac{1}{n}dx[/tex]
plug and chug back into [tex]\int_{2}^{\infty}\frac{1}{nln(n)}dn[/tex] and I got[tex]\int_{2}^{\infty}\frac{1}{nu}ndu[/tex] which becomes [tex]\int_{2}^{\infty}\frac{1}{u}du[/tex] and eazy sailing after that ay?
 
Last edited:


First of all, it's completely normal to make mistakes while studying for long periods of time. It's important to take breaks and give your brain a rest. Now, let's take a look at your work.

Your integration by parts is correct, but your final answer is not. When you cancel out the logarithms, you should have \frac{1}{n} left in the integral, not 1. Also, when you add \frac{1}{2} to the other side, you are essentially saying that the integral converges to \frac{1}{2}, which is not correct.

To determine if the integral converges or diverges, you need to evaluate the integral using a limit. As n approaches infinity, the integral will approach a certain value, and depending on that value, you can determine if the integral converges or diverges. In this case, the integral will approach 0, which means it converges.

So, in conclusion, your integration by parts is correct, but your final answer and reasoning are not. Remember to always use a limit when determining convergence or divergence of an integral. Keep practicing and don't be discouraged by mistakes!
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K