Is There a Real Analysis Method to Solve Complex Integrals?

  • MHB
  • Thread starter nacho-man
  • Start date
  • Tags
    Integral
In summary: However, if you are using a numerical method to try and integrate the function, then it might not give a numerical answer.
  • #1
nacho-man
171
0
This problem came up in worksheet for a real analysis class, so I can only assume they want us to use a method from real analysis.

It would be pretty rigorous to solve otherwise, could anyone point me in the correct direction?

Or do they simply want us to test our differentiating skills?
 

Attachments

  • q5.jpg
    q5.jpg
    3.8 KB · Views: 92
Physics news on Phys.org
  • #2
nacho said:
This problem came up in worksheet for a real analysis class, so I can only assume they want us to use a method from real analysis.

It would be pretty rigorous to solve otherwise, could anyone point me in the correct direction?

Or do they simply want us to test our differentiating skills?

Take into account that...

$\displaystyle \frac{x^{2}}{(x^{2}+9)\ (x^{2}+4)^{2}} = \frac{\frac{9}{25}}{x^{2}+ 9} - \frac{\frac{9}{25}}{x^{2}+4} + \frac{\frac{4}{5}}{(x^{2} + 4)^{2}}$

Kind regards

$\chi$ $\sigma$
 
  • #3
nacho said:
This problem came up in worksheet for a real analysis class, so I can only assume they want us to use a method from real analysis.

It would be pretty rigorous to solve otherwise, could anyone point me in the correct direction?

Or do they simply want us to test our differentiating skills?

Do you need to prove that the integral converges?

chisigma said:
Take into account that...

$\displaystyle \frac{x^{2}}{(x^{2}+9)\ (x^{2}+4)^{2}} = \frac{\frac{9}{25}}{x^{2}+ 9} - \frac{\frac{9}{25}}{x^{2}+4} + \frac{\frac{4}{5}}{(x^{2} + 4)^{2}}$

Kind regards

$\chi$ $\sigma$

Excellent! That's where my mind went, too.
 
  • #4
Ackbach said:
Do you need to prove that the integral converges?
Excellent! That's where my mind went, too.

Thanks for the response sigma,

Uh, as far as I am sure, no we do not. Although, briefly, what would one need to do in order to find convergence for an integral.

Thanks again!
 
  • #5
nacho said:
Thanks for the response sigma,

Uh, as far as I am sure, no we do not. Although, briefly, what would one need to do in order to find convergence for an integral.

Thanks again!

Well, with this integral, you could probably get away with noting two things: 1. The denominator is always strictly positive, and bounded away from zero. Hence, the lower limit of the region of integration will give you no problems. 2. Furthermore, the integrand behaves like $1/x^4$ for large $x$, and hence goes to zero faster than $1/x^2$. Therefore, the upper limit of the region of integration will give you no problems. This is a hand-waving style of argument, but you could probably use it to give you an outline of the proof.
 
  • #6
nacho said:
Thanks for the response sigma,

Uh, as far as I am sure, no we do not. Although, briefly, what would one need to do in order to find convergence for an integral.

Thanks again!

If you are unsure if an improper integral converges, you CAN always just evaluate the integral and see if you get a numerical answer. In this case you will :)
 
  • #7
Haha, well there's that.

Ok so after simplifying the integral, I would have to apply some use Cauchy Residue theorem?
 
  • #8
nacho said:
Haha, well there's that.

Ok so after simplifying the integral, I would have to apply some use Cauchy Residue theorem?

No, an indefinite integral can be found. Substitute [tex]\displaystyle \begin{align*} x = 3\tan{(\theta)} \end{align*}[/tex] into the first term, and [tex]\displaystyle \begin{align*} x = 2\tan{(\theta)} \end{align*}[/tex] into the others.
 
