MHB Is There a Real Analysis Method to Solve Complex Integrals?

[nacho-man]

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?

 

[chisigma]

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

$\chi$ $\sigma$

 

[Ackbach]

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?

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.

 

[nacho-man]

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!

 

[Ackbach]

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.

 

[Prove It]

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 :)

 

[nacho-man]

Haha, well there's that.

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

 

[Prove It]

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 \displaystyle \begin{align*} x = 3\tan{(\theta)} \end{align*} into the first term, and \displaystyle \begin{align*} x = 2\tan{(\theta)} \end{align*} into the others.

 

[Nono713]

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​

 

[Prove It]

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...

 

[alyafey22]

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 .

 

[Nono713]

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.

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.