  • #9
Prove It said:
If you are unsure if an improper integral converges, you CAN always just evaluate the integral and see if you get a numerical answer. In this case you will :)

Just as a quick off-topic question, does this always work? Or are there integrals which do not converge yet give bogus numerical results if you go ahead and still try to numerically evaluate them? I'm asking because I know that messing with infinite series without knowing if they converge is recipe for disaster, not sure if it's the same for integrals..

Might be a stupid question but I've learned to never take anything for granted :p​
 
Last edited:
  • #10
Bacterius said:
Just as a quick off-topic question, does this always work? Or are there integrals which do not converge yet give bogus numerical results if you go ahead and still try to numerically evaluate them? I'm asking because I know that messing with infinite series without knowing if they converge is recipe for disaster, not sure if it's the same for integrals..​

I don't see any reason why this method shouldn't work. If the integral doesn't converge, then why would it give a number for its value? I could be wrong though...

Just for clarification as I don't know if I was clear enough, when I said see if you get a numerical answer, I meant see if you get a number, not if you can integrate it using a numerical method...
 
  • #11
Bacterius said:
Just as a quick off-topic question, does this always work? Or are there integrals which do not converge yet give bogus numerical results if you go ahead and still try to numerically evaluate them? I'm asking because I know that messing with infinite series without knowing if they converge is recipe for disaster, not sure if it's the same for integrals..

Might be a stupid question but I've learned to never take anything for granted :p​

Generally if you follow the correct steps that will work even for series but most of the time you need to test the uniform convergence in an intermediate step . For series I think you are talking about absolute convergence and rearrangement theorem . Assuming an integral or series converge only tells us that the problem has an analytic value .Solving a divergent integral or series should not result in an analytic value unless we are making a mistake somewhere . Of course we first need to define what we mean by divergent and in what sense .
 
  • #12
Prove It said:
I don't see any reason why this method shouldn't work. If the integral doesn't converge, then why would it give a number for its value? I could be wrong though...

Just for clarification as I don't know if I was clear enough, when I said see if you get a numerical answer, I meant see if you get a number, not if you can integrate it using a numerical method...

No sorry this is indeed what I meant, e.g. integrate say an improper integral such as $\int_{-1}^1 \frac{1}{x} ~ \mathrm{d}{x}$ and get, say, $2.5$ as a result even though the integral does not converge. I thought maybe there were exotic integrals which could give misleading results like these if one was not careful enough about establishing convergence.

ZaidAlyafey said:
Generally if you follow the correct steps that will work even for series but most of the time you need to test the uniform convergence in an intermediate step . For series I think you are talking about absolute convergence and rearrangement theorem . Assuming an integral or series converge only tells us that the problem has an analytic value .Solving a divergent integral or series should not result in an analytic value unless we are making a mistake somewhere . Of course we first need to define what we mean by divergent and in what sense .

Ah, that makes sense. Yes, I was thinking about the rearrangement theorem but there are probably many such examples.
 
Last edited:

FAQ: Is There a Real Analysis Method to Solve Complex Integrals?

What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to solve problems related to finding the total or accumulated value of a function over a given interval.

Why is it important to find a simpler way to solve an integral?

Solving integrals can be a complex and time-consuming process. Finding a simpler way to solve an integral can save time and effort, making it easier to solve more complicated problems.

What are some common methods for solving integrals?

Some common methods for solving integrals include using basic integration formulas, integration by substitution, and integration by parts. Other techniques such as partial fractions, trigonometric substitution, and numerical integration can also be used.

How can technology be used to simplify the process of solving integrals?

Technology such as calculators and computer programs can be used to quickly and accurately solve integrals. These tools are especially useful for solving complex integrals that would be difficult or impossible to solve manually.

Are there any tips for finding a simpler way to solve an integral?

Some tips for finding a simpler way to solve an integral include breaking the integral into smaller parts, using symmetry to simplify the problem, and looking for patterns or common techniques that can be applied. It also helps to practice and become familiar with different integration methods.

Back
Top